• 360,182

# Posts Tagged ‘free particle’

## PHY456H1F: Quantum Mechanics II. Lecture 21 (Taught by Prof J.E. Sipe). Scattering theory

Posted by peeterjoot on November 24, 2011

[Click here for a PDF of this post with nicer formatting and figures if the post had any (especially if my latex to wordpress script has left FORMULA DOES NOT PARSE errors.)]

# Disclaimer.

Peeter’s lecture notes from class. May not be entirely coherent.

# Scattering theory.

READING: section 19, section 20 of the text [1].

Here’s (\ref{fig:qmTwoL21:qmTwoL21Fig1}) a simple classical picture of a two particle scattering collision

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig1}
\caption{classical collision of particles.}
\end{figure}

We will focus on point particle elastic collisions (no energy lost in the collision). With particles of mass $m_1$ and $m_2$ we write for the total and reduced mass respectively

\begin{aligned}M = m_1 + m_2\end{aligned} \hspace{\stretch{1}}(2.1)

\begin{aligned}\frac{1}{{\mu}} = \frac{1}{{m_1}} + \frac{1}{{m_2}},\end{aligned} \hspace{\stretch{1}}(2.2)

so that interaction due to a potential $V(\mathbf{r}_1 - \mathbf{r}_2)$ that depends on the difference in position $\mathbf{r} = \mathbf{r}_1 - \mathbf{r}$ has, in the center of mass frame, the Hamiltonian

\begin{aligned}H = \frac{\mathbf{p}^2}{2 \mu} + V(\mathbf{r})\end{aligned} \hspace{\stretch{1}}(2.3)

In the classical picture we would investigate the scattering radius $r_0$ associated with the impact parameter $\rho$ as depicted in figure (\ref{fig:qmTwoL21:qmTwoL21Fig2})

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig2}
\caption{Classical scattering radius and impact parameter.}
\end{figure}

## 1D QM scattering. No potential wave packet time evolution.

Now lets move to the QM picture where we assume that we have a particle that can be represented as a wave packet as in figure (\ref{fig:qmTwoL21:qmTwoL21Fig3})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig3}
\caption{Wave packet for a particle wavefunction $\Re(\psi(x,0))$}
\end{figure}

First without any potential $V(x) = 0$, lets consider the evolution. Our position and momentum space representations are related by

\begin{aligned}\int {\left\lvert{\psi(x, t)}\right\rvert}^2 dx = 1 = \int {\left\lvert{\psi(p, t)}\right\rvert}^2 dp,\end{aligned} \hspace{\stretch{1}}(2.4)

and by Fourier transform

\begin{aligned}\psi(x, t) = \int \frac{dp}{\sqrt{2 \pi \hbar}} \overline{\psi}(p, t) e^{i p x/\hbar}.\end{aligned} \hspace{\stretch{1}}(2.5)

Schr\”{o}dinger’s equation takes the form

\begin{aligned}i \hbar \frac{\partial {\psi(x,t)}}{\partial {t}} = - \frac{\hbar^2}{2 \mu} \frac{\partial^2 {{\psi(x, t)}}}{\partial {{x}}^2},\end{aligned} \hspace{\stretch{1}}(2.6)

or more simply in momentum space

\begin{aligned}i \hbar \frac{\partial {\overline{\psi}(p,t)}}{\partial {t}} = \frac{p^2}{2 \mu} \frac{\partial^2 {{\overline{\psi}(p, t)}}}{\partial {{x}}^2}.\end{aligned} \hspace{\stretch{1}}(2.7)

Rearranging to integrate we have

\begin{aligned}\frac{\partial {\overline{\psi}}}{\partial {t}} = -\frac{i p^2}{2 \mu \hbar} \overline{\psi},\end{aligned} \hspace{\stretch{1}}(2.8)

and integrating

\begin{aligned}\ln \overline{\psi} = -\frac{i p^2 t}{2 \mu \hbar} + \ln C,\end{aligned} \hspace{\stretch{1}}(2.9)

or

\begin{aligned}\overline{\psi} = C e^{-\frac{i p^2 t}{2 \mu \hbar}} = \overline{\psi}(p, 0) e^{-\frac{i p^2 t}{2 \mu \hbar}}.\end{aligned} \hspace{\stretch{1}}(2.10)

Time evolution in momentum space for the free particle changes only the phase of the wavefunction, the momentum probability density of that particle.

Fourier transforming, we find our position space wavefunction to be

\begin{aligned}\psi(x, t) = \int \frac{dp}{\sqrt{2 \pi \hbar}} \overline{\psi}(p, 0) e^{i p x/\hbar} e^{-i p^2 t/2 \mu \hbar}.\end{aligned} \hspace{\stretch{1}}(2.11)

To clean things up, write

\begin{aligned}p = \hbar k,\end{aligned} \hspace{\stretch{1}}(2.12)

for

\begin{aligned}\psi(x, t) = \int \frac{dk}{\sqrt{2 \pi}} a(k, 0) ) e^{i k x} e^{-i \hbar k^2 t/2 \mu},\end{aligned} \hspace{\stretch{1}}(2.13)

where

\begin{aligned}a(k, 0) = \sqrt{\hbar} \overline{\psi}(p, 0).\end{aligned} \hspace{\stretch{1}}(2.14)

Putting

\begin{aligned}a(k, t) = a(k, 0) e^{ -i \hbar k^2/2 \mu},\end{aligned} \hspace{\stretch{1}}(2.15)

we have

\begin{aligned}\psi(x, t) = \int \frac{dk}{\sqrt{2 \pi}} a(k, t) ) e^{i k x} \end{aligned} \hspace{\stretch{1}}(2.16)

Observe that we have

\begin{aligned}\int dk {\left\lvert{ a(k, t)}\right\rvert}^2 = \int dp {\left\lvert{ \overline{\psi}(p, t)}\right\rvert}^2 = 1.\end{aligned} \hspace{\stretch{1}}(2.17)

## A Gaussian wave packet

Suppose that we have, as depicted in figure (\ref{fig:qmTwoL21:qmTwoL21Fig4})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig4}
\caption{Gaussian wave packet.}
\end{figure}

a Gaussian wave packet of the form

\begin{aligned}\psi(x, 0) = \frac{ (\pi \Delta^2)^{1/4}} e^{i k_0 x} e^{- x^2/2 \Delta^2}.\end{aligned} \hspace{\stretch{1}}(2.18)

This is actually a minimum uncertainty packet with

\begin{aligned}\Delta x &= \frac{\Delta}{\sqrt{2}} \\ \Delta p &= \frac{\hbar}{\Delta \sqrt{2}}.\end{aligned} \hspace{\stretch{1}}(2.19)

Taking Fourier transforms we have

\begin{aligned}a(k, 0) &= \left(\frac{\Delta^2}{\pi}\right)^{1/4} e^{-(k - k_0)^2 \Delta^2/2} \\ a(k, t) &= \left(\frac{\Delta^2}{\pi}\right)^{1/4} e^{-(k - k_0)^2 \Delta^2/2} e^{ -i \hbar k^2 t/ 2\mu} \equiv \alpha(k, t)\end{aligned} \hspace{\stretch{1}}(2.21)

For $t > 0$ our wave packet will start moving and spreading as in figure (\ref{fig:qmTwoL21:qmTwoL21Fig5})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig5}
\end{figure}

## With a potential.

Now “switch on” a potential, still assuming a wave packet representation for the particle. With a positive (repulsive) potential as in figure (\ref{fig:qmTwoL21:qmTwoL21Fig6}), at a time long before the interaction of the wave packet with the potential we can visualize the packet as heading towards the barrier.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig6}
\caption{QM wave packet prior to interaction with repulsive potential.}
\end{figure}

After some time long after the interaction, classically for this sort of potential where the particle kinetic energy is less than the barrier “height”, we would have total reflection. In the QM case, we’ve seen before that we will have a reflected and a transmitted portion of the wave packet as depicted in figure (\ref{fig:qmTwoL21:qmTwoL21Fig7})
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig7}
\caption{QM wave packet long after interaction with repulsive potential.}
\end{figure}

Even if the particle kinetic energy is greater than the barrier height, as in figure (\ref{fig:qmTwoL21:qmTwoL21Fig8}), we can still have a reflected component.
\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig8}
\caption{Kinetic energy greater than potential energy.}
\end{figure}

This is even true for a negative potential as depicted in figure (\ref{fig:qmTwoL21:qmTwoL21Fig9})!

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig9}
\caption{qmTwoL21Fig9}
\end{figure}

Consider the probability for the particle to be found anywhere long after the interaction, summing over the transmitted and reflected wave functions, we have

\begin{aligned}1 &= \int {\left\lvert{\psi_r + \psi_t}\right\rvert}^2 \\ &= \int {\left\lvert{\psi_r}\right\rvert}^2 + \int {\left\lvert{\psi_t}\right\rvert}^2 + 2 \Re \int \psi_r^{*} \psi_t\end{aligned}

Observe that long after the interaction the cross terms in the probabilities will vanish because they are non-overlapping, leaving just the probably densities for the transmitted and reflected probably densities independently.

We define

\begin{aligned}T &= \int {\left\lvert{\psi_t(x, t)}\right\rvert}^2 dx \\ R &= \int {\left\lvert{\psi_r(x, t)}\right\rvert}^2 dx.\end{aligned} \hspace{\stretch{1}}(2.23)

The objective of most of our scattering problems will be the calculation of these probabilities and the comparisons of their ratios.

Question. Can we have more than one wave packet reflect off. Yes, we could have multiple wave packets for both the reflected and the transmitted portions. For example, if the potential has some internal structure there could be internal reflections before anything emerges on either side and things could get quite messy.

# Considering the time independent case temporarily.

We are going to work through something that is going to seem at first to be completely unrelated. We will (eventually) see that this can be applied to this problem, so a bit of patience will be required.

We will be using the time independent Schr\”{o}dinger equation

\begin{aligned}- \frac{\hbar^2}{2 \mu} \psi_k''(x) = V(x) \psi_k(x) = E \psi_k(x),\end{aligned} \hspace{\stretch{1}}(3.25)

where we have added a subscript $k$ to our wave function with the intention (later) of allowing this to vary. For “future use” we define for $k > 0$

\begin{aligned}E = \frac{\hbar^2 k^2}{2 \mu}.\end{aligned} \hspace{\stretch{1}}(3.26)

Consider a potential as in figure (\ref{fig:qmTwoL21:qmTwoL21Fig10}), where $V(x) = 0$ for $x > x_2$ and $x < x_1$.

\begin{figure}[htp]
\centering
\includegraphics[totalheight=0.2\textheight]{qmTwoL21Fig10}
\caption{potential zero outside of a specific region.}
\end{figure}

We won't have bound states here (repulsive potential). There will be many possible solutions, but we want to look for a solution that is of the form

\begin{aligned}\psi_k(x) = C e^{i k x}, \qquad x > x_2\end{aligned} \hspace{\stretch{1}}(3.27)

Suppose $x = x_3 > x_2$, we have

\begin{aligned}\psi_k(x_3) = C e^{i k x_3}\end{aligned} \hspace{\stretch{1}}(3.28)

\begin{aligned}{\left.{{\frac{d\psi_k}{dx}}}\right\vert}_{{x = x_3}} = i k C e^{i k x_3} \equiv \phi_k(x_3)\end{aligned} \hspace{\stretch{1}}(3.29)

\begin{aligned}{\left.{{\frac{d^2\psi_k}{dx^2}}}\right\vert}_{{x = x_3}} = -k^2 C e^{i k x_3} \end{aligned} \hspace{\stretch{1}}(3.30)

Defining

\begin{aligned}\phi_k(x) = \frac{d\psi_k}{dx},\end{aligned} \hspace{\stretch{1}}(3.31)

we write Schr\”{o}dinger’s equation as a pair of coupled first order equations

\begin{aligned}\frac{d\psi_k}{dx} &= \phi_k(x) \\ -\frac{\hbar^2}{2 \mu} \frac{d\phi_k(x)}{dx} = - V(x) \psi_k(x) + \frac{\hbar^2 k^2}{2\mu} \psi_k(x).\end{aligned} \hspace{\stretch{1}}(3.32)

At this $x = x_3$ specifically, we “know” both $\phi_k(x_3)$ and $\psi_k(x_3)$ and have

\begin{aligned}{\left.{{\frac{d\psi_k}{dx}}}\right\vert}_{{x_3}} &= \phi_k(x) \\ -\frac{\hbar^2}{2 \mu} {\left.{{\frac{d\phi_k(x)}{dx}}}\right\vert}_{{x_3}} = - V(x_3) \psi_k(x_3) + \frac{\hbar^2 k^2}{2\mu} \psi_k(x_3),\end{aligned} \hspace{\stretch{1}}(3.34)

This allows us to find both

\begin{aligned}{dx}}}\right\vert}_{{x_3}} \\ {dx}}}\right\vert}_{{x_3}} \end{aligned} \hspace{\stretch{1}}(3.36)

then proceed to numerically calculate $\phi_k(x)$ and $\psi_k(x)$ at neighboring points $x = x_3 + \epsilon$. Essentially, this allows us to numerically integrate backwards from $x_3$ to find the wave function at previous points for any sort of potential.

# References

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

## PHY450H1S. Relativistic Electrodynamics Lecture 1 (Taught by Prof. Erich Poppitz). Speed of light and simultaneity.

Posted by peeterjoot on January 11, 2011

# Distance as a clock.

The title of this course is an oxymoron since ELECTRODYNAMICS == RELATIVITY. In classical and quantum physics (non-gravitational) we start by postulating the existence of space and time. These are, in non-gravitational physics, the arena where everything takes place. The space that we work with is the three dimensional Euclidean space $\mathbb{R}^{3}$. One way of describing it is using three coordinates

\begin{aligned}\mathbb{R}^{3} = \{ x, y, z ; x,y,z \in [-\infty,\infty] \}.\end{aligned} \hspace{\stretch{1}}(1.1)

We define a distance between $P$ and $P'$ as

\begin{aligned}{\left\lvert{P P' }\right\rvert} = \sqrt{ (x-x')^2 + (y-y')^2 + (z-z')^2 }\end{aligned} \hspace{\stretch{1}}(1.2)

\underline{time} is a parameter with respect to which positions of \underline{free} particles particles change at a constant rate.

Mathematically, we describe the motion of free particles by giving $(x(t), y(t), z(t))$ : coordinates as functions of t,

\begin{aligned}\frac{d^2 x_i(t)}{dt^2} = 0, i = 1,2,3\end{aligned} \hspace{\stretch{1}}(1.3)

Here $x,y,z$ are the free particle coordinates in an “internal frame”, the frame where $\dot{d}{\mathbf{r}} = 0$ holds for a free particle ($\dot{d}{\mathbf{r}} = d^2 \mathbf{r}/dt^2$) for a free particle $x = v_0 t, y = z = 0$.

# The principle of relativity.

\paragraph{Principle of relativity (Galileo or Einstein)}: “Laws of nature are identical in all inertial frames”.

Equivalently, “Identical experiments in two inertial frames yield identical results”.

\paragraph{What do we mean by laws of nature}. Equations that describe dynamics.

Now we need to get more specific. Identical equations means that the equations have the same form in two inertial frames provided, you express them (the equations) via the coordinates $\mathbf{r},t$ in the given inertial frame.

FIXME: DRAW x,y,z COORDINATE SYSTEM with origin $O$. And another with origin $O'$ where the origin is moving with velocity $v$ in the y direction.

The Galilean relativity principle states that “equations of motion are invariant under Galilean transformations”. What do we mean by transformations? If we have a point $P(t)$ in space with coordinates in both frames that are related. It is pretty clear that the coordinates $x = x'$ and $z = z'$. What about the $y'$ coordinate? For that we have $y' = y - v t$, so that the origins overlap ($O = O'$) at $t=0$.

In Galilean relativity, time is absolute. i.e. It is the same in all inertial frames. It is now a no-brainer to find the velocities of the particle. Taking derivatives we take time derivatives of

\begin{aligned}x' &= x \\ y' &= y - v t \\ z' &= z,\end{aligned} \hspace{\stretch{1}}(2.4)

for

\begin{aligned}v_x' &= v_x \\ v_y' &= v_y - v \\ v_z' &= v_z.\end{aligned} \hspace{\stretch{1}}(2.7)

In vector notation we have

\begin{aligned}\mathbf{r}' &= \mathbf{r} - \mathbf{v}_0 t \\ \mathbf{v}' &= \mathbf{v} - \mathbf{v}_0 \end{aligned} \hspace{\stretch{1}}(2.10)

The principle of relativity says that the dynamical equations are invariant under such transformations.

Take Newton’s law for example applied to two bodies, labeled by their masses $M_1$ and $M_2$.

These bodies may be interacting. For example, with Newtonian gravitation

\begin{aligned}V(\mathbf{r}_1 - \mathbf{r}_2) = -G_N \frac{M_1 M_2}{{\left\lvert{\mathbf{r}_1 - \mathbf{r}_2}\right\rvert}},\end{aligned} \hspace{\stretch{1}}(2.12)

or the Van Der Waals, interaction

\begin{aligned}V(\mathbf{r}_1 - \mathbf{r}_2) = - \text{const} \frac{1}{{{\left\lvert{\mathbf{r}_1 - \mathbf{r}_2}\right\rvert}^6}},\end{aligned} \hspace{\stretch{1}}(2.13)

Our interaction is via a gradient ${\partial {f(\mathbf{r})}}/{\partial {\mathbf{r}}} = ( {\partial {f}}/{\partial {x}}, {\partial {f}}/{\partial {y}}, {\partial {f}}/{\partial {z}} )$

\begin{aligned}M_1 \dot{d}{\mathbf{r}}_1 &= -\frac{\partial {}}{\partial {\mathbf{r}_1}} V(\mathbf{r}_1 - \mathbf{r}_2) \\ M_2 \dot{d}{\mathbf{r}}_2 &= -\frac{\partial {}}{\partial {\mathbf{r}_2}} V(\mathbf{r}_1 - \mathbf{r}_2)\end{aligned} \hspace{\stretch{1}}(2.14)

In the unprimed frame, these are “the laws of physics”. Consider a primed frame $O' : \mathbf{r}_i' = \mathbf{r}_i - \mathbf{v}_0 t$ (for $i=1,2$). Taking derivatives we have $\mathbf{v}_i' = \mathbf{v}_i' + \mathbf{v}_0$, and $\dot{\mathbf{v}}_i' = \dot{\mathbf{v}}_i'$.

We note that the distance between the two particles is unchanged in the primed coordinate system

\begin{aligned}\mathbf{r}_1' - \mathbf{r}_2' = \mathbf{r}_1 - \mathbf{v}_0 t -( \mathbf{r}_2 - \mathbf{v}_0 t ) = \mathbf{r}_1 - \mathbf{r}_2\end{aligned} \hspace{\stretch{1}}(2.16)

Similarly

\begin{aligned}\frac{\partial {}}{\partial {\mathbf{r}_i}} = \frac{\partial {}}{\partial {(\mathbf{r}_i' + \mathbf{v}_0 t)}} = \frac{\partial {}}{\partial {\mathbf{r}_i'}}\end{aligned} \hspace{\stretch{1}}(2.17)

Observe that the interaction 2.14 is unchanged by this change in coordinates.

# Enter electromagnetism.

If the only interactions are $1/r$ gravity and $1/r$ Coulomb, Galilean relativity holds. Electromagnetism came along and Maxwell’s prediction that electromagnetic waves exist and propagate with speed

\begin{aligned}c \approx 3 \times 10^8 m/s\end{aligned} \hspace{\stretch{1}}(3.18)

(Note that in SI units $c = 1/\sqrt{ \epsilon_0 \mu_0 }$).

It was proposed that the speed of light was the speed in a medium (the “aether”) through which electrodynamic waves propagate. The idea was that the oscillations of this medium constitute electromagnetic waves. Then “c” would be the speed of light with respect to that medium. This medium would fill all space.

FIXME: PICTURE of gradient field, with aether velocity at different points. Superimposed on this is a picture of the Earth’s orbit, so that the velocity of the aether could be measured at different points of the earth’s orbit by measuring the speed of light at different points in the orbit.

FIXME: PICTURE of interferometer.

We can study this effect by rotating this platform to measure at different points of the day and the year.

We note that the speed of the earth is approximately $v_{+} = 150 \times 10^6 km/ 10^7 s \approx 15 km/s$. The shift of fringes would then be $v_{+} \approx (v_{+}/c)^2 \approx 10^{-8}$. What Einstein did was to elevate the principle of relativity to one that applies to electromagnetism, but replacing the transformation relating frames to the Lorentz transformation, a transformation observed by Lorentz and Poincare that leave Maxwell’s equations invariant. Einstein did this by postulating that the speed of light is a constant in all frames, and we will see how this is the case.

\paragraph{Student’s question.} Isn’t this true only outside of matter? In matter we have electromagnetic wave propagation at speeds less than $c$.

\paragraph{A:} (paraphrasing) We can consider the in-matter case to be a special case, treating collections of discreet particles as continuous approximations. It’s only as a side effect of these approximations that one produces the in-matter Maxwell’s equation, and we will consider the “vacuum” Maxwell equation as always true, provided the points of interest do not fall exactly on any specific particle.

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

Posted by peeterjoot on November 30, 2010

# Problem 1.

## Statement

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).

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

## Solution

### (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)

Writing

\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.

## Statement

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)

and

\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.

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.

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

Posted by peeterjoot on November 16, 2010

# Problem 1.

A particle of mass $m$ is free to move along the x-direction such that $V(X)=0$. Express the time evolution operator $U(t,t_0)$ defined by Eq. (2.166) using the momentum eigenstates ${\lvert {p} \rangle}$ with delta-function normalization. Find ${\langle {x} \rvert} U(t,t0) {\lvert {x'} \rangle}$, where ${\lvert {x} \rangle}$ and ${\lvert {x'} \rangle}$ are position eigenstates. What is the physical meaning of this expression?

## Momentum matrix element.

We can expand the time evolution operator in series

\begin{aligned}U(t,t_0) &= e^{-i H(t-t_0)/\hbar} \\ &= e^{ -i P^2 (t-t_0)/ 2m \hbar } \\ &= 1 + \sum_{k=1}^\infty \frac{1}{{k!}} \left( -i \frac{P^2 (t-t_0)}{2m \hbar} \right)^k.\end{aligned}

We can now evaluate the momentum matrix element ${\langle {p} \rvert} U(t,t_0) {\lvert {p'} \rangle}$, which will essentially require the value of ${\langle {p} \rvert} P^{2k} {\lvert {p'} \rangle}$. That is

\begin{aligned}{\langle {p} \rvert} P^{2k} {\lvert {p'} \rangle}&= {\langle {p} \rvert} P^{2k-1} P {\lvert {p'} \rangle} \\ &= {\langle {p} \rvert} P^{2k-1} {\lvert {p'} \rangle} p' \\ &= \cdots \\ &= \left\langle{{p}} \vert {{p'}}\right\rangle (p')^{2k}.\end{aligned}

The momentum matrix element is therefore reduced to

\begin{aligned}{\langle {p} \rvert} U(t,t_0) {\lvert {p'} \rangle}&=\left\langle{{p}} \vert {{p'}}\right\rangle \exp\left( -i \frac{p^2 (t-t_0)}{2m \hbar} \right)= \delta(p-p') \exp\left( -i \frac{p^2 (t-t_0)}{2m \hbar} \right)\end{aligned} \hspace{\stretch{1}}(1.1)

## Position matrix element.

For the position matrix element we have a similar sum

\begin{aligned}{\langle {x} \rvert} U(t,t_0) {\lvert {x'} \rangle} &= \left\langle{{x}} \vert {{x'}}\right\rangle + \sum_{k=1}^\infty \frac{1}{{k!}} {\langle {x} \rvert} \left( -i \frac{P^2 (t-t_0)}{2m \hbar} \right)^k {\lvert {x'} \rangle},\end{aligned}

and require ${\langle {x} \rvert} P^{2k} {\lvert {x'} \rangle}$ to continue. That is

\begin{aligned}{\langle {x} \rvert} P^{2k} {\lvert {x'} \rangle}&=\int dx''{\langle {x} \rvert} P^{2k-1} {\lvert {x''} \rangle}{\langle {x''} \rvert} P {\lvert {x'} \rangle} \\ &=\int dx''{\langle {x} \rvert} P^{2k-1} {\lvert {x''} \rangle} \delta(x''-x') (-i\hbar) \frac{d}{dx'} \\ &={\langle {x} \rvert} P^{2k-1} {\lvert {x'} \rangle} (-i\hbar) \frac{d}{dx'} \\ &= \cdots \\ &= \left\langle{{x}} \vert {{x'}}\right\rangle \left( (-i\hbar) \frac{d}{dx'} \right)^{2k}\end{aligned}

Our position matrix element is therefore the differential operator

\begin{aligned}{\langle {x} \rvert} U(t,t_0) {\lvert {x'} \rangle} &=\left\langle{{x}} \vert {{x'}}\right\rangle \exp\left( \frac{i (t-t_0)\hbar}{2m} \frac{d^2}{d{x'}^2} \right)=\delta(x-x') \exp\left( \frac{i (t-t_0)\hbar}{2m} \frac{d^2}{d{x'}^2} \right)\end{aligned} \hspace{\stretch{1}}(1.2)

## Physical interpretation of the position matrix element operator.

Finally, we need to determine the physical meaning of such a matrix element operator.

With the delta function that this matrix element operator includes it really only takes on a meaning with a convolution integral. The simplest such integral would be

\begin{aligned}\int dx' {\langle {x} \rvert} U {\lvert {x'} \rangle} \left\langle{{x'}} \vert {{\phi_0}}\right\rangle &={\langle {x} \rvert} U {\lvert {\phi_0} \rangle} \\ &=\left\langle{{x}} \vert {{\phi(t)}}\right\rangle \\ &=\phi(x,t),\end{aligned}

or

\begin{aligned}\phi(x,t) = \int dx' {\langle {x} \rvert} U {\lvert {x'} \rangle} \phi(x',0)\end{aligned}

The LHS has a physical meaning, and in the absolute square

\begin{aligned}\int_{x_0}^{x_0+ \Delta x} {\left\lvert{\phi(x,t)}\right\rvert}^2 dx,\end{aligned} \hspace{\stretch{1}}(1.3)

provides the probability that the particle will be found in the region $[x_0, x_0+ \Delta x]$.

If we ignore the absolute square requirement and think of the (presumed normalized) wave function $\phi(x,t)$ more loosely as representing a probability directly, then we can in turn give a meaning to the matrix element ${\langle {x} \rvert} U {\lvert {x'} \rangle}$ for the time evolution operator. This provides an operator valued weighting function that provides us with the probability that a particle initially at position $x'$ will be at position $x$ at time $t$. This probability is indirect since we need to absolute square and sum over a finite interval to obtain the probability of finding the particle in that interval.

Observe that the integral on the RHS of 1.3 is a summation over all $x'$, so we can think of this as adding the probabilities that the particle was at each point to arrive at the total probability for finding it at the new location $x$. The time evolution operator matrix element provides the weighting in this conditional probability.

In 1.2 we found that the time evolution operators matrix element is differential operator in the position representation. In the general case this means that this probability weighting is not just numeric since the operation of the matrix element initial time wave function can produce wave functions for additional states. In some special cases, we may find that this weighting is strictly numeric, and one such example would be the Gaussian wave packet $\phi(x',0) = e^{-a{x'}^2}$. Application of the differential operations would then produce polynomial weighted multiples of the original Gaussian. In this special case we would be able to write

\begin{aligned}\phi(x,t) = \int dx' {\langle {x} \rvert} U {\lvert {x'} \rangle} \phi(x',0) = \int dx' K(x,x',t) \phi(x',0) \end{aligned}

Where $K(x,x',t)$ is a polynomial valued function (and is in fact another exponential), and now just provides a numerical weighting for the conditional probability for the particle to move from $x'$ to $x$ in time $t$. In [1], this $K(x,x',t)$ is called the Propagator function. It is perhaps justifiable to also call our similar operator valued matrix element a Propagator.