Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Posts Tagged ‘PHY356’

reorganization of PHY356 (Quantum I) notes.

Posted by peeterjoot on December 22, 2012

Looking for something in my old notes while studying for my optics exam, I found that my old phy356 notes were very disorganized.  That was my first attempt to compile my class notes into a bookish unit, and it was not done terribly well.  Here’s a reorganized version:

The new structure is

Basic formalism

  •    Dirac Adjoint notes
  •    Lecture notes: Review
  •    Problems

Commutator and time evolution

  •    Rotations using matrix exponentials
  •    On commutation of exponentials
  •    Desai Chapter II notes
  •    Unitary exponential sandwich
  •    Lecture notes: Review
  •    Problems

Dynamical equations

  •    Problems

Free particles

  •    Notes for Desai chapter IV
  •    Lecture notes: Review
  •    Problems

Spin 1/2

  •    Lecture notes: Review
  •    Lecture: Orbital and Intrinsic Momentum
  •    Problems

Gauge invariance, angular momentum and spin

  •    Interaction with orbital angular momentum


  •    Lecture: Stern Gerlach

Lecture: Making Sense of Quantum Mechanics

Bound state problems

  •    Hydrogen like atom, and Laguerre polynomials
  •    Lecture: Bound-state problems
  •    Lecture: Hydrogen atom
  •    Problems

Harmonic oscillator

  •    Lecture: Harmonic oscillator
  •    Problems

Coherent states

  •    interaction with a electric field

Rotations and angular momentum

  •    Lecture: Rotations and Angular Momentum
  •    Notes for Desai Chapter 26

Exam prep and review

  •    Questions from the Dec 2007 PHY355H1F exam
  •    Questions from the Dec 2008 PHY355H1F exam
  •    A problem on spherical harmonics (2010 final exam reflection)


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Updated notes compilations for phy356 and phy456 (QM I & II)

Posted by peeterjoot on July 1, 2012

Here’s two updates of class notes compilations for Quantum Mechanics

The QM I notes updates are strictly cosmetic (the book template is updated to that of classicthesis since it was originally posted). The chapters in QM II are reorganized a bit, grouping things by topic instead by lecture dates.

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Collection of PHY356H1F course notes from 2010 Quantum Physics I course at UofT.

Posted by peeterjoot on January 10, 2011

Available to anybody inclined, I have now assembled all of my reading notes, problems, and lecture notes from the PHY356H1F course that I recently completed at the University of Toronto.

I have separated this from any of my other notes collections since it was fairly self contained. This includes the following individual postings:

Jan 9, 2011 A problem on spherical harmonics.

Jan 4, 2011 Some worked problems from old PHY356 exams.
Some worked problems from old PHY356 exams.

Dec 9, 2010 Notes for Desai Chapter 26.
Notes for Desai Chapter 26.

Dec 7, 2010 PHY356F lecture notes.
PHY356F lecture notes.

Nov 25, 2010 PHY356 Problem Set 5.
PHY356 Problem Set 5.

Nov 24, 2010 Hydrogen like atom, and Laguerre polynomials.
Playing with the math of Laguerre polynomials for the Hydrogen like atom.

Nov 20, 2010 Desai Chapter 10 notes and problems.
Desai Chapter 10 notes and problems.

Nov 20, 2010 Desai Chapter 10 notes and problems.
Desai Chapter 10 notes and problems.

Nov 19, 2010 Desai Chapter 9 notes and problems.
Desai Chapter 9 notes and problems.

Nov 16, 2010 PHY356 Problem Set 4.
PHY356 Problem Set 4.

Oct 31, 2010 Believed to be typos in Desai’s QM Text
Believed typos, some confirmed, in Desai’s QM book.

Oct 23, 2010 PHY356 Problem Set III.
My submission for problem set III.

Oct 23, 2010 PHY356 Problem Set II.
A couple more problems from my QM1 course.

Oct 22, 2010 Classical Electrodynamic gauge interaction.
Momentum and Energy transformation to derive Lorentz force law from a free particle Hamiltonian.

Oct 18, 2010 Notes and problems for Desai Chapter VI.
Notes and problems for Desai Chapter VI.

Oct 18, 2010 Notes and problems for Desai Chapter V.
Problems from Desai Chapter V.

Oct 10, 2010 Notes and problems for Desai chapter IV.
Chapter IV Notes and problems for Desai’s “Quantum Mechanics and Introductory Field Theory” text.

Oct 7, 2010 PHY356 Problem Set I.
A couple problems from my QM1 course.

Oct 1, 2010 Notes and problems for Desai chapter III.
Chapter III Notes and problems for Desai’s “Quantum Mechanics and Introductory Field Theory” text.

Sept 27, 2010 Unitary exponential sandwich
Unitary transformation using anticommutators.

Sept 19, 2010 Desai Chapter II notes and problems.
Chapter II Notes and problems for Desai’s “Quantum Mechanics and Introductory Field Theory” text.

July 23, 2010 Dirac Notation Ponderings.
Chapter 1 solutions and some associated notes.

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A problem on spherical harmonics.

Posted by peeterjoot on January 10, 2011

[Click here for a PDF of this post with nicer formatting]


One of the PHY356 exam questions from the final I recall screwing up on, and figuring it out after the fact on the drive home. The question actually clarified a difficulty I’d had, but unfortunately I hadn’t had the good luck to perform such a question, to help figure this out before the exam.

From what I recall the question provided an initial state, with some degeneracy in m, perhaps of the following form

\begin{aligned}{\lvert {\phi(0)} \rangle} = \sqrt{\frac{1}{7}} {\lvert { 12 } \rangle}+\sqrt{\frac{2}{7}} {\lvert { 10 } \rangle}+\sqrt{\frac{4}{7}} {\lvert { 20 } \rangle},\end{aligned} \hspace{\stretch{1}}(1.1)

and a Hamiltonian of the form

\begin{aligned}H = \alpha L_z\end{aligned} \hspace{\stretch{1}}(1.2)

From what I recall of the problem, I am going to reattempt it here now.

Evolved state.

One part of the question was to calculate the evolved state. Application of the time evolution operator gives us

\begin{aligned}{\lvert {\phi(t)} \rangle} = e^{-i \alpha L_z t/\hbar} \left(\sqrt{\frac{1}{7}} {\lvert { 12 } \rangle}+\sqrt{\frac{2}{7}} {\lvert { 10 } \rangle}+\sqrt{\frac{4}{7}} {\lvert { 20 } \rangle} \right).\end{aligned} \hspace{\stretch{1}}(1.3)

Now we note that L_z {\lvert {12} \rangle} = 2 \hbar {\lvert {12} \rangle}, and L_z {\lvert { l 0} \rangle} = 0 {\lvert {l 0} \rangle}, so the exponentials reduce this nicely to just

\begin{aligned}{\lvert {\phi(t)} \rangle} = \sqrt{\frac{1}{7}} e^{ -2 i \alpha t } {\lvert { 12 } \rangle}+\sqrt{\frac{2}{7}} {\lvert { 10 } \rangle}+\sqrt{\frac{4}{7}} {\lvert { 20 } \rangle}.\end{aligned} \hspace{\stretch{1}}(1.4)

Probabilities for L_z measurement outcomes.

I believe we were also asked what the probabilities for the outcomes of a measurement of L_z at this time would be. Here is one place that I think that I messed up, and it is really a translation error, attempting to get from the english description of the problem to the math description of the same. I’d had trouble with this process a few times in the problems, and managed to blunder through use of language like “measure”, and “outcome”, but don’t think I really understood how these were used properly.

What are the outcomes that we measure? We measure operators, but the result of a measurement is the eigenvalue associated with the operator. What are the eigenvalues of the L_z operator? These are the m \hbar values, from the operation L_z {\lvert {l m} \rangle} = m \hbar {\lvert {l m} \rangle}. So, given this initial state, there are really two outcomes that are possible, since we have two distinct eigenvalues. These are 2 \hbar and 0 for m = 2, and m= 0 respectively.

A measurement of the “outcome” 2 \hbar, will be the probability associated with the amplitude \left\langle{{ 1 2 }} \vert {{\phi(t)}}\right\rangle (ie: the absolute square of this value). That is

\begin{aligned}{\left\lvert{ \left\langle{{ 1 2 }} \vert {{\phi(t) }}\right\rangle }\right\rvert}^2 = \frac{1}{7}.\end{aligned} \hspace{\stretch{1}}(1.5)

Now, the only other outcome for a measurement of L_z for this state is a measurement of 0 \hbar, and the probability of this is then just 1 - \frac{1}{7} = \frac{6}{7}. On the exam, I think I listed probabilities for three outcomes, with values \frac{1}{7}, \frac{2}{7}, \frac{4}{7} respectively, but in retrospect that seems blatently wrong.

Probabilities for \mathbf{L}^2 measurement outcomes.

What are the probabilities for the outcomes for a measurement of \mathbf{L}^2 after this? The first question is really what are the outcomes. That’s really a question of what are the possible eigenvalues of \mathbf{L}^2 that can be measured at this point. Recall that we have

\begin{aligned}\mathbf{L}^2 {\lvert {l m} \rangle} = \hbar^2 l (l + 1) {\lvert {l m} \rangle}\end{aligned} \hspace{\stretch{1}}(1.6)

So for a state that has only l=1,2 contributions before the measurement, the eigenvalues that can be observed for the \mathbf{L}^2 operator are respectively 2 \hbar^2 and 6 \hbar^2 respectively.

For the l=2 case, our probability is 4/7, leaving 3/7 as the probability for measurement of the l=1 (2 \hbar^2) eigenvalue. We can compute this two ways, and it seems worthwhile to consider both. This first method makes use of the fact that the L_z operator leaves the state vector intact, but it also seems like a bit of a cheat. Consider instead two possible results of measurement after the L_z observation. When an L_z measurement of 0 \hbar is performed our state will be left with only the m=0 kets. That is

\begin{aligned}{\lvert {\psi_a} \rangle} = \frac{1}{{\sqrt{3}}} \left( {\lvert {10} \rangle} + \sqrt{2} {\lvert {20} \rangle} \right),\end{aligned} \hspace{\stretch{1}}(1.7)

whereas, when a 2 \hbar measurement of L_z is performed our state would then only have the m=2 contribution, and would be

\begin{aligned}{\lvert {\psi_b} \rangle} = e^{-2 i \alpha t} {\lvert {12 } \rangle}.\end{aligned} \hspace{\stretch{1}}(1.8)

We have two possible ways of measuring the 2 \hbar^2 eigenvalue for \mathbf{L}^2. One is when our state was {\lvert {\psi_a} \rangle} (, and the resulting state has a {\lvert {10} \rangle} component, and the other is after the m=2 measurement, where our state is left with a {\lvert {12} \rangle} component.

The resulting probability is then a conditional probability result

\begin{aligned}\frac{6}{7} {\left\lvert{ \left\langle{{10}} \vert {{\psi_a}}\right\rangle }\right\rvert}^2 + \frac{1}{7} {\left\lvert{ \left\langle{{12 }} \vert {{\psi_b}}\right\rangle}\right\rvert}^2 = \frac{3}{7}\end{aligned} \hspace{\stretch{1}}(1.9)

The result is the same, as expected, but this is likely a more convicing argument.

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Some worked problems from old PHY356 exams.

Posted by peeterjoot on January 9, 2011

[Click here for a PDF of this post with nicer formatting]


Some of the old exam questions that I did for preparation for the exam I liked, and thought I’d write up some of them for potential future reference.

Questions from the Dec 2007 PHY355H1F exam.

1b. Parity operator.

\paragraph{Q:} If \Pi is the parity operator, defined by \Pi {\lvert {x} \rangle} = {\lvert {-x} \rangle}, where {\lvert {x} \rangle} is the eigenket of the position operator X with eigenvalue x), and P is the momentum operator conjugate to X, show (carefully) that \Pi P \Pi = -P.


Consider the matrix element {\langle {-x'} \rvert} \left[{\Pi},{P}\right] {\lvert {x} \rangle}. This is

\begin{aligned}{\langle {-x'} \rvert} \left[{\Pi},{P}\right] {\lvert {x} \rangle}&={\langle {-x'} \rvert} \Pi P - P \Pi {\lvert {x} \rangle} \\ &={\langle {-x'} \rvert} \Pi P {\lvert {x} \rangle} - {\langle {-x} \rvert} P \Pi {\lvert {x} \rangle} \\ &={\langle {x'} \rvert} P {\lvert {x} \rangle} - {\langle {-x} \rvert} P {\lvert {-x} \rangle} \\ &=- i \hbar \left(\delta(x'-x) \frac{\partial {}}{\partial {x}}-\underbrace{\delta(-x -(-x'))}_{= \delta(x'-x) = \delta(x-x')} \frac{\partial {}}{\partial {-x}}\right) \\ &=- 2 i \hbar \delta(x'-x) \frac{\partial {}}{\partial {x}} \\ &=2 {\langle {x'} \rvert} P {\lvert {x} \rangle} \\ &=2 {\langle {-x'} \rvert} \Pi P {\lvert {x} \rangle} \\ \end{aligned}

We’ve taken advantage of the Hermitian property of P and \Pi here, and can rearrange for

\begin{aligned}{\langle {-x'} \rvert} \Pi P - P \Pi - 2 \Pi P {\lvert {x} \rangle} = 0\end{aligned} \hspace{\stretch{1}}(2.1)

Since this is true for all {\langle {-x} \rvert} and {\lvert {x} \rangle} we have

\begin{aligned}\Pi P + P \Pi = 0.\end{aligned} \hspace{\stretch{1}}(2.2)

Right multiplication by \Pi and rearranging we have

\begin{aligned}\Pi P \Pi = - P \Pi \Pi = - P.\end{aligned} \hspace{\stretch{1}}(2.3)

1f. Free particle propagator.

\paragraph{Q:} For a free particle moving in one-dimension, the propagator (i.e. the coordinate representation of the evolution operator),

\begin{aligned}G(x,x';t) = {\langle {x} \rvert} U(t) {\lvert {x'} \rangle}\end{aligned} \hspace{\stretch{1}}(2.4)

is given by

\begin{aligned}G(x,x';t) = \sqrt{\frac{m}{2 \pi i \hbar t}} e^{i m (x-x')^2/ (2 \hbar t)}.\end{aligned} \hspace{\stretch{1}}(2.5)


This problem is actually fairly straightforward, but it is nice to work it having had a similar problem set question where we were asked about this time evolution operator matrix element (ie: what it’s physical meaning is). Here we have a concrete example of the form of this matrix operator.

Proceeding directly, we have

\begin{aligned}{\langle {x} \rvert} U {\lvert {x'} \rangle}&=\int \left\langle{x} \vert {p'}\right\rangle {\langle {p'} \rvert} U {\lvert {p} \rangle} \left\langle{p} \vert {x'}\right\rangle dp dp' \\ &=\int u_{p'}(x) {\langle {p'} \rvert} e^{-i P^2 t/(2 m \hbar)} {\lvert {p} \rangle} u_p^{*}(x') dp dp' \\ &=\int u_{p'}(x) e^{-i p^2 t/(2 m \hbar)} \delta(p-p') u_p^{*}(x') dp dp' \\ &=\int u_{p}(x) e^{-i p^2 t/(2 m \hbar)} u_p^{*}(x') dp \\ &=\frac{1}{(\sqrt{2 \pi \hbar})^2} \int e^{i p (x-x')/\hbar} e^{-i p^2 t/(2 m \hbar)} dp \\ &=\frac{1}{2 \pi \hbar} \int e^{i p (x-x')/\hbar} e^{-i p^2 t/(2 m \hbar)} dp \\ &=\frac{1}{2 \pi} \int e^{i k (x-x')} e^{-i \hbar k^2 t/(2 m)} dk \\ &=\frac{1}{2 \pi} \int dk e^{- \left(k^2 \frac{ i \hbar t}{2m} - i k (x-x')\right)} \\ &=\frac{1}{2 \pi} \int dk e^{- \frac{ i \hbar t}{2m}\left(k - i \frac{2m}{i \hbar t}\frac{(x-x')}{2} \right)^2- \frac{i^2 2 m (x-x')^2}{4 i \hbar t} } \\ &=\frac{1}{2 \pi}  \sqrt{\pi} \sqrt{\frac{2m}{i \hbar t}}e^{\frac{ i m (x-x')^2}{2 \hbar t}},\end{aligned}

which is the desired result. Now, let’s look at how this would be used. We can express our time evolved state using this matrix element by introducing an identity

\begin{aligned}\left\langle{{x}} \vert {{\psi(t)}}\right\rangle &={\langle {x} \rvert} U {\lvert {\psi(0)} \rangle} \\ &=\int dx' {\langle {x} \rvert} U {\lvert {x'} \rangle} \left\langle{{x'}} \vert {{\psi(0)}}\right\rangle \\ &=\sqrt{\frac{m}{2 \pi i \hbar t}} \int dx' e^{i m (x-x')^2/ (2 \hbar t)}\left\langle{{x'}} \vert {{\psi(0)}}\right\rangle \\ \end{aligned}

This gives us

\begin{aligned}\psi(x, t)=\sqrt{\frac{m}{2 \pi i \hbar t}} \int dx' e^{i m (x-x')^2/ (2 \hbar t)} \psi(x', 0)\end{aligned} \hspace{\stretch{1}}(2.6)

However, note that our free particle wave function at time zero is

\begin{aligned}\psi(x, 0) = \frac{e^{i p x/\hbar}}{\sqrt{2 \pi \hbar}}\end{aligned} \hspace{\stretch{1}}(2.7)

So the convolution integral 2.6 does not exist. We likely have to require that the solution be not a pure state, but instead a superposition of a set of continuous states (a wave packet in position or momentum space related by Fourier transforms). That is

\begin{aligned}\psi(x, 0) &= \frac{1}{{\sqrt{2 \pi \hbar}}} \int \hat{\psi}(p, 0) e^{i p x/\hbar} dp \\ \hat{\psi}(p, 0) &= \frac{1}{{\sqrt{2 \pi \hbar}}} \int \psi(x'', 0) e^{-i p x''/\hbar} dx''\end{aligned} \hspace{\stretch{1}}(2.8)

The time evolution of this wave packet is then determined by the propagator, and is

\begin{aligned}\psi(x,t) =\sqrt{\frac{m}{2 \pi i \hbar t}} \frac{1}{{\sqrt{2 \pi \hbar}}} \int dx' dpe^{i m (x-x')^2/ (2 \hbar t)}\hat{\psi}(p, 0) e^{i p x'/\hbar} ,\end{aligned} \hspace{\stretch{1}}(2.10)

or in terms of the position space wave packet evaluated at time zero

\begin{aligned}\psi(x,t) =\sqrt{\frac{m}{2 \pi i \hbar t}}\frac{1}{{2 \pi}}\int dx' dx'' dke^{i m (x-x')^2/ (2 \hbar t)}e^{i k (x' - x'')} \psi(x'', 0)\end{aligned} \hspace{\stretch{1}}(2.11)

We see that the propagator also ends up with a Fourier transform structure, and we have

\begin{aligned}\psi(x,t) &= \int dx' U(x, x' ; t) \psi(x', 0) \\ U(x, x' ; t) &=\sqrt{\frac{m}{2 \pi i \hbar t}}\frac{1}{{2 \pi}}\int du dke^{i m (x - x' - u)^2/ (2 \hbar t)}e^{i k u }\end{aligned} \hspace{\stretch{1}}(2.12)

Does that Fourier transform exist? I’d not be surprised if it ended up with a delta function representation. I’ll hold off attempting to evaluate and reduce it until another day.

4. Hydrogen atom.

This problem deals with the hydrogen atom, with an initial ket

\begin{aligned}{\lvert {\psi(0)} \rangle} = \frac{1}{{\sqrt{3}}} {\lvert {100} \rangle}+\frac{1}{{\sqrt{3}}} {\lvert {210} \rangle}+\frac{1}{{\sqrt{3}}} {\lvert {211} \rangle},\end{aligned} \hspace{\stretch{1}}(2.14)


\begin{aligned}\left\langle{\mathbf{r}} \vert {{100}}\right\rangle = \Phi_{100}(\mathbf{r}),\end{aligned} \hspace{\stretch{1}}(2.15)


\paragraph{Q: (a)}

If no measurement is made until time t = t_0,

\begin{aligned}t_0 = \frac{\pi \hbar}{ \frac{3}{4} (13.6 \text{eV}) } = \frac{ 4 \pi \hbar }{ 3 E_I},\end{aligned} \hspace{\stretch{1}}(2.16)

what is the ket {\lvert {\psi(t)} \rangle} just before the measurement is made?


Our time evolved state is

\begin{aligned}{\lvert {\psi{t_0}} \rangle} = \frac{1}{{\sqrt{3}}} e^{-i E_1 t_0 /\hbar } {\lvert {100} \rangle}+\frac{1}{{\sqrt{3}}} e^{- i E_2 t_0/\hbar } ({\lvert {210} \rangle} + {\lvert {211} \rangle}).\end{aligned} \hspace{\stretch{1}}(2.17)

Also observe that this initial time was picked to make the exponential values come out nicely, and we have

\begin{aligned}\frac{E_n t_0 }{\hbar} &= - \frac{E_I \pi \hbar }{\frac{3}{4} E_I n^2 \hbar} \\ &= - \frac{4 \pi }{ 3 n^2 },\end{aligned}

so our time evolved state is just

\begin{aligned}{\lvert {\psi(t_0)} \rangle} = \frac{1}{{\sqrt{3}}} e^{-i 4 \pi / 3} {\lvert {100} \rangle}+\frac{1}{{\sqrt{3}}} e^{- i \pi / 3 } ({\lvert {210} \rangle} + {\lvert {211} \rangle}).\end{aligned} \hspace{\stretch{1}}(2.18)

\paragraph{Q: (b)}

Suppose that at time t_0 an L_z measurement is made, and the outcome 0 is recorded. What is the appropriate ket \psi_{\text{after}}(t_0) right after the measurement?


A measurement with outcome 0, means that the L_z operator measurement found the state at that point to be the eigenstate for L_z eigenvalue 0. Recall that if {\lvert {\phi} \rangle} is an eigenstate of L_z we have

\begin{aligned}L_z {\lvert {\phi} \rangle} = m \hbar {\lvert {\phi} \rangle},\end{aligned} \hspace{\stretch{1}}(2.19)

so a measurement of L_z with outcome zero means that we have m=0. Our measurement of L_z at time t_0 therefore filters out all but the m=0 states and our new state is proportional to the projection over all m=0 states as follows

\begin{aligned}{\lvert {\psi_{\text{after}}(t_0)} \rangle}&\propto \left( \sum_{n l} {\lvert {n l 0} \rangle}{\langle {n l 0} \rvert} \right) {\lvert {\psi(t_0)} \rangle}  \\ &\propto \left( {\lvert {1 0 0} \rangle}{\langle {1 0 0} \rvert} +{\lvert {2 1 0} \rangle}{\langle {2 1 0} \rvert} \right) {\lvert {\psi(t_0)} \rangle}  \\ &= \frac{1}{{\sqrt{3}}} e^{-i 4 \pi / 3} {\lvert {100} \rangle}+\frac{1}{{\sqrt{3}}} e^{- i \pi / 3 } {\lvert {210} \rangle} \end{aligned}

A final normalization yields

\begin{aligned}{\lvert {\psi_{\text{after}}(t_0)} \rangle}= \frac{1}{{\sqrt{2}}} ({\lvert {210} \rangle} - {\lvert {100} \rangle})\end{aligned} \hspace{\stretch{1}}(2.20)

\paragraph{Q: (c)}

Right after this L_z measurement, what is {\left\lvert{\psi_{\text{after}}(t_0)}\right\rvert}^2?


Our amplitude is

\begin{aligned}\left\langle{\mathbf{r}} \vert {{\psi_{\text{after}}(t_0)}}\right\rangle&= \frac{1}{{\sqrt{2}}} (\left\langle{\mathbf{r}} \vert {{210}}\right\rangle - \left\langle{\mathbf{r}} \vert {{100}}\right\rangle) \\ &= \frac{1}{{\sqrt{2 \pi a_0^3}}}\left(\frac{r}{4\sqrt{2} a_0} e^{-r/2a_0} \cos\theta-e^{-r/a_0}\right) \\ &= \frac{1}{{\sqrt{2 \pi a_0^3}}}e^{-r/2 a_0} \left(\frac{r}{4\sqrt{2} a_0} \cos\theta-e^{-r/2 a_0}\right),\end{aligned}

so the probability density is

\begin{aligned}{\left\lvert{\left\langle{\mathbf{r}} \vert {{\psi_{\text{after}}(t_0)}}\right\rangle}\right\rvert}^2= \frac{1}{{2 \pi a_0^3}}e^{-r/a_0} \left(\frac{r}{4\sqrt{2} a_0} \cos\theta-e^{-r/2 a_0}\right)^2 \end{aligned} \hspace{\stretch{1}}(2.21)

\paragraph{Q: (d)}

If then a position measurement is made immediately, which if any components of the expectation value of \mathbf{R} will be nonvanishing? Justify your answer.


The expectation value of this vector valued operator with respect to a radial state {\lvert {\psi} \rangle} = \sum_{nlm} a_{nlm} {\lvert {nlm} \rangle} can be expressed as

\begin{aligned}\left\langle{\mathbf{R}}\right\rangle = \sum_{i=1}^3 \mathbf{e}_i \sum_{nlm, n'l'm'} a_{nlm}^{*} a_{n'l'm'} {\langle {nlm} \rvert} X_i{\lvert {n'l'm'} \rangle},\end{aligned} \hspace{\stretch{1}}(2.22)

where X_1 = X = R \sin\Theta \cos\Phi, X_2 = Y = R \sin\Theta \sin\Phi, X_3 = Z = R \cos\Phi.

Consider one of the matrix elements, and expand this by introducing an identity twice

\begin{aligned}{\langle {nlm} \rvert} X_i {\lvert {n'l'm'} \rangle}&=\int r^2 \sin\theta dr d\theta d\phi{r'}^2 \sin\theta' dr' d\theta' d\phi'\left\langle{{nlm}} \vert {{r \theta \phi}}\right\rangle {\langle {r \theta \phi} \rvert} X_i {\lvert {r' \theta' \phi' } \rangle}\left\langle{{r' \theta' \phi'}} \vert {{n'l'm'}}\right\rangle \\ &=\int r^2 \sin\theta dr d\theta d\phi{r'}^2 \sin\theta' dr' d\theta' d\phi'R_{nl}(r) Y_{lm}^{*}(\theta,\phi)\delta^3(\mathbf{x} - \mathbf{x}') x_iR_{n'l'}(r') Y_{l'm'}(\theta',\phi')\\ &=\int r^2 \sin\theta dr d\theta d\phi{r'}^2 \sin\theta' dr' d\theta' d\phi'R_{nl}(r) Y_{lm}^{*}(\theta,\phi) \\ &\qquad{r'}^2 \sin\theta' \delta(r-r') \delta(\theta - \theta') \delta(\phi-\phi')x_iR_{n'l'}(r') Y_{l'm'}(\theta',\phi')\\ &=\int r^2 \sin\theta dr d\theta d\phidr' d\theta' d\phi'R_{nl}(r) Y_{lm}^{*}(\theta,\phi) \delta(r-r') \delta(\theta - \theta') \delta(\phi-\phi')x_iR_{n'l'}(r') Y_{l'm'}(\theta',\phi')\\ &=\int r^2 \sin\theta dr d\theta d\phiR_{nl}(r) R_{n'l'}(r) Y_{lm}^{*}(\theta,\phi) Y_{l'm'}(\theta,\phi)x_i\\ \end{aligned}

Because our state has only m=0 contributions, the only \phi dependence for the X and Y components of \mathbf{R} come from those components themselves. For X, we therefore integrate \int_0^{2\pi} \cos\phi d\phi = 0, and for Y we integrate \int_0^{2\pi} \sin\phi d\phi = 0, and these terms vanish. Our expectation value for \mathbf{R} for this state, therefore lies completely on the z axis.

Questions from the Dec 2008 PHY355H1F exam.

1b. Trace invariance for unitary transformation.

\paragraph{Q:} Show that the trace of an operator is invariant under unitary transforms, i.e. if A' = U^\dagger A U, where U is a unitary operator, prove \text{Tr}(A') = \text{Tr}(A).


The bulk of this question is really to show that commutation of operators leaves the trace invariant (unless this is assumed). To show that we start with the definition of the trace

\begin{aligned}\text{Tr}(AB) &= \sum_n {\langle {n} \rvert} A B {\lvert {n} \rangle} \\ &= \sum_{n m} {\langle {n} \rvert} A {\lvert {m} \rangle} {\langle {m} \rvert} B {\lvert {n} \rangle} \\ &= \sum_{n m} {\langle {m} \rvert} B {\lvert {n} \rangle} {\langle {n} \rvert} A {\lvert {m} \rangle} \\ &= \sum_{m} {\langle {m} \rvert} B A {\lvert {m} \rangle}.\end{aligned}

Thus we have

\begin{aligned}\text{Tr}(A B) = \text{Tr}( B A ).\end{aligned} \hspace{\stretch{1}}(3.23)

For the unitarily transformed operator we have

\begin{aligned}\text{Tr}(A') &= \text{Tr}( U^\dagger A U ) \\ &= \text{Tr}( U^\dagger (A U) ) \\ &= \text{Tr}( (A U) U^\dagger ) \\ &= \text{Tr}( A (U U^\dagger) ) \\ &= \text{Tr}( A ) \qquad \square\end{aligned}

1d. Determinant of an exponential operator in terms of trace.

\paragraph{Q:} If A is an Hermitian operator, show that

\begin{aligned}\text{Det}( \exp A ) = \exp ( \text{Tr}(A) )\end{aligned} \hspace{\stretch{1}}(3.24)

where the Determinant (\text{Det}) of an operator is the product of all its eigenvectors.


The eigenvalues clue in the question provides the starting point. We write the exponential in its series form

\begin{aligned}e^A = 1 + \sum_{k=1}^\infty \frac{1}{{k!}} A^k\end{aligned} \hspace{\stretch{1}}(3.25)

Now, suppose that we have the following eigenvalue relationships for A

\begin{aligned}A {\lvert {n} \rangle} = \lambda_n {\lvert {n} \rangle}.\end{aligned} \hspace{\stretch{1}}(3.26)

From this the exponential is

\begin{aligned}e^A {\lvert {n} \rangle} &= {\lvert {n} \rangle} + \sum_{k=1}^\infty \frac{1}{{k!}} A^k {\lvert {n} \rangle} \\ &= {\lvert {n} \rangle} + \sum_{k=1}^\infty \frac{1}{{k!}} (\lambda_n)^k {\lvert {n} \rangle} \\ &= e^{\lambda_n} {\lvert {n} \rangle}.\end{aligned}

We see that the eigenstates of e^A are those of A, with eigenvalues e^{\lambda_n}.

By the definition of the determinant given we have

\begin{aligned}\text{Det}( e^A ) &= \Pi_n e^{\lambda_n} \\ &= e^{\sum_n \lambda_n} \\ &= e^{\text{Tr}ace(A)} \qquad \square\end{aligned}

1e. Eigenvectors of the Harmonic oscillator creation operator.

\paragraph{Q:} Prove that the only eigenvector of the Harmonic oscillator creation operator is {\lvert {\text{null}} \rangle}.


Recall that the creation (raising) operator was given by

\begin{aligned}a^\dagger = \sqrt{\frac{m \omega}{2 \hbar}} X - \frac{ i }{\sqrt{2 m \omega \hbar} } P= \frac{1}{{ \alpha \sqrt{2} }} X - \frac{ i \alpha }{\sqrt{2} \hbar } P,\end{aligned} \hspace{\stretch{1}}(3.27)

where \alpha = \sqrt{\hbar/m \omega}. Now assume that a^\dagger {\lvert {\phi} \rangle} = \lambda {\lvert {\phi} \rangle} so that

\begin{aligned}{\langle {x} \rvert} a^\dagger {\lvert {\phi} \rangle} = {\langle {x} \rvert} \lambda {\lvert {\phi} \rangle}.\end{aligned} \hspace{\stretch{1}}(3.28)

Write \left\langle{{x}} \vert {{\phi}}\right\rangle = \phi(x), and expand the LHS using 3.27 for

\begin{aligned}\lambda \phi(x) &= {\langle {x} \rvert} a^\dagger {\lvert {\phi} \rangle}  \\ &= {\langle {x} \rvert} \left( \frac{1}{{ \alpha \sqrt{2} }} X - \frac{ i \alpha }{\sqrt{2} \hbar } P \right) {\lvert {\phi} \rangle} \\ &= \frac{x \phi(x)}{ \alpha \sqrt{2} } - \frac{ i \alpha }{\sqrt{2} \hbar } (-i\hbar)\frac{\partial {}}{\partial {x}} \phi(x) \\ &= \frac{x \phi(x)}{ \alpha \sqrt{2} } - \frac{ \alpha }{\sqrt{2} } \frac{\partial {\phi(x)}}{\partial {x}}.\end{aligned}

As usual write \xi = x/\alpha, and rearrange. This gives us

\begin{aligned}\frac{\partial {\phi}}{\partial {\xi}} +\sqrt{2} \lambda \phi - \xi \phi = 0.\end{aligned} \hspace{\stretch{1}}(3.29)

Observe that this can be viewed as a homogeneous LDE of the form

\begin{aligned}\frac{\partial {\phi}}{\partial {\xi}} - \xi \phi = 0,\end{aligned} \hspace{\stretch{1}}(3.30)

augmented by a forcing term \sqrt{2}\lambda \phi. The homogeneous equation has the solution \phi = A e^{\xi^2/2}, so for the complete equation we assume a solution

\begin{aligned}\phi(\xi) = A(\xi) e^{\xi^2/2}.\end{aligned} \hspace{\stretch{1}}(3.31)

Since \phi' = (A' + A \xi) e^{\xi^2/2}, we produce a LDE of

\begin{aligned}0 &= (A' + A \xi -\xi A + \sqrt{2} \lambda A ) e^{\xi^2/2} \\ &= (A' + \sqrt{2} \lambda A ) e^{\xi^2/2},\end{aligned}


\begin{aligned}0 = A' + \sqrt{2} \lambda A.\end{aligned} \hspace{\stretch{1}}(3.32)

This has solution A = B e^{-\sqrt{2} \lambda \xi}, so our solution for 3.29 is

\begin{aligned}\phi(\xi) = B e^{\xi^2/2 - \sqrt{2} \lambda \xi} = B' e^{ (\xi - \lambda \sqrt{2} )^2/2}.\end{aligned} \hspace{\stretch{1}}(3.33)

This wave function is an imaginary Gaussian with minimum at \xi = \lambda\sqrt{2}. It is also unnormalizable since we require B' = 0 for any \lambda if \int {\left\lvert{\phi}\right\rvert}^2 < \infty. Since \left\langle{{\xi}} \vert {{\phi}}\right\rangle = \phi(\xi) = 0, we must also have {\lvert {\phi} \rangle} = 0, completing the exercise.

2. Two level quantum system.

Consider a two-level quantum system, with basis states \{{\lvert {a} \rangle}, {\lvert {b} \rangle}\}. Suppose that the Hamiltonian for this system is given by

\begin{aligned}H = \frac{\hbar \Delta}{2} ( {\lvert {b} \rangle}{\langle {b} \rvert}- {\lvert {a} \rangle}{\langle {a} \rvert})+ i \frac{\hbar \Omega}{2} ( {\lvert {a} \rangle}{\langle {b} \rvert}- {\lvert {b} \rangle}{\langle {a} \rvert})\end{aligned} \hspace{\stretch{1}}(3.34)

where \Delta and \Omega are real positive constants.

\paragraph{Q: (a)} Find the energy eigenvalues and the normalized energy eigenvectors (expressed in terms of the \{{\lvert {a} \rangle}, {\lvert {b} \rangle}\} basis). Write the time evolution operator U(t) = e^{-i H t/\hbar} using these eigenvectors.


The eigenvalue part of this problem is probably easier to do in matrix form. Let

\begin{aligned}{\lvert {a} \rangle} &= \begin{bmatrix}1 \\ 0\end{bmatrix} \\ {\lvert {b} \rangle} &= \begin{bmatrix}0 \\ 1\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.35)

Our Hamiltonian is then

\begin{aligned}H = \frac{\hbar}{2} \begin{bmatrix}-\Delta & i \Omega \\ -i \Omega & \Delta\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.37)

Computing \det{H - \lambda I} = 0, we get

\begin{aligned}\lambda = \pm \frac{\hbar}{2} \sqrt{ \Delta^2 + \Omega^2 }.\end{aligned} \hspace{\stretch{1}}(3.38)

Let \delta = \sqrt{ \Delta^2 + \Omega^2 }. Our normalized eigenvectors are found to be

\begin{aligned}{\lvert {\pm} \rangle} = \frac{1}{{\sqrt{ 2 \delta (\delta \pm \Delta)} }}\begin{bmatrix}i \Omega \\ \Delta \pm \delta\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.39)

In terms of {\lvert {a} \rangle} and {\lvert {b} \rangle}, we then have

\begin{aligned}{\lvert {\pm} \rangle} = \frac{1}{{\sqrt{ 2 \delta (\delta \pm \Delta)} }}\left(i \Omega {\lvert {a} \rangle}+ (\Delta \pm \delta) {\lvert {b} \rangle} \right).\end{aligned} \hspace{\stretch{1}}(3.40)

Note that our Hamiltonian has a simple form in this basis. That is

\begin{aligned}H = \frac{\delta \hbar}{2} ({\lvert {+} \rangle}{\langle {+} \rvert} - {\lvert {-} \rangle}{\langle {-} \rvert} )\end{aligned} \hspace{\stretch{1}}(3.41)

Observe that once we do the diagonalization, we have a Hamiltonian that appears to have the form of a scaled projector for an open Stern-Gerlach aparatus.

Observe that the diagonalized Hamiltonian operator makes the time evolution operator’s form also simple, which is, by inspection

\begin{aligned}U(t) = e^{-i t \frac{\delta}{2}} {\lvert {+} \rangle}{\langle {+} \rvert} + e^{i t \frac{\delta}{2}} {\lvert {-} \rangle}{\langle {-} \rvert}.\end{aligned} \hspace{\stretch{1}}(3.42)

Since we are asked for this in terms of {\lvert {a} \rangle}, and {\lvert {b} \rangle}, the projectors {\lvert {\pm} \rangle}{\langle {\pm} \rvert} are required. These are

\begin{aligned}{\lvert {\pm} \rangle}{\langle {\pm} \rvert} &= \frac{1}{{2 \delta (\delta \pm \Delta)}}\Bigl( i \Omega {\lvert {a} \rangle} + (\Delta \pm \delta) {\lvert {b} \rangle} \Bigr)\Bigl( -i \Omega {\langle {a} \rvert} + (\Delta \pm \delta) {\langle {b} \rvert} \Bigr) \\ \end{aligned}

\begin{aligned}{\lvert {\pm} \rangle}{\langle {\pm} \rvert} = \frac{1}{{2 \delta (\delta \pm \Delta)}}\Bigl(\Omega^2 {\lvert {a} \rangle}{\langle {a} \rvert}+(\delta \pm \delta)^2 {\lvert {b} \rangle}{\langle {b} \rvert}+i \Omega (\Delta \pm \delta) ({\lvert {a} \rangle}{\langle {b} \rvert}-{\lvert {b} \rangle}{\langle {a} \rvert})\Bigr)\end{aligned} \hspace{\stretch{1}}(3.43)

Substitution into 3.42 and a fair amount of algebra leads to

\begin{aligned}U(t) = \cos(\delta t/2) \Bigl( {\lvert {a} \rangle}{\langle {a} \rvert} + {\lvert {b} \rangle}{\langle {b} \rvert} \Bigr)+ i \frac{\Omega}{\delta} \sin(\delta t/2) \Bigl( {\lvert {a} \rangle}{\langle {a} \rvert} - {\lvert {b} \rangle}{\langle {b} \rvert} -i ({\lvert {a} \rangle}{\langle {b} \rvert} - {\lvert {b} \rangle}{\langle {a} \rvert} )\Bigr).\end{aligned} \hspace{\stretch{1}}(3.44)

Note that while a big cumbersome, we can also verify that we can recover the original Hamiltonian from 3.41 and 3.43.

\paragraph{Q: (b)}

Suppose that the initial state of the system at time t = 0 is {\lvert {\phi(0)} \rangle}= {\lvert {b} \rangle}. Find an expression for the state at some later time t > 0, {\lvert {\phi(t)} \rangle}.


Most of the work is already done. Computation of {\lvert {\phi(t)} \rangle} = U(t) {\lvert {\phi(0)} \rangle} follows from 3.44

\begin{aligned}{\lvert {\phi(t)} \rangle} =\cos(\delta t/2) {\lvert {b} \rangle}- i \frac{\Omega}{\delta} \sin(\delta t/2) \Bigl( {\lvert {b} \rangle} +i {\lvert {a} \rangle}\Bigr).\end{aligned} \hspace{\stretch{1}}(3.45)

\paragraph{Q: (c)}

Suppose that an observable, specified by the operator X = {\lvert {a} \rangle}{\langle {b} \rvert} + {\lvert {b} \rangle}{\langle {a} \rvert}, is measured for this system. What is the probabilbity that, at time t, the result 1 is obtained? Plot this probability as a function of time, showing the maximum and minimum values of the function, and the corresponding values of t.


The language of questions like these attempt to bring some physics into the mathematics. The phrase “the result 1 is obtained”, is really a statement that the operator X, after measurement is found to have the eigenstate with numeric value 1.

We can calcuate the eigenvectors for this operator easily enough and find them to be \pm 1. For the positive eigenvalue we can also compute the eigenstate to be

\begin{aligned}{\lvert {X+} \rangle} = \frac{1}{{\sqrt{2}}} \Bigl( {\lvert {a} \rangle} + {\lvert {b} \rangle} \Bigr).\end{aligned} \hspace{\stretch{1}}(3.46)

The question of what the probability for this measurement is then really a question asking for the computation of the amplitude

\begin{aligned}{\left\lvert{\frac{1}{{\sqrt{2}}}\left\langle{{ (a + b)}} \vert {{\phi(t)}}\right\rangle}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(3.47)

From 3.45 we find this probability to be

\begin{aligned}{\left\lvert{\frac{1}{{\sqrt{2}}}\left\langle{{ (a + b)}} \vert {{\phi(t)}}\right\rangle}\right\rvert}^2&=\frac{1}{{2}} \left(\left(\cos(\delta t/2) + \frac{\Omega}{\delta} \sin(\delta t/2)\right)^2+ \frac{ \Omega^2 \sin^2(\delta t/2)}{\delta^2}\right) \\ &=\frac{1}{{4}} \left( 1 + 3 \frac{\Omega^2}{\delta^2} + \frac{\Delta^2}{\delta^2} \cos (\delta t) + 2 \frac{ \Omega}{\delta} \sin(\delta t) \right)\end{aligned}

We have a simple superposition of two sinusuiods out of phase, periodic with period 2 \pi/\delta. I’d attempted a rough sketch of this on paper, but won’t bother scanning it here or describing it further.

\paragraph{Q: (d)}

Suppose an experimenter has control over the values of the parameters \Delta and \Omega. Explain how she might prepare the state ({\lvert {a} \rangle} + {\lvert {b} \rangle})/\sqrt{2}.


For this part of the question I wasn’t sure what approach to take. I thought perhaps this linear combination of states could be made to equal one of the energy eigenstates, and if one could prepare the system in that state, then for certain values of \delta and \Delta one would then have this desired state.

To get there I note that we can express the states {\lvert {a} \rangle}, and {\lvert {b} \rangle} in terms of the eigenstates by inverting

\begin{aligned}\begin{bmatrix}{\lvert {+} \rangle} \\ {\lvert {-} \rangle} \\ \end{bmatrix}=\frac{1}{{\sqrt{2\delta}}}\begin{bmatrix}\frac{i \Omega}{\sqrt{\delta + \Delta}} & \sqrt{\delta + \Delta} \\ \frac{i \Omega}{\sqrt{\delta - \Delta}} & -\sqrt{\delta - \Delta}\end{bmatrix}\begin{bmatrix}{\lvert {a} \rangle} \\ {\lvert {b} \rangle} \\ \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.48)

Skipping all the algebra one finds

\begin{aligned}\begin{bmatrix}{\lvert {a} \rangle} \\ {\lvert {b} \rangle} \\ \end{bmatrix}=\begin{bmatrix}-i\sqrt{\delta - \Delta} & -i\sqrt{\delta + \Delta} \\ \frac{\Omega}{\sqrt{\delta - \Delta}} &-\frac{\Omega}{\sqrt{\delta + \Delta}} \end{bmatrix}\begin{bmatrix}{\lvert {+} \rangle} \\ {\lvert {-} \rangle} \\ \end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(3.49)

Unfortunately, this doesn’t seem helpful. I find

\begin{aligned}\frac{1}{{\sqrt{2}}} ( {\lvert {a} \rangle} + {\lvert {b} \rangle} ) = \frac{{\lvert {+} \rangle}}{\sqrt{\delta - \Delta}}( \Omega - i (\delta - \Delta) )-\frac{{\lvert {-} \rangle}}{\sqrt{\delta + \Delta}}( \Omega + i (\delta + \Delta) )\end{aligned} \hspace{\stretch{1}}(3.50)

There’s no obvious way to pick \Omega and \Delta to leave just {\lvert {+} \rangle} or {\lvert {-} \rangle}. When I did this on paper originally I got a different answer for this sum, but looking at it now, I can’t see how I managed to get that answer (it had no factors of i in the result as the one above does).

3. One dimensional harmonic oscillator.

Consider a one-dimensional harmonic oscillator with the Hamiltonian

\begin{aligned}H = \frac{1}{{2m}}P^2 + \frac{1}{{2}} m \omega^2 X^2\end{aligned} \hspace{\stretch{1}}(3.51)

Denote the ground state of the system by {\lvert {0} \rangle}, the first excited state by {\lvert {1} \rangle} and so on.

\paragraph{Q: (a)}
Evaluate {\langle {n} \rvert} X {\lvert {n} \rangle} and {\langle {n} \rvert} X^2 {\lvert {n} \rangle} for arbitrary {\lvert {n} \rangle}.


Writing X in terms of the raising and lowering operators we have

\begin{aligned}X = \frac{\alpha}{\sqrt{2}} (a^\dagger + a),\end{aligned} \hspace{\stretch{1}}(3.52)

so \left\langle{{X}}\right\rangle is proportional to

\begin{aligned}{\langle {n} \rvert} a^\dagger + a {\lvert {n} \rangle} = \sqrt{n+1} \left\langle{{n}} \vert {{n+1}}\right\rangle + \sqrt{n} \left\langle{{n}} \vert {{n-1}}\right\rangle = 0.\end{aligned} \hspace{\stretch{1}}(3.53)

For \left\langle{{X^2}}\right\rangle we have

\begin{aligned}\left\langle{{X^2}}\right\rangle&=\frac{\alpha^2}{2}{\langle {n} \rvert} (a^\dagger + a)(a^\dagger + a) {\lvert {n} \rangle} \\ &=\frac{\alpha^2}{2}{\langle {n} \rvert} (a^\dagger + a) \left( \sqrt{n+1} {\lvert {n+1} \rangle} + \sqrt{n-1} {\lvert {n-1} \rangle}\right)  \\ &=\frac{\alpha^2}{2}{\langle {n} \rvert} \Bigl( (n+1) {\lvert {n} \rangle} + \sqrt{n(n-1)} {\lvert {n-2} \rangle}+ \sqrt{(n+1)(n+2)} {\lvert {n+2} \rangle} + n {\lvert {n} \rangle} \Bigr).\end{aligned}

We are left with just

\begin{aligned}\left\langle{{X^2}}\right\rangle = \frac{\hbar}{2 m \omega} (2n + 1).\end{aligned} \hspace{\stretch{1}}(3.54)

\paragraph{Q: (b)}

Suppose that at t=0 the system is prepared in the state

\begin{aligned}{\lvert {\psi(0)} \rangle} = \frac{1}{{\sqrt{2}}} ( {\lvert {0} \rangle} + i {\lvert {1} \rangle} ).\end{aligned} \hspace{\stretch{1}}(3.55)

If a measurement of position X were performaed immediately, sketch the propability distribution P(x) that a particle would be found within dx of x. Justify how you construct the sketch.


The probability that we started in state {\lvert {\psi(0)} \rangle} and ended up in position x is governed by the amplitude \left\langle{{x}} \vert {{\psi(0)}}\right\rangle, and the probability of being within an interval \Delta x, surrounding the point x is given by

\begin{aligned}\int_{x'=x-\Delta x/2}^{x+\Delta x/2} {\left\lvert{ \left\langle{{x'}} \vert {{\psi(0)}}\right\rangle }\right\rvert}^2 dx'.\end{aligned} \hspace{\stretch{1}}(3.56)

In the limit as \Delta x \rightarrow 0, this is just the squared amplitude itself evaluated at the point x, so we are interested in the quantity

\begin{aligned}{\left\lvert{ \left\langle{{x}} \vert {{\psi(0)}}\right\rangle }\right\rvert}^2  = \frac{1}{{2}} {\left\lvert{ \left\langle{{x}} \vert {{0}}\right\rangle + i \left\langle{{x}} \vert {{1}}\right\rangle }\right\rvert}^2.\end{aligned} \hspace{\stretch{1}}(3.57)

We are given these wave functions in the supplemental formulas. Namely,

\begin{aligned}\left\langle{{x}} \vert {{0}}\right\rangle &= \psi_0(x) = \frac{e^{-x^2/2\alpha^2}}{ \sqrt{\alpha \sqrt{\pi}}} \\ \left\langle{{x}} \vert {{1}}\right\rangle &= \psi_1(x) = \frac{e^{-x^2/2\alpha^2} 2 x }{ \alpha \sqrt{2 \alpha \sqrt{\pi}}}.\end{aligned} \hspace{\stretch{1}}(3.58)

Substituting these into 3.57 we have

\begin{aligned}{\left\lvert{ \left\langle{{x}} \vert {{\psi(0)}}\right\rangle }\right\rvert}^2 = \frac{1}{{2}} e^{-x^2/\alpha^2}\frac{1}{{ \alpha \sqrt{\pi}}}{\left\lvert{ 1 + \frac{2 i x}{\alpha \sqrt{2} } }\right\rvert}^2=\frac{e^{-x^2/\alpha^2}}{ 2\alpha \sqrt{\pi}}\left( 1 + \frac{2 x^2}{\alpha^2 } \right).\end{aligned} \hspace{\stretch{1}}(3.60)

This \href{^(-x^2)+(1+

\paragraph{Q: (c)}

Now suppose the state given in (b) above were allowed to evolve for a time t, determine the expecation value of X and \Delta X at that time.


Our time evolved state is

\begin{aligned}U(t) {\lvert {\psi(0)} \rangle} = \frac{1}{{\sqrt{2}}}\left(e^{-i \hbar \omega \left( 0 + \frac{1}{{2}} \right) t/\hbar } {\lvert {0} \rangle}+ i e^{-i \hbar \omega \left( 1 + \frac{1}{{2}} \right) t/\hbar } {\lvert {0} \rangle}\right)=\frac{1}{{\sqrt{2}}}\left(e^{-i \omega t/2 } {\lvert {0} \rangle}+ i e^{- 3 i \omega t/2 } {\lvert {1} \rangle}\right).\end{aligned} \hspace{\stretch{1}}(3.61)

The position expectation is therefore

\begin{aligned}{\langle {\psi(t)} \rvert} X {\lvert {\psi(t)} \rangle}&= \frac{\alpha}{2 \sqrt{2}}\left(e^{i \omega t/2 } {\langle {0} \rvert}- i e^{ 3 i \omega t/2 } {\langle {1} \rvert}\right)(a^\dagger + a)\left(e^{-i \omega t/2 } {\lvert {0} \rangle}+ i e^{- 3 i \omega t/2 } {\lvert {1} \rangle}\right) \\ \end{aligned}

We have already demonstrated that {\langle {n} \rvert} X {\lvert {n} \rangle} = 0, so we must only expand the cross terms, but those are just {\langle {0} \rvert} a^\dagger + a {\lvert {1} \rangle} = 1. This leaves

\begin{aligned}{\langle {\psi(t)} \rvert} X {\lvert {\psi(t)} \rangle}= \frac{\alpha}{2 \sqrt{2}}\left( -i e^{i \omega t} + i e^{-i \omega t} \right)=\sqrt{\frac{\hbar}{2 m \omega}} \cos(\omega t)\end{aligned} \hspace{\stretch{1}}(3.62)

For the squared position expectation

\begin{aligned}{\langle {\psi(t)} \rvert} X^2 {\lvert {\psi(t)} \rangle}&= \frac{\alpha^2}{4 (2)}\left(e^{i \omega t/2 } {\langle {0} \rvert}- i e^{ 3 i \omega t/2 } {\langle {1} \rvert}\right)(a^\dagger + a)^2\left(e^{-i \omega t/2 } {\lvert {0} \rangle}+ i e^{- 3 i \omega t/2 } {\lvert {1} \rangle}\right) \\ &=\frac{1}{{2}} ( {\langle {0} \rvert} X^2 {\lvert {0} \rangle} + {\langle {1} \rvert} X^2 {\lvert {1} \rangle} )+ i \frac{\alpha^2 }{8} ( - e^{ i \omega t} {\langle {1} \rvert} (a^\dagger + a)^2 {\lvert {0} \rangle}+ e^{ -i \omega t} {\langle {0} \rvert} (a^\dagger + a)^2 {\lvert {1} \rangle})\end{aligned}

Noting that (a^\dagger + a) {\lvert {0} \rangle} = {\lvert {1} \rangle}, and (a^\dagger + a)^2 {\lvert {0} \rangle} = (a^\dagger + a){\lvert {1} \rangle} = \sqrt{2} {\lvert {2} \rangle} + {\lvert {0} \rangle}, so we see the last two terms are zero. The first two we can evaluate using our previous result 3.54 which was \left\langle{{X^2}}\right\rangle = \frac{\alpha^2}{2} (2n + 1). This leaves

\begin{aligned}{\langle {\psi(t)} \rvert} X^2 {\lvert {\psi(t)} \rangle} = \alpha^2 \end{aligned} \hspace{\stretch{1}}(3.63)

Since \left\langle{{X}}\right\rangle^2 = \alpha^2 \cos^2(\omega t)/2, we have

\begin{aligned}(\Delta X)^2 = \left\langle{{X^2}}\right\rangle - \left\langle{{X}}\right\rangle^2 = \alpha^2 \left(1 - \frac{1}{{2}} \cos^2(\omega t) \right)\end{aligned} \hspace{\stretch{1}}(3.64)

\paragraph{Q: (d)}

Now suppose that initially the system were prepared in the ground state {\lvert {0} \rangle}, and then the resonance frequency is changed abrubtly from \omega to \omega' so that the Hamiltonian becomes

\begin{aligned}H = \frac{1}{{2m}}P^2 + \frac{1}{{2}} m {\omega'}^2 X^2.\end{aligned} \hspace{\stretch{1}}(3.65)

Immediately, an energy measurement is performed ; what is the probability of obtaining the result E = \hbar \omega' (3/2)?


This energy measurement E = \hbar \omega' (3/2) = \hbar \omega' (1 + 1/2), corresponds to an observation of state {\lvert {1'} \rangle}, after an initial observation of {\lvert {0} \rangle}. The probability of such a measurement is

\begin{aligned}{\left\lvert{ \left\langle{{1'}} \vert {{0}}\right\rangle }\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(3.66)

Note that

\begin{aligned}\left\langle{{1'}} \vert {{0}}\right\rangle &=\int dx \left\langle{{1'}} \vert {{x}}\right\rangle\left\langle{{x}} \vert {{0}}\right\rangle \\ &=\int dx \psi_{1'}^{*} \psi_0(x) \\ \end{aligned}

The wave functions above are

\begin{aligned}\phi_{1'}(x) &= \frac{ 2 x e^{-x^2/2 {\alpha'}^2 }}{ \alpha' \sqrt{ 2 \alpha' \sqrt{\pi} } } \\ \phi_{0}(x) &= \frac{ e^{-x^2/2 {\alpha}^2 } } { \sqrt{ \alpha \sqrt{\pi} } } \end{aligned} \hspace{\stretch{1}}(3.67)

Putting the pieces together we have

\begin{aligned}\left\langle{{1'}} \vert {{0}}\right\rangle &=\frac{2 }{ \alpha' \sqrt{ 2 \alpha' \alpha \pi } }\int dxx e^{-\frac{x^2}{2}\left( \frac{1}{{{\alpha'}^2}} + \frac{1}{{\alpha^2}} \right) }\end{aligned} \hspace{\stretch{1}}(3.69)

Since this is an odd integral kernel over an even range, this evaluates to zero, and we conclude that the probability of measuring the specified energy is zero when the system is initially prepared in the ground state associated with the original Hamiltonian. Intuitively this makes some sense, if one thinks of the Fourier coefficient problem: one cannot construct an even function from linear combinations of purely odd functions.

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Harmonic Oscillator position and momentum Hamiltonian operators

Posted by peeterjoot on December 18, 2010

[Click here for a PDF of this post with nicer formatting]


Hamiltonian problem from Chapter 9 of [1].

Problem 1.


Assume x(t) and p(t) to be Heisenberg operators with x(0) = x_0 and p(0) = p_0. For a Hamiltonian corresponding to the harmonic oscillator show that

\begin{aligned}x(t) &= x_0 \cos \omega t + \frac{p_0}{m \omega} \sin \omega t \\ p(t) &= p_0 \cos \omega t - m \omega x_0 \sin \omega t.\end{aligned} \hspace{\stretch{1}}(3.1)


Recall that the Hamiltonian operators were defined by factoring out the time evolution from a set of states

\begin{aligned}{\langle {\alpha(t) } \rvert} A {\lvert { \beta(t) } \rangle}={\langle {\alpha(0) } \rvert} e^{i H t/\hbar} A e^{-i H t/\hbar} {\lvert { \beta(0) } \rangle}.\end{aligned} \hspace{\stretch{1}}(3.3)

So one way to complete the task is to compute these exponential sandwiches. Recall from the appendix of chapter 10, that we have

\begin{aligned}e^A B e^{-A}= B + \left[{A},{B}\right]+ \frac{1}{{2!}} \left[{A},{\left[{A},{B}\right]}\right] + \cdots\end{aligned} \hspace{\stretch{1}}(3.4)

Perhaps there is also some smarter way to do this, but lets first try the obvious way.

Let’s summarize the variables we will work with

\begin{aligned}\alpha &= \sqrt{\frac{m \omega}{\hbar}} \\ X &= \frac{1}{{\alpha \sqrt{2}}} ( a + a^\dagger ) \\ P &= -i \hbar \frac{\alpha}{\sqrt{2}} ( a - a^\dagger ) \\ H &= \hbar \omega ( a^\dagger a + 1/2 ) \\ \left[{a},{a^\dagger}\right] &= 1 \end{aligned} \hspace{\stretch{1}}(3.5)

The operator in the exponential sandwich is

\begin{aligned}A = i H t/\hbar = i \omega t ( a^\dagger a + 1/2 )\end{aligned} \hspace{\stretch{1}}(3.10)

Note that the constant 1/2 factor will commute with all operators, which reduces the computation required

\begin{aligned}\antisymmetric{i H t/\hbar} {B } = (i\omega t) \left[{a^\dagger a},{B}\right]\end{aligned} \hspace{\stretch{1}}(3.11)

For B = X, or B = P, we’ll want some intermediate results

\begin{aligned}\left[{a^\dagger a},{a}\right]&=a^\dagger a a - a a^\dagger a \\  &=a^\dagger a a - (a^\dagger a + 1) a \\  &=-a,\end{aligned}


\begin{aligned}\left[{a^\dagger a},{a^\dagger}\right]&=a^\dagger a a^\dagger - a^\dagger a^\dagger a \\  &=a^\dagger a a^\dagger - a^\dagger (a a^\dagger -1) \\  &=a^\dagger\end{aligned}

Using these we can evaluate the commutators for the position and momentum operators. For position we have

\begin{aligned}\left[{i H t /\hbar },{X}\right] &= (i \omega t) \frac{1}{{\alpha \sqrt{2}}} \left[{a^\dagger a},{a+ a^\dagger}\right] \\ &= (i \omega t) \frac{1}{{\alpha \sqrt{2}}} (-a + a^\dagger ) \\ &= \frac{\omega t}{\alpha^2} \frac{-i \hbar \alpha}{ \sqrt{2}} (a - a^\dagger ).\end{aligned}

Since \alpha^2 \hbar = m \omega, we have

\begin{aligned}\left[{i H t /\hbar },{X}\right] = (\omega t) \frac{P}{m \omega }.\end{aligned} \hspace{\stretch{1}}(3.12)

For the momentum operator we have

\begin{aligned}\left[{i H t /\hbar },{P}\right] &= (i \omega t) \frac{-i \hbar \alpha}{ \sqrt{2}} \left[{a^\dagger a},{a- a^\dagger}\right] \\ &= (i \omega t) \frac{i \hbar \alpha}{ \sqrt{2}} (a + a^\dagger) \\ &= (\omega t) (\hbar \alpha^2) X\end{aligned}

So we have

\begin{aligned}\left[{i H t /\hbar },{P}\right] = (-\omega t) (m \omega ) X\end{aligned} \hspace{\stretch{1}}(3.13)

The expansion of the exponential series of nested commutators can now be written down by inspection and we get

\begin{aligned}X_H = X + (\omega t) \frac{P}{m \omega} - \frac{(\omega t)^2}{2!} X - \frac{(\omega t)^3}{3!} \frac{P}{m \omega} + \cdots\end{aligned} \hspace{\stretch{1}}(3.14)

\begin{aligned}P_H = P - (\omega t) (m \omega)X - \frac{(\omega t)^2}{2!} P + \frac{(\omega t)^3}{3!} (m \omega)X + \cdots\end{aligned} \hspace{\stretch{1}}(3.15)

Collection of terms gives us the desired answer

\begin{aligned}X_H = X \cos(\omega t) + \frac{P}{m \omega} \sin(\omega t)\end{aligned} \hspace{\stretch{1}}(3.16)

\begin{aligned}P_H = P \cos(\omega t) - (m \omega) X \sin(\omega t)\end{aligned} \hspace{\stretch{1}}(3.17)


[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

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My submission for PHY356 (Quantum Mechanics I) Problem Set 5.

Posted by peeterjoot on December 7, 2010

[Click here for a PDF of this post with nicer formatting]

Grading notes.

The pdf version above has been adjusted after seeing the comments from the grading. [Click here for the PDF for the original submission, as found below, uncorrected.



A particle of mass m moves along the x-direction such that V(X)=\frac{1}{{2}}KX^2. Is the state

\begin{aligned}u(\xi) = B \xi e^{+\xi^2/2},\end{aligned} \hspace{\stretch{1}}(2.1)

where \xi is given by Eq. (9.60), B is a constant, and time t=0, an energy eigenstate of the system? What is probability per unit length for measuring the particle at position x=0 at t=t_0>0? Explain the physical meaning of the above results.


Is this state an energy eigenstate?

Recall that \xi = \alpha x, \alpha = \sqrt{m\omega/\hbar}, and K = m \omega^2. With this variable substitution Schr\”{o}dinger’s equation for this harmonic oscillator potential takes the form

\begin{aligned}\frac{d^2 u}{d\xi^2} - \xi^2 u = \frac{2 E }{\hbar\omega} u\end{aligned} \hspace{\stretch{1}}(2.2)

While we can blindly substitute a function of the form \xi e^{\xi^2/2} into this to get

\begin{aligned}\frac{1}{{B}} \left(\frac{d^2 u}{d\xi^2} - \xi^2 u\right)&=\frac{d}{d\xi} \left( 1 + \xi^2 \right) e^{\xi^2/2} - \xi^3 e^{\xi^2/2} \\ &=\left( 2 \xi + \xi + \xi^3 \right) e^{\xi^2/2} - \xi^3 e^{\xi^2/2} \\ &=3 \xi e^{\xi^2/2}\end{aligned}

and formally make the identification E = 3 \omega \hbar/2 = (1 + 1/2) \omega \hbar, this isn’t a normalizable wavefunction, and has no physical relevance, unless we set B = 0.

By changing the problem, this state could be physically relevant. We’d require a potential of the form

\begin{aligned}V(x) =\left\{\begin{array}{l l}f(x) & \quad \mbox{if latex x < a$} \\ \frac{1}{{2}} K x^2 & \quad \mbox{if $latex a < x b$} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.3)$

For example, f(x) = V_1, g(x) = V_2, for constant V_1, V_2. For such a potential, within the harmonic well, a general solution of the form

\begin{aligned}u(x,t) = \sum_n H_n(\xi) \Bigl(A_n e^{-\xi^2/2} + B_n e^{\xi^2/2} \Bigr) e^{-i E_n t/\hbar},\end{aligned} \hspace{\stretch{1}}(2.4)

is possible since normalization would not prohibit non-zero B_n values in that situation. For the wave function to be a physically relevant, we require it to be (absolute) square integrable, and must also integrate to unity over the entire interval.

Probability per unit length at x=0.

We cannot answer the question for the probability that the particle is found at the specific x=0 position at t=t_0 (that probability is zero in a continuous space), but we can answer the question for the probability that a particle is found in an interval surrounding a specific point at this time. By calculating the average of the probability to find the particle in an interval, and dividing by that interval’s length, we arrive at plausible definition of probability per unit length for an interval surrounding x = x_0

\begin{aligned}P = \text{Probability per unit length near latex x = x_0$} =\lim_{\epsilon \rightarrow 0} \frac{1}{{\epsilon}} \int_{x_0 – \epsilon/2}^{x_0 + \epsilon/2} {\left\lvert{ \Psi(x, t_0) }\right\rvert}^2 dx = {\left\lvert{\Psi(x_0, t_0)}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(2.5)$

By this definition, the probability per unit length is just the probability density itself, evaluated at the point of interest.

Physically, for an interval small enough that the probability density is constant in magnitude over that interval, this probability per unit length times the length of this small interval, represents the probability that we will find the particle in that interval.

Probability per unit length for the non-normalizable state given.

It seems possible, albeit odd, that this question is asking for the probability per unit length for the non-normalizable E_1 wavefunction 2.1. Since normalization requires B=0, that probability density is simply zero (or undefined, depending on one’s point of view).

Probability per unit length for some more interesting harmonic oscillator states.

Suppose we form the wavefunction for a superposition of all the normalizable states

\begin{aligned}u(x,t) = \sum_n A_n H_n(\xi) e^{-\xi^2/2} e^{-i E_n t/\hbar}\end{aligned} \hspace{\stretch{1}}(2.6)

Here it is assumed that the A_n coefficients yield unit probability

\begin{aligned}\int {\left\lvert{u(x,0)}\right\rvert}^2 dx = \sum_n {\left\lvert{A_n}\right\rvert}^2 = 1\end{aligned} \hspace{\stretch{1}}(2.7)

For the impure state of 2.6 we have for the probability density

\begin{aligned}{\left\lvert{u}\right\rvert}^2&=\sum_{m,n}A_n A_m^{*} H_n(\xi) H_m(\xi) e^{-\xi^2} e^{-i (E_n - E_m)t_0/\hbar} \\ &=\sum_n{\left\lvert{A_n}\right\rvert}^2 (H_n(\xi))^2 e^{-\xi^2}+\sum_{m \ne n}A_n A_m^{*} H_n(\xi) H_m(\xi) e^{-\xi^2} e^{-i (E_n - E_m)t_0/\hbar} \\ &=\sum_n{\left\lvert{A_n}\right\rvert}^2 (H_n(\xi))^2 e^{-\xi^2}+\sum_{m \ne n}A_n A_m^{*} H_n(\xi) H_m(\xi) e^{-\xi^2} e^{-i (E_n - E_m)t_0/\hbar} \\ &=\sum_n{\left\lvert{A_n}\right\rvert}^2 (H_n(\xi))^2 e^{-\xi^2}+\sum_{m < n}H_n(\xi) H_m(\xi)\left(A_n A_m^{*}e^{-\xi^2} e^{-i (E_n - E_m)t_0/\hbar}+A_m A_n^{*}e^{-\xi^2} e^{-i (E_m - E_n)t_0/\hbar}\right) \\ &=\sum_n{\left\lvert{A_n}\right\rvert}^2 (H_n(\xi))^2 e^{-\xi^2}+2 \sum_{m < n}H_n(\xi) H_m(\xi)e^{-\xi^2}\Re \left(A_n A_m^{*}e^{-i (E_n - E_m)t_0/\hbar}\right) \\ &=\sum_n{\left\lvert{A_n}\right\rvert}^2 (H_n(\xi))^2 e^{-\xi^2}  \\ &\quad+2 \sum_{m < n}H_n(\xi) H_m(\xi)e^{-\xi^2}\left(\Re ( A_n A_m^{*} ) \cos( (n - m)\omega t_0)+\Im ( A_n A_m^{*} ) \sin( (n - m)\omega t_0)\right) \\ \end{aligned}

Evaluation at the point x = 0, we have

\begin{aligned}{\left\lvert{u(0,t_0)}\right\rvert}^2=\sum_n{\left\lvert{A_n}\right\rvert}^2 (H_n(0))^2 +2 \sum_{m < n} H_n(0) H_m(0) \left( \Re ( A_n A_m^{*} ) \cos( (n - m)\omega t_0) +\Im ( A_n A_m^{*} ) \sin( (n - m)\omega t_0)\right)\end{aligned} \hspace{\stretch{1}}(2.8)

It is interesting that the probability per unit length only has time dependence for a mixed state.

For a pure state and its wavefunction u(x,t) = N_n H_n(\xi) e^{-\xi^2/2} e^{-i E_n t/\hbar} we have just

\begin{aligned}{\left\lvert{u(0,t_0)}\right\rvert}^2=N_n^2 (H_n(0))^2 = \frac{\alpha}{\sqrt{\pi} 2^n n!} H_n(0)^2\end{aligned} \hspace{\stretch{1}}(2.9)

This is zero for odd n. For even n is appears that (H_n(0))^2 may equal 2^n (this is true at least up to n=4). If that’s the case, we have for non-mixed states, with even numbered energy quantum numbers, at x=0 a probability per unit length value of {\left\lvert{u(0,t_0)}\right\rvert}^2 = \frac{\alpha}{\sqrt{\pi} n!}.

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My submission for PHY356 (Quantum Mechanics I) Problem Set 4.

Posted by peeterjoot on December 7, 2010

[Click here for a PDF of this post with nicer formatting]

Grading notes.

The pdf version above has been adjusted with some grading commentary. [Click here for the PDF for the original submission, as found below.

Problem 1.


Is it possible to derive the eigenvalues and eigenvectors presented in Section 8.2 from those in Section 8.1.2? What does this say about the potential energy operator in these two situations?

For reference 8.1.2 was a finite potential barrier, V(x) = V_0, {\left\lvert{x}\right\rvert} > a, and zero in the interior of the well. This had trigonometric solutions in the interior, and died off exponentially past the boundary of the well.

On the other hand, 8.2 was a delta function potential V(x) = -g \delta(x), which had the solution u(x) = \sqrt{\beta} e^{-\beta {\left\lvert{x}\right\rvert}}, where \beta = m g/\hbar^2.


The pair of figures in the text [1] for these potentials doesn’t make it clear that there are possibly any similarities. The attractive delta function potential isn’t illustrated (although the delta function is, but with opposite sign), and the scaling and the reference energy levels are different. Let’s illustrate these using the same reference energy level and sign conventions to make the similarities more obvious.

\caption{8.1.2 Finite Well potential (with energy shifted downwards by V_0)}

\caption{8.2 Delta function potential.}

The physics isn’t changed by picking a different point for the reference energy level, so let’s compare the two potentials, and their solutions using V(x) = 0 outside of the well for both cases. The method used to solve the finite well problem in the text is hard to follow, so re-doing this from scratch in a slightly tidier way doesn’t hurt.

Schr\”{o}dinger’s equation for the finite well, in the {\left\lvert{x}\right\rvert} > a region is

\begin{aligned}-\frac{\hbar^2}{2m} u'' = E u = - E_B u,\end{aligned} \hspace{\stretch{1}}(2.1)

where a positive bound state energy E_B = -E > 0 has been introduced.


\begin{aligned}\beta = \sqrt{\frac{2 m E_B}{\hbar^2}},\end{aligned} \hspace{\stretch{1}}(2.2)

the wave functions outside of the well are

\begin{aligned}u(x) =\left\{\begin{array}{l l}u(-a) e^{\beta(x+a)} &\quad \mbox{latex x a$} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.3)$

Within the well Schr\”{o}dinger’s equation is

\begin{aligned}-\frac{\hbar^2}{2m} u'' - V_0 u = E u = - E_B u,\end{aligned} \hspace{\stretch{1}}(2.4)


\begin{aligned}\frac{\hbar^2}{2m} u'' = - \frac{2m}{\hbar^2} (V_0 - E_B) u,\end{aligned} \hspace{\stretch{1}}(2.5)

Noting that the bound state energies are the E_B < V_0 values, let \alpha^2 = 2m (V_0 - E_B)/\hbar^2, so that the solutions are of the form

\begin{aligned}u(x) = A e^{i\alpha x} + B e^{-i\alpha x}.\end{aligned} \hspace{\stretch{1}}(2.6)

As was done for the wave functions outside of the well, the normalization constants can be expressed in terms of the values of the wave functions on the boundary. That provides a pair of equations to solve

\begin{aligned}\begin{bmatrix}u(a) \\ u(-a)\end{bmatrix}=\begin{bmatrix}e^{i \alpha a} & e^{-i \alpha a} \\ e^{-i \alpha a} & e^{i \alpha a}\end{bmatrix}\begin{bmatrix}A \\ B\end{bmatrix}.\end{aligned} \hspace{\stretch{1}}(2.7)

Inverting this and substitution back into 2.6 yields

\begin{aligned}u(x) &=\begin{bmatrix}e^{i\alpha x} & e^{-i\alpha x}\end{bmatrix}\begin{bmatrix}A \\ B\end{bmatrix} \\ &=\begin{bmatrix}e^{i\alpha x} & e^{-i\alpha x}\end{bmatrix}\frac{1}{{e^{2 i \alpha a} - e^{-2 i \alpha a}}}\begin{bmatrix}e^{i \alpha a} & -e^{-i \alpha a} \\ -e^{-i \alpha a} & e^{i \alpha a}\end{bmatrix}\begin{bmatrix}u(a) \\ u(-a)\end{bmatrix} \\ &=\begin{bmatrix}\frac{\sin(\alpha (a + x))}{\sin(2 \alpha a)} &\frac{\sin(\alpha (a - x))}{\sin(2 \alpha a)}\end{bmatrix}\begin{bmatrix}u(a) \\ u(-a)\end{bmatrix}.\end{aligned}

Expanding the last of these matrix products the wave function is close to completely specified.

\begin{aligned}u(x) =\left\{\begin{array}{l l}u(-a) e^{\beta(x+a)} & \quad \mbox{latex x < -a$} \\ u(a) \frac{\sin(\alpha (a + x))}{\sin(2 \alpha a)} +u(-a) \frac{\sin(\alpha (a – x))}{\sin(2 \alpha a)} & \quad \mbox{$latex {\left\lvert{x}\right\rvert} a$} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.8)$

There are still two unspecified constants u(\pm a) and the constraints on E_B have not been determined (both \alpha and \beta are functions of that energy level). It should be possible to eliminate at least one of the u(\pm a) by computing the wavefunction normalization, and since the well is being narrowed the \alpha term will not be relevant. Since only the vanishingly narrow case where a \rightarrow 0, x \in [-a,a] is of interest, the wave function in that interval approaches

\begin{aligned}u(x) \rightarrow \frac{1}{{2}} (u(a) + u(-a)) + \frac{x}{2} ( u(a) - u(-a) ) \rightarrow \frac{1}{{2}} (u(a) + u(-a)).\end{aligned} \hspace{\stretch{1}}(2.9)

Since no discontinuity is expected this is just u(a) = u(-a). Let’s write \lim_{a\rightarrow 0} u(a) = A for short, and the limited width well wave function becomes

\begin{aligned}u(x) =\left\{\begin{array}{l l}A e^{\beta x} & \quad \mbox{latex x 0$} \\ \end{array}\right.\end{aligned} \hspace{\stretch{1}}(2.10)$

This is now the same form as the delta function potential, and normalization also gives A = \sqrt{\beta}.

One task remains before the attractive delta function potential can be considered a limiting case for the finite well, since the relation between a, V_0, and g has not been established. To do so integrate the Schr\”{o}dinger equation over the infinitesimal range [-a,a]. This was done in the text for the delta function potential, and that provided the relation

\begin{aligned}\beta = \frac{mg}{\hbar^2}\end{aligned} \hspace{\stretch{1}}(2.11)

For the finite well this is

\begin{aligned}\int_{-a}^a -\frac{\hbar^2}{2m} u'' - V_0 \int_{-a}^a u = -E_B \int_{-a}^a u \\ \end{aligned} \hspace{\stretch{1}}(2.12)

In the limit as a \rightarrow 0 this is

\begin{aligned}\frac{\hbar^2}{2m} (u'(a) - u'(-a)) + V_0 2 a u(0) = 2 E_B a u(0).\end{aligned} \hspace{\stretch{1}}(2.13)

Some care is required with the V_0 a term since a \rightarrow 0 as V_0 \rightarrow \infty, but the E_B term is unambiguously killed, leaving

\begin{aligned}\frac{\hbar^2}{2m} u(0) (-2\beta e^{-\beta a}) = -V_0 2 a u(0).\end{aligned} \hspace{\stretch{1}}(2.14)

The exponential vanishes in the limit and leaves

\begin{aligned}\beta = \frac{m (2 a) V_0}{\hbar^2}\end{aligned} \hspace{\stretch{1}}(2.15)

Comparing to 2.11 from the attractive delta function completes the problem. The conclusion is that when the finite well is narrowed with a \rightarrow 0, also letting V_0 \rightarrow \infty such that the absolute area of the well g = (2 a) V_0 is maintained, the finite potential well produces exactly the attractive delta function wave function and associated bound state energy.

Problem 2.


For the hydrogen atom, determine {\langle {nlm} \rvert}(1/R){\lvert {nlm} \rangle} and 1/{\langle {nlm} \rvert}R{\lvert {nlm} \rangle} such that (nlm)=(211) and R is the radial position operator (X^2+Y^2+Z^2)^{1/2}. What do these quantities represent physically and are they the same?


Both of the computation tasks for the hydrogen like atom require expansion of a braket of the following form

\begin{aligned}{\langle {nlm} \rvert} A(R) {\lvert {nlm} \rangle},\end{aligned} \hspace{\stretch{1}}(3.16)

where A(R) = R = (X^2 + Y^2 + Z^2)^{1/2} or A(R) = 1/R.

The spherical representation of the identity resolution is required to convert this braket into integral form

\begin{aligned}\mathbf{1} = \int r^2 \sin\theta dr d\theta d\phi {\lvert { r \theta \phi} \rangle}{\langle { r \theta \phi} \rvert},\end{aligned} \hspace{\stretch{1}}(3.17)

where the spherical wave function is given by the braket \left\langle{{ r \theta \phi}} \vert {{nlm}}\right\rangle = R_{nl}(r) Y_{lm}(\theta,\phi).

Additionally, the radial form of the delta function will be required, which is

\begin{aligned}\delta(\mathbf{x} - \mathbf{x}') = \frac{1}{{r^2 \sin\theta}} \delta(r - r') \delta(\theta - \theta') \delta(\phi - \phi')\end{aligned} \hspace{\stretch{1}}(3.18)

Two applications of the identity operator to the braket yield

\begin{aligned}\rvert} A(R) {\lvert {nlm} \rangle} \\ &={\langle {nlm} \rvert} \mathbf{1} A(R) \mathbf{1} {\lvert {nlm} \rangle} \\ &=\int dr d\theta d\phi dr' d\theta' d\phi'r^2 \sin\theta {r'}^2 \sin\theta' \left\langle{{nlm}} \vert {{ r \theta \phi}}\right\rangle{\langle { r \theta \phi} \rvert} A(R) {\lvert { r' \theta' \phi'} \rangle}\left\langle{{ r' \theta' \phi'}} \vert {{nlm}}\right\rangle \\ &=\int dr d\theta d\phi dr' d\theta' d\phi'r^2 \sin\theta {r'}^2 \sin\theta' R_{nl}(r) Y_{lm}^{*}(\theta, \phi){\langle { r \theta \phi} \rvert} A(R) {\lvert { r' \theta' \phi'} \rangle}R_{nl}(r') Y_{lm}(\theta', \phi') \\ \end{aligned}

To continue an assumption about the matrix element {\langle { r \theta \phi} \rvert} A(R) {\lvert { r' \theta' \phi'} \rangle} is required. It seems reasonable that this would be

\begin{aligned}{\langle { r \theta \phi} \rvert} A(R) {\lvert { r' \theta' \phi'} \rangle} = \\ \delta(\mathbf{x} - \mathbf{x}') A(r) = \frac{1}{{r^2 \sin\theta}} \delta(r-r') \delta(\theta -\theta')\delta(\phi-\phi') A(r).\end{aligned} \hspace{\stretch{1}}(3.19)

The braket can now be written completely in integral form as

\begin{aligned}\rvert} A(R) {\lvert {nlm} \rangle} \\ &=\int dr d\theta d\phi dr' d\theta' d\phi'r^2 \sin\theta {r'}^2 \sin\theta' R_{nl}(r) Y_{lm}^{*}(\theta, \phi)\frac{1}{{r^2 \sin\theta}} \delta(r-r') \delta(\theta -\theta')\delta(\phi-\phi') A(r)R_{nl}(r') Y_{lm}(\theta', \phi') \\ &=\int dr d\theta d\phi {r'}^2 \sin\theta' dr' d\theta' d\phi'R_{nl}(r) Y_{lm}^{*}(\theta, \phi)\delta(r-r') \delta(\theta -\theta')\delta(\phi-\phi') A(r)R_{nl}(r') Y_{lm}(\theta', \phi') \\ \end{aligned}

Application of the delta functions then reduces the integral, since the only \theta, and \phi dependence is in the (orthonormal) Y_{lm} terms they are found to drop out

\begin{aligned}{\langle {nlm} \rvert} A(R) {\lvert {nlm} \rangle}&=\int dr d\theta d\phi r^2 \sin\theta R_{nl}(r) Y_{lm}^{*}(\theta, \phi)A(r)R_{nl}(r) Y_{lm}(\theta, \phi) \\ &=\int dr r^2 R_{nl}(r) A(r)R_{nl}(r) \underbrace{\int\sin\theta d\theta d\phi Y_{lm}^{*}(\theta, \phi)Y_{lm}(\theta, \phi) }_{=1}\\ \end{aligned}

This leaves just the radial wave functions in the integral

\begin{aligned}{\langle {nlm} \rvert} A(R) {\lvert {nlm} \rangle}=\int dr r^2 R_{nl}^2(r) A(r)\end{aligned} \hspace{\stretch{1}}(3.20)

As a consistency check, observe that with A(r) = 1, this integral evaluates to 1 according to equation (8.274) in the text, so we can think of (r R_{nl}(r))^2 as the radial probability density for functions of r.

The problem asks specifically for these expectation values for the {\lvert {211} \rangle} state. For that state the radial wavefunction is found in (8.277) as

\begin{aligned}R_{21}(r) = \left(\frac{Z}{2a_0}\right)^{3/2} \frac{ Z r }{a_0 \sqrt{3}} e^{-Z r/2 a_0}\end{aligned} \hspace{\stretch{1}}(3.21)

The braket can now be written explicitly

\begin{aligned}{\langle {21m} \rvert} A(R) {\lvert {21m} \rangle}=\frac{1}{{24}} \left(\frac{ Z }{a_0 } \right)^5\int_0^\inftydr r^4 e^{-Z r/ a_0}A(r)\end{aligned} \hspace{\stretch{1}}(3.22)

Now, let’s consider the two functions A(r) separately. First for A(r) = r we have

\begin{aligned}{\langle {21m} \rvert} R {\lvert {21m} \rangle}&=\frac{1}{{24}} \left(\frac{ Z }{a_0 } \right)^5\int_0^\inftydr r^5 e^{-Z r/ a_0} \\ &=\frac{ a_0 }{ 24 Z } \int_0^\inftydu u^5 e^{-u} \\ \end{aligned}

The last integral evaluates to 120, leaving

\begin{aligned}{\langle {21m} \rvert} R {\lvert {21m} \rangle}=\frac{ 5 a_0 }{ Z }.\end{aligned} \hspace{\stretch{1}}(3.23)

The expectation value associated with this {\lvert {21m} \rangle} state for the radial position is found to be proportional to the Bohr radius. For the hydrogen atom where Z=1 this average value for repeated measurements of the physical quantity associated with the operator R is found to be 5 times the Bohr radius for n=2, l=1 states.

Our problem actually asks for the inverse of this expectation value, and for reference this is

\begin{aligned}1/ {\langle {21m} \rvert} R {\lvert {21m} \rangle}=\frac{ Z }{ 5 a_0 } \end{aligned} \hspace{\stretch{1}}(3.24)

Performing the same task for A(R) = 1/R

\begin{aligned}{\langle {21m} \rvert} 1/R {\lvert {21m} \rangle}&=\frac{1}{{24}} \left(\frac{ Z }{a_0 } \right)^5\int_0^\inftydr r^3e^{-Z r/ a_0} \\ &=\frac{1}{{24}} \frac{ Z }{ a_0 } \int_0^\inftydu u^3e^{-u}.\end{aligned}

This last integral has value 6, and we have the second part of the computational task complete

\begin{aligned}{\langle {21m} \rvert} 1/R {\lvert {21m} \rangle} = \frac{1}{{4}} \frac{ Z }{ a_0 } \end{aligned} \hspace{\stretch{1}}(3.25)

The question of whether or not 3.24, and 3.25 are equal is answered. They are not.

Still remaining for this problem is the question of the what these quantities represent physically.

The quantity {\langle {nlm} \rvert} R {\lvert {nlm} \rangle} is the expectation value for the radial position of the particle measured from the center of mass of the system. This is the average outcome for many measurements of this radial distance when the system is prepared in the state {\lvert {nlm} \rangle} prior to each measurement.

Interestingly, the physical quantity that we associate with the operator R has a different measurable value than the inverse of the expectation value for the inverted operator 1/R. Regardless, we have a physical (observable) quantity associated with the operator 1/R, and when the system is prepared in state {\lvert {21m} \rangle} prior to each measurement, the average outcome of many measurements of this physical quantity produces this value {\langle {21m} \rvert} 1/R {\lvert {21m} \rangle} = Z/n^2 a_0, a quantity inversely proportional to the Bohr radius.

ASIDE: Comparing to the general case.

As a confirmation of the results obtained, we can check 3.24, and 3.25 against the general form of the expectation values \left\langle{{R^s}}\right\rangle for various powers s of the radial position operator. These can be found in locations such as which gives for Z=1 (without proof), and in [2] (where these and harder looking ones expectation values are left as an exercise for the reader to prove). Both of those give:

\begin{aligned}\left\langle{{R}}\right\rangle &= \frac{a_0}{2} ( 3 n^2 -l (l+1) ) \\ \left\langle{{1/R}}\right\rangle &= \frac{1}{n^2 a_0} \end{aligned} \hspace{\stretch{1}}(3.26)

It is curious to me that the general expectation values noted in 3.26 we have a l quantum number dependence for \left\langle{{R}}\right\rangle, but only the n quantum number dependence for \left\langle{{1/R}}\right\rangle. It is not obvious to me why this would be the case.


[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

[2] R. Liboff. Introductory quantum mechanics. Cambridge: Addison-Wesley Press, Inc, 2003.

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Desai Chapter 10 notes.

Posted by peeterjoot on December 7, 2010

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Chapter 10 notes for [1].


In 10.3 (interaction with a electric field), Green’s functions are introduced to solve the first order differential equation

\begin{aligned}\frac{da}{dt} + i \omega_0 a = - i \omega_0 \lambda(t)\end{aligned} \hspace{\stretch{1}}(2.1)

A simpler way is to use the usual trick of assuming that we can take the constant term in the homogeneous solution and allow it to vary with time.

Since our homogeneous solution is of the form

\begin{aligned}a_H(t) = a_H(0) e^{-i\omega_0 t},\end{aligned} \hspace{\stretch{1}}(2.2)

we can look for a specific solution to the forcing term equation of the form

\begin{aligned}a_S(t) = f(t) e^{-i\omega_0 t}\end{aligned} \hspace{\stretch{1}}(2.3)

We get

\begin{aligned}f' = -i \omega_0 \lambda(t) e^{i \omega_0 t}\end{aligned} \hspace{\stretch{1}}(2.4)

which can be integrate directly to find the non-homogeneous solution

\begin{aligned}a_S(t) = a_S(t_0) e^{-i \omega_0 (t - t_0)} - i \omega_0 \int_{t_0}^t \lambda(t') e^{-i \omega_0 (t-t')} dt'\end{aligned} \hspace{\stretch{1}}(2.5)

Setting t_0 = -\infty, with a requirement that a_S(-\infty) = 0 and adding in a general homogeneous solution one then has 10.92 without the complications of Green’s functions or the associated contour integrals. I suppose the author wanted to introduce this as a general purpose tool and this was a simple way to do so.

His introduction of Green’s functions this way I didn’t personally find very clear. Specifically, he doesn’t actually define what a Green’s function is, and the Appendix 20.13 he refers to only discusses the subtlies of the associated Contour integration. I didn’t understand where equation 10.83 came from in the first place.

Something like the following would have been helpful (the type of argument found in [2])

Given a linear operator L, such that L u(x) = f(x), we search for the Green’s function G(x,s) such that L G(x,s) = \delta(x-s). For such a function we have

\begin{aligned}\int L G(x,s) f(s) ds &= \int \delta(x-s) f(s) ds \\ &= f(x)\end{aligned}

and by linearity we also have

\begin{aligned}f(x) &=\int L G(x,s) f(s) ds \\ &= L \int G(x,s) f(s) ds \\ \end{aligned}

and can therefore identify u(x) = \int G(x,s) f(s) ds as the desired solution to L u(x) = f(x) once the Green’s function G(x,s) associated with operator L has been determined.


[1] BR Desai. Quantum mechanics with basic field theory. Cambridge University Press, 2009.

[2] Wikipedia. Green’s function — wikipedia, the free encyclopedia [online]. 2010. [Online; accessed 20-November-2010].

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My submission for PHY356 (Quantum Mechanics I) Problem Set 3.

Posted by peeterjoot on November 30, 2010

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Problem 1.


A particle of mass m is free to move along the x-direction such that V(X)=0. The state of the system is represented by the wavefunction Eq. (4.74)

\begin{aligned}\psi(x,t) = \frac{1}{{\sqrt{2\pi}}} \int_{-\infty}^\infty dk e^{i k x} e^{- i \omega t} f(k)\end{aligned} \hspace{\stretch{1}}(1.1)

with f(k) given by Eq. (4.59).

\begin{aligned}f(k) &= N e^{-\alpha k^2}\end{aligned} \hspace{\stretch{1}}(1.2)

Note that I’ve inserted a 1/\sqrt{2\pi} factor above that isn’t in the text, because otherwise \psi(x,t) will not be unit normalized (assuming f(k) is normalized in wavenumber space).

(a) What is the group velocity associated with this state?
(b) What is the probability for measuring the particle at position x=x_0>0 at time t=t_0>0?
(c) What is the probability per unit length for measuring the particle at position x=x_0>0 at time t=t_0>0?
(d) Explain the physical meaning of the above results.


(a). group velocity.

To calculate the group velocity we need to know the dependence of \omega on k.

Let’s step back and consider the time evolution action on \psi(x,0). For the free particle case we have

\begin{aligned}H = \frac{\mathbf{p}^2}{2m} = -\frac{\hbar^2}{2m} \partial_{xx}.\end{aligned} \hspace{\stretch{1}}(1.3)

Writing N' = N/\sqrt{2\pi} we have

\begin{aligned}-\frac{i t}{\hbar} H \psi(x,0) &= \frac{i t \hbar }{2m} N' \int_{-\infty}^\infty dk (i k)^2 e^{i k x - \alpha k^2} \\ &= N' \int_{-\infty}^\infty dk \frac{-i t \hbar k^2}{2m} e^{i k x - \alpha k^2}\end{aligned}

Each successive application of -iHt/\hbar will introduce another power of -it\hbar k^2/2 m, so once we sum all the terms of the exponential series U(t) = e^{-iHt/\hbar} we have

\begin{aligned}\psi(x,t) =N' \int_{-\infty}^\infty dk \exp\left( \frac{-i t \hbar k^2}{2m} + i k x - \alpha k^2 \right).\end{aligned} \hspace{\stretch{1}}(1.4)

Comparing with 1.1 we find

\begin{aligned}\omega(k) = \frac{\hbar k^2}{2m}.\end{aligned} \hspace{\stretch{1}}(1.5)

This completes this section of the problem since we are now able to calculate the group velocity

\begin{aligned}v_g = \frac{\partial {\omega(k)}}{\partial {k}} = \frac{\hbar k}{m}.\end{aligned} \hspace{\stretch{1}}(1.6)

(b). What is the probability for measuring the particle at position x=x_0>0 at time t=t_0>0?

In order to evaluate the probability, it looks desirable to evaluate the wave function integral 1.4.
Writing 2 \beta = i/(\alpha + i t \hbar/2m ), the exponent of that integral is

\begin{aligned}-k^2 \left( \alpha + \frac{i t \hbar }{2m} \right) + i k x&=-\left( \alpha + \frac{i t \hbar }{2m} \right) \left( k^2 - \frac{i k x }{\alpha + \frac{i t \hbar }{2m} } \right) \\ &=-\frac{i}{2\beta} \left( (k - x \beta )^2 - x^2 \beta^2 \right)\end{aligned}

The x^2 portion of the exponential

\begin{aligned}\frac{i x^2 \beta^2}{2\beta} = \frac{i x^2 \beta}{2} = - \frac{x^2 }{4 (\alpha + i t \hbar /2m)}\end{aligned}

then comes out of the integral. We can also make a change of variables q = k - x \beta to evaluate the remainder of the Gaussian and are left with

\begin{aligned}\psi(x,t) =N' \sqrt{ \frac{\pi}{\alpha + i t \hbar/2m} } \exp\left( - \frac{x^2 }{4 (\alpha + i t \hbar /2m)} \right).\end{aligned} \hspace{\stretch{1}}(1.7)

Observe that from 1.2 we can compute N = (2 \alpha/\pi)^{1/4}, which could be substituted back into 1.7 if desired.

Our probability density is

\begin{aligned}{\left\lvert{ \psi(x,t) }\right\rvert}^2 &=\frac{1}{{2 \pi}} N^2 {\left\lvert{ \frac{\pi}{\alpha + i t \hbar/2m} }\right\rvert} \exp\left( - \frac{x^2}{4} \left( \frac{1}{{(\alpha + i t \hbar /2m)}} + \frac{1}{{(\alpha - i t \hbar /2m)}} \right) \right) \\ &=\frac{1}{{2 \pi}} \sqrt{\frac{2 \alpha}{\pi} } \frac{\pi}{\sqrt{\alpha^2 + (t \hbar/2m)^2 }} \exp\left( - \frac{x^2}{4} \frac{1}{{\alpha^2 + (t \hbar/2m)^2 }} \left( \alpha - i t \hbar /2m + \alpha + i t \hbar /2m \right)\right) \\ &=\end{aligned}

With a final regrouping of terms, this is

\begin{aligned}{\left\lvert{ \psi(x,t) }\right\rvert}^2 =\sqrt{\frac{ \alpha }{ 2 \pi (\alpha^2 + (t \hbar/2m)^2 }) }\exp\left( - \frac{x^2}{2} \frac{\alpha}{\alpha^2 + (t \hbar/2m)^2 } \right).\end{aligned} \hspace{\stretch{1}}(1.8)

As a sanity check we observe that this integrates to unity for all t as desired. The probability that we find the particle at position x > x_0 is then

\begin{aligned}P_{x>x_0}(t) = \sqrt{\frac{ \alpha }{ 2 \pi (\alpha^2 + (t \hbar/2m)^2 }) }\int_{x=x_0}^\infty dx \exp\left( - \frac{x^2}{2} \frac{\alpha}{\alpha^2 + (t \hbar/2m)^2 } \right)\end{aligned} \hspace{\stretch{1}}(1.9)

The only simplification we can make is to rewrite this in terms of the complementary error function

\begin{aligned}\text{erfc}(x) = \frac{2}{\sqrt{\pi}} \int_x^\infty e^{-t^2} dt.\end{aligned} \hspace{\stretch{1}}(1.10)


\begin{aligned}\beta(t) = \frac{\alpha}{\alpha^2 + (t \hbar/2m)^2 },\end{aligned} \hspace{\stretch{1}}(1.11)

we have

\begin{aligned}P_{x>x_0}(t_0) = \frac{1}{{2}} \text{erfc} \left( \sqrt{\beta(t_0)/2} x_0 \right)\end{aligned} \hspace{\stretch{1}}(1.12)

Sanity checking this result, we note that since \text{erfc}(0) = 1 the probability for finding the particle in the x>0 range is 1/2 as expected.

(c). What is the probability per unit length for measuring the particle at position x=x_0>0 at time t=t_0>0?

This unit length probability is thus

\begin{aligned}P_{x>x_0+1/2}(t_0) - P_{x>x_0-1/2}(t_0) &=\frac{1}{{2}} \text{erfc}\left( \sqrt{\frac{\beta(t_0)}{2}} \left(x_0+\frac{1}{{2}} \right) \right) -\frac{1}{{2}} \text{erfc}\left( \sqrt{\frac{\beta(t_0)}{2}} \left(x_0-\frac{1}{{2}} \right) \right) \end{aligned} \hspace{\stretch{1}}(1.13)

(d). Explain the physical meaning of the above results.

To get an idea what the group velocity means, observe that we can write our wavefunction 1.1 as

\begin{aligned}\psi(x,t) = \frac{1}{{\sqrt{2\pi}}} \int_{-\infty}^\infty dk e^{i k (x - v_g t)} f(k)\end{aligned} \hspace{\stretch{1}}(1.14)

We see that the phase coefficient of the Gaussian f(k) “moves” at the rate of the group velocity v_g. Also recall that in the text it is noted that the time dependent term 1.11 can be expressed in terms of position and momentum uncertainties (\Delta x)^2, and (\Delta p)^2 = \hbar^2 (\Delta k)^2. That is

\begin{aligned}\frac{1}{{\beta(t)}} = (\Delta x)^2 + \frac{(\Delta p)^2}{m^2} t^2 \equiv (\Delta x(t))^2\end{aligned} \hspace{\stretch{1}}(1.15)

This makes it evident that the probability density flattens and spreads over time with the rate equal to the uncertainty of the group velocity \Delta p/m = \Delta v_g (since v_g = \hbar k/m). It is interesting that something as simple as this phase change results in a physically measurable phenomena. We see that a direct result of this linear with time phase change, we are less able to find the particle localized around it’s original time x = 0 position as more time elapses.

Problem 2.


A particle with intrinsic angular momentum or spin s=1/2 is prepared in the spin-up with respect to the z-direction state {\lvert {f} \rangle}={\lvert {z+} \rangle}. Determine

\begin{aligned}\left({\langle {f} \rvert} \left( S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2}\end{aligned} \hspace{\stretch{1}}(2.16)


\begin{aligned}\left({\langle {f} \rvert} \left( S_x - {\langle {f} \rvert} S_x {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2}\end{aligned} \hspace{\stretch{1}}(2.17)

and explain what these relations say about the system.

Solution: Uncertainty of S_z with respect to {\lvert {z+} \rangle}

Noting that S_z {\lvert {f} \rangle} = S_z {\lvert {z+} \rangle} = \hbar/2 {\lvert {z+} \rangle} we have

\begin{aligned}{\langle {f} \rvert} S_z {\lvert {f} \rangle} = \frac{\hbar}{2} \end{aligned} \hspace{\stretch{1}}(2.18)

The average outcome for many measurements of the physical quantity associated with the operator S_z when the system has been prepared in the state {\lvert {f} \rangle} = {\lvert {z+} \rangle} is \hbar/2.

\begin{aligned}\Bigl(S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1} \Bigr) {\lvert {f} \rangle}&= \frac{\hbar}{2} {\lvert {f} \rangle} -\frac{\hbar}{2} {\lvert {f} \rangle} = 0\end{aligned} \hspace{\stretch{1}}(2.19)

We could also compute this from the matrix representations, but it is slightly more work.

Operating once more with S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1} on the zero ket vector still gives us zero, so we have zero in the root for 2.16

\begin{aligned}\left({\langle {f} \rvert} \left( S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2} = 0\end{aligned} \hspace{\stretch{1}}(2.20)

What does 2.20 say about the state of the system? Given many measurements of the physical quantity associated with the operator V = (S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1})^2, where the initial state of the system is always {\lvert {f} \rangle} = {\lvert {z+} \rangle}, then the average of the measurements of the physical quantity associated with V is zero. We can think of the operator V^{1/2} = S_z - {\langle {f} \rvert} S_z {\lvert {f} \rangle} \mathbf{1} as a representation of the observable, “how different is the measured result from the average {\langle {f} \rvert} S_z {\lvert {f} \rangle}”.

So, given a system prepared in state {\lvert {f} \rangle} = {\lvert {z+} \rangle}, and performance of repeated measurements capable of only examining spin-up, we find that the system is never any different than its initial spin-up state. We have no uncertainty that we will measure any difference from spin-up on average, when the system is prepared in the spin-up state.

Solution: Uncertainty of S_x with respect to {\lvert {z+} \rangle}

For this second part of the problem, we note that we can write

\begin{aligned}{\lvert {f} \rangle} = {\lvert {z+} \rangle} = \frac{1}{{\sqrt{2}}} ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ).\end{aligned} \hspace{\stretch{1}}(2.21)

So the expectation value of S_x with respect to this state is

\begin{aligned}{\langle {f} \rvert} S_x {\lvert {f} \rangle}&=\frac{1}{{2}}( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) S_x ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) \\ &=\hbar ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) ( {\lvert {x+} \rangle} - {\lvert {x-} \rangle} ) \\ &=\hbar ( 1 + 0 + 0 -1 ) \\ &= 0\end{aligned}

After repeated preparation of the system in state {\lvert {f} \rangle}, the average measurement of the physical quantity associated with operator S_x is zero. In terms of the eigenstates for that operator {\lvert {x+} \rangle} and {\lvert {x-} \rangle} we have equal probability of measuring either given this particular initial system state.

For the variance calculation, this reduces our problem to the calculation of {\langle {f} \rvert} S_x^2 {\lvert {f} \rangle}, which is

\begin{aligned}{\langle {f} \rvert} S_x^2 {\lvert {f} \rangle} &=\frac{1}{{2}} \left( \frac{\hbar}{2} \right)^2 ( {\lvert {x+} \rangle} + {\lvert {x-} \rangle} ) ( (+1)^2 {\lvert {x+} \rangle} + (-1)^2 {\lvert {x-} \rangle} ) \\ &=\left( \frac{\hbar}{2} \right)^2,\end{aligned}

so for 2.22 we have

\begin{aligned}\left({\langle {f} \rvert} \left( S_x - {\langle {f} \rvert} S_x {\lvert {f} \rangle} \mathbf{1} \right)^2 {\lvert {f} \rangle} \right)^{1/2} = \frac{\hbar}{2}\end{aligned} \hspace{\stretch{1}}(2.22)

The average of the absolute magnitude of the physical quantity associated with operator S_x is found to be \hbar/2 when repeated measurements are performed given a system initially prepared in state {\lvert {f} \rangle} = {\lvert {z+} \rangle}. We saw that the average value for the measurement of that physical quantity itself was zero, showing that we have equal probabilities of measuring either \pm \hbar/2 for this experiment. A measurement that would show the system was in the x-direction spin-up or spin-down states would find that these states are equi-probable.

Grading comments.

I lost one mark on the group velocity response. Instead of 3.23 he wanted

\begin{aligned}v_g = {\left. \frac{\partial {\omega(k)}}{\partial {k}} \right\vert}_{k = k_0}= \frac{\hbar k_0}{m} = 0\end{aligned} \hspace{\stretch{1}}(3.23)

since f(k) peaks at k=0.

I’ll have to go back and think about that a bit, because I’m unsure of the last bits of the reasoning there.

I also lost 0.5 and 0.25 (twice) because I didn’t explicitly state that the probability that the particle is at x_0, a specific single point, is zero. I thought that was obvious and didn’t have to be stated, but it appears expressing this explicitly is what he was looking for.

Curiously, one thing that I didn’t loose marks on was, the wrong answer for the probability per unit length. What he was actually asking for was the following

\begin{aligned}\lim_{\epsilon \rightarrow 0} \frac{1}{{\epsilon}} \int_{x_0 - \epsilon/2}^{x_0 + \epsilon/2} {\left\lvert{ \Psi(x_0, t_0) }\right\rvert}^2 dx = {\left\lvert{\Psi(x_0, t_0)}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(3.24)

That’s a whole lot more sensible seeming quantity to calculate than what I did, but I don’t think that I can be faulted too much since the phrase was never used in the text nor in the lectures.

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