Peeter Joot's (OLD) Blog.

Math, physics, perl, and programming obscurity.

Posts Tagged ‘number operator’

An updated compilation of notes, for ‘PHY452H1S Basic Statistical Mechanics’, Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 27, 2013

Here’s my second update of my notes compilation for this course, including all of the following:

March 27, 2013 Fermi gas

March 26, 2013 Fermi gas thermodynamics

March 26, 2013 Fermi gas thermodynamics

March 23, 2013 Relativisitic generalization of statistical mechanics

March 21, 2013 Kittel Zipper problem

March 18, 2013 Pathria chapter 4 diatomic molecule problem

March 17, 2013 Gibbs sum for a two level system

March 16, 2013 open system variance of N

March 16, 2013 probability forms of entropy

March 14, 2013 Grand Canonical/Fermion-Bosons

March 13, 2013 Quantum anharmonic oscillator

March 12, 2013 Grand canonical ensemble

March 11, 2013 Heat capacity of perturbed harmonic oscillator

March 10, 2013 Langevin small approximation

March 10, 2013 Addition of two one half spins

March 10, 2013 Midterm II reflection

March 07, 2013 Thermodynamic identities

March 06, 2013 Temperature

March 05, 2013 Interacting spin

plus everything detailed in the description of my first update and before.

Posted in Math and Physics Learning. | Tagged: , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , | 1 Comment »

PHY452H1S Basic Statistical Mechanics. Lecture 15: Grand Canonical/Fermion-Bosons. Taught by Prof. Arun Paramekanti

Posted by peeterjoot on March 14, 2013

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


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

Grand Canonical/Fermion-Bosons

Was mentioned that three dimensions confines us to looking at either Fermions or Bosons, and that two dimensions is a rich subject (interchange of two particles isn’t the same as one particle cycling around the other ending up in the same place — how is that different than a particle cycling around another in a two dimensional space?)


  1. Fermion. Antisymmetric under exchange. n_k = 0, 1
  2. Boson. Symmetric under exchange. n_k = 0, 1, 2, \cdots

In either case our energies are

\begin{aligned}\epsilon_k = \frac{\hbar^2 k^2}{2m},\end{aligned} \hspace{\stretch{1}}(1.2.1)

For Fermions we’ll have occupation filling of the form fig. 1.1, where there can be only one particle at any given site (an energy level for that value of momentum). For Bosonic systems as in fig. 1.2, we don’t have a restriction of only one particle for each state, and can have any given number of particles for each value of momentum.


Fig 1.1: Fermionic energy level filling for free particle in a box



Fig 1.2: Bosonic free particle in a box energy level filling


Our Hamiltonian is

\begin{aligned}H = \sum_k \hat{n}_k \epsilon_k,\end{aligned} \hspace{\stretch{1}}(1.2.2)

where we have a number operator

\begin{aligned}N = \sum \hat{n}_k,\end{aligned} \hspace{\stretch{1}}(1.2.3)

such that

\begin{aligned}\left[{N},{H}\right] = 0.\end{aligned} \hspace{\stretch{1}}(1.2.4)

\begin{aligned}Z_{\mathrm{G}} = \sum_{N=0}^\infty e^{\beta \mu N}\sum_{n_k, \sum n_k = N} e^{-\beta \sum_k n_k \epsilon_k}.\end{aligned} \hspace{\stretch{1}}(1.2.5)

While the second sum is constrained, because we are summing over all n_k, this is essentially an unconstrained sum, so we can write

\begin{aligned}Z_{\mathrm{G}} &= \sum_{n_k}e^{\beta \mu \sum_k n_k}e^{-\beta \sum_k n_k \epsilon_k} \\ &= \sum_{n_k} \left( \prod_k e^{-\beta(\epsilon_k - \mu) n_k} \right) \\ &= \prod_{n} \left( \sum_{n_k} e^{-\beta(\epsilon_k - \mu) n_k} \right).\end{aligned} \hspace{\stretch{1}}(1.2.6)


\begin{aligned}\sum_{n_k = 0}^1 e^{-\beta(\epsilon_k - \mu) n_k} = 1 + e^{-\beta(\epsilon_k - \mu)}\end{aligned} \hspace{\stretch{1}}(1.2.7)


\begin{aligned}\sum_{n_k = 0}^\infty e^{-\beta(\epsilon_k - \mu) n_k} = \frac{1}{{1 - e^{-\beta(\epsilon_k - \mu)}}}\end{aligned} \hspace{\stretch{1}}(1.2.8)

(\epsilon_k - \mu \ge 0).

Our grand partition functions are then

\begin{aligned}Z_{\mathrm{G}}^f = \prod_k \left( 1 + e^{-\beta(\epsilon_k - \mu)} \right)\end{aligned} \hspace{\stretch{1}}(1.0.9a)

\begin{aligned}Z_{\mathrm{G}}^b = \prod_k \frac{1}{{ 1 - e^{-\beta(\epsilon_k - \mu)} }}\end{aligned} \hspace{\stretch{1}}(1.0.9b)

We can use these to compute the average number of particles

\begin{aligned}\left\langle{{n_k^f}}\right\rangle = \frac{1 \times 0 + e^{-\beta(\epsilon_k - \mu)} \times 1}{ 1 + e^{-\beta(\epsilon_k - \mu)} }=\frac{1}{{ 1 + e^{-\beta(\epsilon_k - \mu)} }}\end{aligned} \hspace{\stretch{1}}(1.0.10)

\begin{aligned}\left\langle{{n_k^b}}\right\rangle = \frac{1 \times 0 + e^{-\beta(\epsilon_k - \mu)} \times 1+e^{-2 \beta(\epsilon_k - \mu)} \times 2 + \cdots}{ 1+e^{-\beta(\epsilon_k - \mu)} +e^{-2 \beta(\epsilon_k - \mu)} }\end{aligned} \hspace{\stretch{1}}(1.0.11)

This chemical potential over temperature exponential

\begin{aligned}e^{\beta \mu} \equiv z,\end{aligned} \hspace{\stretch{1}}(1.0.12)

is called the fugacity. The denominator has the form

\begin{aligned}D = 1 + z e^{-\beta \epsilon_k}+ z^2 e^{-2 \beta \epsilon_k},\end{aligned} \hspace{\stretch{1}}(1.0.13)

so we see that

\begin{aligned}z \frac{\partial {D}}{\partial {z}} = z e^{-\beta \epsilon_k}+ 2 z^2 e^{-2 \beta \epsilon_k}+ 3 z^3 e^{-3 \beta \epsilon_k}+ \cdots\end{aligned} \hspace{\stretch{1}}(1.0.14)

Thus the numerator is

\begin{aligned}N = z \frac{\partial {D}}{\partial {z}},\end{aligned} \hspace{\stretch{1}}(1.0.15)


\begin{aligned}\left\langle{{n_k^b}}\right\rangle &= \frac{z \frac{\partial {D_k}}{\partial {z}} }{D_k} \\ &= z \frac{\partial {}}{\partial {z}} \ln D_k \\ &= \cdots \\ &= \frac{1}{{ e^{\beta(\epsilon_k - \mu)} - 1}}\end{aligned} \hspace{\stretch{1}}(1.0.16)

What is the density \rho?

For Fermions

\begin{aligned}\rho = \frac{N}{V} =\frac{1}{{V}} \sum_{\mathbf{k}}\frac{1}{{ e^{\beta(\epsilon_\mathbf{k} - \mu)} + 1}}\end{aligned} \hspace{\stretch{1}}(1.0.17)

Using a “particle in a box” quantization where k_\alpha = 2 \pi m_\alpha/L, in a d-dimensional space, we can approximate this as

\begin{aligned}\boxed{\rho = \int \frac{d^d k}{(2 \pi)^d}\frac{1}{{ e^{\beta(\epsilon_k - \mu)} - 1}}.}\end{aligned} \hspace{\stretch{1}}(1.0.18)

This integral is actually difficult to evaluate. For T \rightarrow 0 (\beta \rightarrow \infty, where

\begin{aligned}n_k = \Theta(\mu - \epsilon_k).\end{aligned} \hspace{\stretch{1}}(1.0.19)

This is illustrated in, where we also show the smearing that occurs as temperature increases fig. 1.3.


Fig 1.3: Occupation numbers for different energies



\begin{aligned}E_{\mathrm{F}} = \mu(T = 0),\end{aligned} \hspace{\stretch{1}}(1.0.20)

we want to ask what is the radius of the ball for which

\begin{aligned}\epsilon_k = E_{\mathrm{F}}\end{aligned} \hspace{\stretch{1}}(1.0.21)


\begin{aligned}E_{\mathrm{F}} = \frac{\hbar^2 k_{\mathrm{F}}^2}{2m},\end{aligned} \hspace{\stretch{1}}(1.0.22)

so that

\begin{aligned}k_{\mathrm{F}} = \sqrt{\frac{2 m E_{\mathrm{F}}}{\hbar^2}},\end{aligned} \hspace{\stretch{1}}(1.0.23)

so that our density where \epsilon_k = \mu is

\begin{aligned}\rho &= \int_{k \le k_{\mathrm{F}}} \frac{d^3 k}{(2 \pi)^3} \times 1 \\ &= \frac{1}{{(2\pi)^3}} 4 \pi \int^{k_{\mathrm{F}}} k^2 dk \\ &= \frac{4 \pi}{3} k_{\mathrm{F}}^3 \frac{1}{{(2 \pi)^3}},\end{aligned} \hspace{\stretch{1}}(1.0.24)

so that

\begin{aligned}k_{\mathrm{F}} = (6 \pi^2 \rho)^{1/3},\end{aligned} \hspace{\stretch{1}}(1.0.25)

Our chemical potential at zero temperature is then

\begin{aligned}\mu(T = 0) = \frac{\hbar^2}{2m} (6 \pi^2 \rho)^{2/3}.\end{aligned} \hspace{\stretch{1}}(1.0.26)

\begin{aligned}\rho^{-1/3} = \mbox{interparticle spacing}.\end{aligned} \hspace{\stretch{1}}(1.0.27)

We can convince ourself that the chemical potential must have the form fig. 1.4.


Fig 1.4: Large negative chemical potential at high temperatures


Given large negative chemical potential at high temperatures our number distribution will have the form

\begin{aligned}\left\langle{{n_k}}\right\rangle = e^{-\beta (\epsilon_k - \mu)} \propto e^{-\beta \epsilon_k}\end{aligned} \hspace{\stretch{1}}(1.0.28)

We see that the classical Boltzmann distribution is recovered for high temperatures.

We can also calculate the chemical potential at high temperatures. We’ll find that this has the form

\begin{aligned}e^{\beta \mu} = \frac{4}{3} \rho \lambda_T^3,\end{aligned} \hspace{\stretch{1}}(1.0.29)

where this quantity \lambda_T is called the Thermal de Broglie wavelength.

\begin{aligned}\lambda_T = \sqrt{\frac{ 2 \pi \hbar^2}{m k_{\mathrm{B}} T}}.\end{aligned} \hspace{\stretch{1}}(1.0.30)

Posted in Math and Physics Learning. | Tagged: , , , , , , , , , , , , , , | Leave a Comment »

PHY456H1F, Quantum Mechanics II. My solutions to problem set 1 (ungraded).

Posted by peeterjoot on September 19, 2011

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

Harmonic oscillator.


\begin{aligned}H_0 = \frac{P^2}{2m} + \frac{1}{{2}} m \omega^2 X^2\end{aligned} \hspace{\stretch{1}}(1.1)

Since it’s been a while let’s compute the raising and lowering factorization that was used so extensively for this problem.

It was of the form

\begin{aligned}H_0 = (a X - i b P)(a X + i b P) + \cdots\end{aligned} \hspace{\stretch{1}}(1.2)

Why this factorization has an imaginary in it is a good question. It’s not one that is given any sort of rationale in the text ([1]).

It’s clear that we want a = \sqrt{m/2} \omega and b = 1/\sqrt{2m}. The difference is then

\begin{aligned}H_0 - (a X - i b P)(a X + i b P)=- i a b \left[{X},{P}\right]  = - i \frac{\omega}{2} \left[{X},{P}\right]\end{aligned} \hspace{\stretch{1}}(1.3)

That commutator is an i\hbar value, but what was the sign? Let’s compute so we don’t get it wrong

\begin{aligned}\left[{x},{ p}\right] \psi&= -i \hbar \left[{x},{\partial_x}\right] \psi \\ &= -i \hbar ( x \partial_x \psi - \partial_x (x \psi) ) \\ &= -i \hbar ( - \psi ) \\ &= i \hbar \psi\end{aligned}

So we have

\begin{aligned}H_0 =\left(\omega \sqrt{\frac{m}{2}} X - i \sqrt{\frac{1}{2m}} P\right)\left(\omega \sqrt{\frac{m}{2}} X + i \sqrt{\frac{1}{2m}} P\right)+ \frac{\hbar \omega}{2}\end{aligned} \hspace{\stretch{1}}(1.4)

Factoring out an \hbar \omega produces the form of the Hamiltonian that we used before

\begin{aligned}H_0 =\hbar \omega \left(\left(\sqrt{\frac{m \omega}{2 \hbar}} X - i \sqrt{\frac{1}{2m \hbar \omega}} P\right)\left(\sqrt{\frac{m \omega}{2 \hbar}} X + i \sqrt{\frac{1}{2m \hbar \omega}} P\right)+ \frac{1}{{2}}\right).\end{aligned} \hspace{\stretch{1}}(1.5)

The factors were labeled the uppering (a^\dagger) and lowering (a) operators respectively, and written

\begin{aligned}H_0 &= \hbar \omega \left( a^\dagger a + \frac{1}{{2}} \right) \\ a &= \sqrt{\frac{m \omega}{2 \hbar}} X + i \sqrt{\frac{1}{2m \hbar \omega}} P \\ a^\dagger &= \sqrt{\frac{m \omega}{2 \hbar}} X - i \sqrt{\frac{1}{2m \hbar \omega}} P.\end{aligned} \hspace{\stretch{1}}(1.6)

Observe that we can find the inverse relations

\begin{aligned}X &= \sqrt{ \frac{\hbar}{2 m \omega} } \left( a + a^\dagger \right) \\ P &= i \sqrt{ \frac{m \hbar \omega}{2} } \left( a^\dagger  - a \right)\end{aligned} \hspace{\stretch{1}}(1.9)

What is a good reason that we chose this particular factorization? For example, a quick computation shows that we could have also picked

\begin{aligned}H_0 = \hbar \omega \left( a a^\dagger - \frac{1}{{2}} \right).\end{aligned} \hspace{\stretch{1}}(1.11)

I don’t know that answer. That said, this second factorization is useful in that it provides the commutator relation between the raising and lowering operators, since subtracting 1.11 and 1.6 yields

\begin{aligned}\left[{a},{a^\dagger}\right] = 1.\end{aligned} \hspace{\stretch{1}}(1.12)

If we suppose that we have eigenstates for the operator a^\dagger a of the form

\begin{aligned}a^\dagger a {\lvert {n} \rangle} = \lambda_n {\lvert {n} \rangle},\end{aligned} \hspace{\stretch{1}}(1.13)

then the problem of finding the eigensolution of H_0 reduces to solving this problem. Because a^\dagger a commutes with 1/2, an eigenstate of a^\dagger a is also an eigenstate of H_0. Utilizing 1.12 we then have

\begin{aligned}a^\dagger a ( a {\lvert {n} \rangle} )&= (a a^\dagger - 1 ) a {\lvert {n} \rangle} \\ &= a (a^\dagger a - 1 ) {\lvert {n} \rangle} \\ &= a (\lambda_n - 1 ) {\lvert {n} \rangle} \\ &= (\lambda_n - 1 ) a {\lvert {n} \rangle},\end{aligned}

so we see that a {\lvert {n} \rangle} is an eigenstate of a^\dagger a with eigenvalue \lambda_n - 1.

Similarly for the raising operator

\begin{aligned}a^\dagger a ( a^\dagger {\lvert {n} \rangle} )&=a^\dagger (a  a^\dagger) {\lvert {n} \rangle} ) \\ &=a^\dagger (a^\dagger a + 1) {\lvert {n} \rangle} ) \\ &=a^\dagger (\lambda_n + 1) {\lvert {n} \rangle} ),\end{aligned}

and find that a^\dagger {\lvert {n} \rangle} is also an eigenstate of a^\dagger a with eigenvalue \lambda_n + 1.

Supposing that there is a lowest energy level (because the potential V(x) = m \omega x^2 /2 has a lower bound of zero) then the state {\lvert {0} \rangle} for which the energy is the lowest when operated on by a we have

\begin{aligned}a {\lvert {0} \rangle} = 0\end{aligned} \hspace{\stretch{1}}(1.14)


\begin{aligned}a^\dagger a {\lvert {0} \rangle} = 0,\end{aligned} \hspace{\stretch{1}}(1.15)


\begin{aligned}\lambda_0 = 0.\end{aligned} \hspace{\stretch{1}}(1.16)

This seems like a small bit of slight of hand, since it sneakily supplies an integer value to \lambda_0 where up to this point 0 was just a label.

If the eigenvalue equation we are trying to solve for the Hamiltonian is

\begin{aligned}H_0 {\lvert {n} \rangle} = E_n {\lvert {n} \rangle}.\end{aligned} \hspace{\stretch{1}}(1.17)

Then we must then have

\begin{aligned}E_n = \hbar \omega \left(\lambda_n + \frac{1}{{2}} \right) = \hbar \omega \left(n + \frac{1}{{2}} \right)\end{aligned} \hspace{\stretch{1}}(1.18)

Part (a)

We’ve now got enough context to attempt the first part of the question, calculation of

\begin{aligned}{\langle {n} \rvert} X^4 {\lvert {n} \rangle}\end{aligned} \hspace{\stretch{1}}(1.19)

We’ve calculated things like this before, such as

\begin{aligned}{\langle {n} \rvert} X^2 {\lvert {n} \rangle}&=\frac{\hbar}{2 m \omega} {\langle {n} \rvert} (a + a^\dagger)^2 {\lvert {n} \rangle}\end{aligned}

To continue we need an exact relation between {\lvert {n} \rangle} and {\lvert {n \pm 1} \rangle}. Recall that a {\lvert {n} \rangle} was an eigenstate of a^\dagger a with eigenvalue n - 1. This implies that the eigenstates a {\lvert {n} \rangle} and {\lvert {n-1} \rangle} are proportional

\begin{aligned}a {\lvert {n} \rangle} = c_n {\lvert {n - 1} \rangle},\end{aligned} \hspace{\stretch{1}}(1.20)


\begin{aligned}{\langle {n} \rvert} a^\dagger a {\lvert {n} \rangle} &= {\left\lvert{c_n}\right\rvert}^2 \left\langle{{n - 1}} \vert {{n-1}}\right\rangle = {\left\lvert{c_n}\right\rvert}^2 \\ n \left\langle{{n}} \vert {{n}}\right\rangle &= \\ n &=\end{aligned}

so that

\begin{aligned}a {\lvert {n} \rangle} = \sqrt{n} {\lvert {n - 1} \rangle}.\end{aligned} \hspace{\stretch{1}}(1.21)

Similarly let

\begin{aligned}a^\dagger {\lvert {n} \rangle} = b_n {\lvert {n + 1} \rangle},\end{aligned} \hspace{\stretch{1}}(1.22)


\begin{aligned}{\langle {n} \rvert} a a^\dagger {\lvert {n} \rangle} &= {\left\lvert{b_n}\right\rvert}^2 \left\langle{{n - 1}} \vert {{n-1}}\right\rangle = {\left\lvert{b_n}\right\rvert}^2 \\ {\langle {n} \rvert} (1 + a^\dagger a) {\lvert {n} \rangle} &= \\ 1 + n &=\end{aligned}

so that

\begin{aligned}a^\dagger {\lvert {n} \rangle} = \sqrt{n+1} {\lvert {n + 1} \rangle}.\end{aligned} \hspace{\stretch{1}}(1.23)

We can now return to 1.19, and find

\begin{aligned}{\langle {n} \rvert} X^4 {\lvert {n} \rangle}&=\frac{\hbar^2}{4 m^2 \omega^2} {\langle {n} \rvert} (a + a^\dagger)^4 {\lvert {n} \rangle}\end{aligned}

Consider half of this braket

\begin{aligned}(a + a^\dagger)^2 {\lvert {n} \rangle}&=\left( a^2 + (a^\dagger)^2 + a^\dagger a + a a^\dagger \right) {\lvert {n} \rangle} \\ &=\left( a^2 + (a^\dagger)^2 + a^\dagger a + (1 + a^\dagger a) \right) {\lvert {n} \rangle} \\ &=\left( a^2 + (a^\dagger)^2 + 1 + 2 a^\dagger a \right) {\lvert {n} \rangle} \\ &=\sqrt{n-1}\sqrt{n-2} {\lvert {n-2} \rangle}+\sqrt{n+1}\sqrt{n+2} {\lvert {n + 2} \rangle}+{\lvert {n} \rangle}+  2 n {\lvert {n} \rangle}\end{aligned}

Squaring, utilizing the Hermitian nature of the X operator

\begin{aligned}{\langle {n} \rvert} X^4 {\lvert {n} \rangle}=\frac{\hbar^2}{4 m^2 \omega^2}\left((n-1)(n-2) + (n+1)(n+2) + (1 + 2n)^2\right)=\frac{\hbar^2}{4 m^2 \omega^2}\left( 6 n^2 + 4 n + 5 \right)\end{aligned} \hspace{\stretch{1}}(1.24)

Part (b)

Find the ground state energy of the Hamiltonian H = H_0 + \gamma X^2 for \gamma > 0.

The new Hamiltonian has the form

\begin{aligned}H = \frac{P^2}{2m} + \frac{1}{{2}} m \left(\omega^2 + \frac{2 \gamma}{m} \right) X^2 =\frac{P^2}{2m} + \frac{1}{{2}} m {\omega'}^2 X^2,\end{aligned} \hspace{\stretch{1}}(1.25)


\begin{aligned}\omega' = \sqrt{ \omega^2 + \frac{2 \gamma}{m} }\end{aligned} \hspace{\stretch{1}}(1.26)

The energy states of the Hamiltonian are thus

\begin{aligned}E_n = \hbar \sqrt{ \omega^2 + \frac{2 \gamma}{m} } \left( n + \frac{1}{{2}} \right)\end{aligned} \hspace{\stretch{1}}(1.27)

and the ground state of the modified Hamiltonian H is thus

\begin{aligned}E_0 = \frac{\hbar}{2} \sqrt{ \omega^2 + \frac{2 \gamma}{m} }\end{aligned} \hspace{\stretch{1}}(1.28)

Part (c)

Find the ground state energy of the Hamiltonian H = H_0 - \alpha X.

With a bit of play, this new Hamiltonian can be factored into

\begin{aligned}H= \hbar \omega \left( b^\dagger b + \frac{1}{{2}} \right) - \frac{\alpha^2}{2 m \omega^2}= \hbar \omega \left( b b^\dagger - \frac{1}{{2}} \right) - \frac{\alpha^2}{2 m \omega^2},\end{aligned} \hspace{\stretch{1}}(1.29)


\begin{aligned}b &= \sqrt{\frac{m \omega}{2\hbar}} X + \frac{i P}{\sqrt{2 m \hbar \omega}} - \frac{\alpha}{\omega \sqrt{ 2 m \hbar \omega }} \\ b^\dagger &= \sqrt{\frac{m \omega}{2\hbar}} X - \frac{i P}{\sqrt{2 m \hbar \omega}} - \frac{\alpha}{\omega \sqrt{ 2 m \hbar \omega }}.\end{aligned} \hspace{\stretch{1}}(1.30)

From 1.29 we see that we have the same sort of commutator relationship as in the original Hamiltonian

\begin{aligned}\left[{b},{b^\dagger}\right] = 1,\end{aligned} \hspace{\stretch{1}}(1.32)

and because of this, all the preceding arguments follow unchanged with the exception that the energy eigenstates of this Hamiltonian are shifted by a constant

\begin{aligned}H {\lvert {n} \rangle} = \left( \hbar \omega \left( n + \frac{1}{{2}} \right) - \frac{\alpha^2}{2 m \omega^2} \right) {\lvert {n} \rangle},\end{aligned} \hspace{\stretch{1}}(1.33)

where the {\lvert {n} \rangle} states are simultaneous eigenstates of the b^\dagger b operator

\begin{aligned}b^\dagger b {\lvert {n} \rangle} = n {\lvert {n} \rangle}.\end{aligned} \hspace{\stretch{1}}(1.34)

The ground state energy is then

\begin{aligned}E_0 = \frac{\hbar \omega }{2} - \frac{\alpha^2}{2 m \omega^2}.\end{aligned} \hspace{\stretch{1}}(1.35)

This makes sense. A translation of the entire position of the system should not effect the energy level distribution of the system, but we have set our reference potential differently, and have this constant energy adjustment to the entire system.

Hydrogen atom and spherical harmonics.

We are asked to show that for any eigenkets of the hydrogen atom {\lvert {\Phi_{nlm}} \rangle} we have

\begin{aligned}{\langle {\Phi_{nlm}} \rvert} X {\lvert {\Phi_{nlm}} \rangle} ={\langle {\Phi_{nlm}} \rvert} Y {\lvert {\Phi_{nlm}} \rangle} ={\langle {\Phi_{nlm}} \rvert} Z {\lvert {\Phi_{nlm}} \rangle}.\end{aligned} \hspace{\stretch{1}}(2.36)

The summary sheet provides us with the wavefunction

\begin{aligned}\left\langle{\mathbf{r}} \vert {{\Phi_{nlm}}}\right\rangle = \frac{2}{n^2 a_0^{3/2}} \sqrt{\frac{(n-l-1)!}{(n+l)!)^3}} F_{nl}\left( \frac{2r}{n a_0} \right) Y_l^m(\theta, \phi),\end{aligned} \hspace{\stretch{1}}(2.37)

where F_{nl} is a real valued function defined in terms of Lagueere polynomials. Working with the expectation of the X operator to start with we have

\begin{aligned}{\langle {\Phi_{nlm}} \rvert} X {\lvert {\Phi_{nlm}} \rangle} &=\int \left\langle{{\Phi_{nlm}}} \vert {{\mathbf{r}'}}\right\rangle {\langle {\mathbf{r}'} \rvert} X {\lvert {\mathbf{r}} \rangle} \left\langle{\mathbf{r}} \vert {{\Phi_{nlm}}}\right\rangle d^3 \mathbf{r} d^3 \mathbf{r}' \\ &=\int \left\langle{{\Phi_{nlm}}} \vert {{\mathbf{r}'}}\right\rangle \delta(\mathbf{r} - \mathbf{r}') r \sin\theta \cos\phi \left\langle{\mathbf{r}} \vert {{\Phi_{nlm}}}\right\rangle d^3 \mathbf{r} d^3 \mathbf{r}' \\ &=\int \Phi_{nlm}^{*}(\mathbf{r}) r \sin\theta \cos\phi \Phi_{nlm}(\mathbf{r}) d^3 \mathbf{r} \\ &\sim\int r^2 dr {\left\lvert{ F_{nl}\left(\frac{2 r}{ n a_0} \right)}\right\rvert}^2 r \int \sin\theta d\theta d\phi{Y_l^m}^{*}(\theta, \phi) \sin\theta \cos\phi Y_l^m(\theta, \phi) \\ \end{aligned}

Recalling that the only \phi dependence in Y_l^m is e^{i m \phi} we can perform the d\phi integration directly, which is

\begin{aligned}\int_{\phi=0}^{2\pi} \cos\phi d\phi e^{-i m \phi} e^{i m \phi} = 0.\end{aligned} \hspace{\stretch{1}}(2.38)

We have the same story for the Y expectation which is

\begin{aligned}{\langle {\Phi_{nlm}} \rvert} X {\lvert {\Phi_{nlm}} \rangle} \sim\int r^2 dr {\left\lvert{F_{nl}\left( \frac{2 r}{ n a_0} \right)}\right\rvert}^2 r \int \sin\theta d\theta d\phi{Y_l^m}^{*}(\theta, \phi) \sin\theta \sin\phi Y_l^m(\theta, \phi).\end{aligned} \hspace{\stretch{1}}(2.39)

Our \phi integral is then just

\begin{aligned}\int_{\phi=0}^{2\pi} \sin\phi d\phi e^{-i m \phi} e^{i m \phi} = 0,\end{aligned} \hspace{\stretch{1}}(2.40)

also zero. The Z expectation is a slightly different story. There we have

\begin{aligned}\begin{aligned}{\langle {\Phi_{nlm}} \rvert} Z {\lvert {\Phi_{nlm}} \rangle} &\sim\int dr {\left\lvert{F_{nl}\left( \frac{2 r}{ n a_0} \right)}\right\rvert}^2 r^3  \\ &\quad \int_0^{2\pi} d\phi\int_0^\pi \sin \theta d\theta\left( \sin\theta \right)^{-2m}\left( \frac{d^{l - m}}{d (\cos\theta)^{l-m}} \sin^{2l}\theta \right)^2\cos\theta.\end{aligned}\end{aligned} \hspace{\stretch{1}}(2.41)

Within this last integral we can make the substitution

\begin{aligned}u &= \cos\theta \\ \sin\theta d\theta &= - d(\cos\theta) = -du \\ u &\in [1, -1],\end{aligned} \hspace{\stretch{1}}(2.42)

and the integral takes the form

\begin{aligned}-\int_{-1}^1 (-du) \frac{1}{{(1 - u^2)^m}} \left( \frac{d^{l-m}}{d u^{l -m }} (1 - u^2)^l\right)^2 u.\end{aligned} \hspace{\stretch{1}}(2.45)

Here we have the product of two even functions, times one odd function (u), over a symmetric interval, so the end result is zero, completing the problem.

I wasn’t able to see how to exploit the parity result suggested in the problem, but it wasn’t so bad to show these directly.

Angular momentum operator.

Working with the appropriate expressions in Cartesian components, confirm that L_i {\lvert {\psi} \rangle} = 0 for each component of angular momentum L_i, if \left\langle{\mathbf{r}} \vert {{\psi}}\right\rangle = \psi(\mathbf{r}) is in fact only a function of r = {\left\lvert{\mathbf{r}}\right\rvert}.

In order to proceed, we will have to consider a matrix element, so that we can operate on {\lvert {\psi} \rangle} in position space. For that matrix element, we can proceed to insert complete states, and reduce the problem to a question of wavefunctions. That is

\begin{aligned}{\langle {\mathbf{r}} \rvert} L_i {\lvert {\psi} \rangle}&=\int d^3 \mathbf{r}' {\langle {\mathbf{r}} \rvert} L_i {\lvert {\mathbf{r}'} \rangle} \left\langle{{\mathbf{r}'}} \vert {{\psi}}\right\rangle \\ &=\int d^3 \mathbf{r}' {\langle {\mathbf{r}} \rvert} \epsilon_{i a b} X_a P_b {\lvert {\mathbf{r}'} \rangle} \left\langle{{\mathbf{r}'}} \vert {{\psi}}\right\rangle \\ &=-i \hbar \epsilon_{i a b} \int d^3 \mathbf{r}' x_a {\langle {\mathbf{r}} \rvert} \frac{\partial {\psi(\mathbf{r}')}}{\partial {X_b}} {\lvert {\mathbf{r}'} \rangle}  \\ &=-i \hbar \epsilon_{i a b} \int d^3 \mathbf{r}' x_a \frac{\partial {\psi(\mathbf{r}')}}{\partial {x_b}} \left\langle{\mathbf{r}} \vert {{\mathbf{r}'}}\right\rangle  \\ &=-i \hbar \epsilon_{i a b} \int d^3 \mathbf{r}' x_a \frac{\partial {\psi(\mathbf{r}')}}{\partial {x_b}} \delta^3(\mathbf{r} - \mathbf{r}') \\ &=-i \hbar \epsilon_{i a b} x_a \frac{\partial {\psi(\mathbf{r})}}{\partial {x_b}} \end{aligned}

With \psi(\mathbf{r}) = \psi(r) we have

\begin{aligned}{\langle {\mathbf{r}} \rvert} L_i {\lvert {\psi} \rangle}&=-i \hbar \epsilon_{i a b} x_a \frac{\partial {\psi(r)}}{\partial {x_b}}  \\ &=-i \hbar \epsilon_{i a b} x_a \frac{\partial {r}}{\partial {x_b}} \frac{d\psi(r)}{dr}  \\ &=-i \hbar \epsilon_{i a b} x_a \frac{1}{{2}} 2 x_b \frac{1}{{r}} \frac{d\psi(r)}{dr}  \\ \end{aligned}

We are left with an sum of a symmetric product x_a x_b with the antisymmetric tensor \epsilon_{i a b} so this is zero for all i \in [1,3].


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

Posted in Math and Physics Learning. | Tagged: , , , , , , , , , , | Leave a Comment »

PHY356F: Quantum Mechanics I. Lecture 11 notes. Harmonic Oscillator.

Posted by peeterjoot on November 30, 2010

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


Why study this problem?

It is relevant to describing the oscillation of molecules, quantum states of light, vibrations of the lattice structure of a solid, and so on.

FIXME: projected picture of masses on springs, with a ladle shaped well, approximately Harmonic about the minimum of the bucket.

The problem to solve is the one dimensional Hamiltonian

\begin{aligned}V(X) &= \frac{1}{{2}} K X^2 \\ K &= m \omega^2 \\ H &= \frac{P^2}{2m} + V(X)\end{aligned} \hspace{\stretch{1}}(8.168)

where m is the mass, \omega is the frequency, X is the position operator, and P is the momentum operator. Of these quantities, \omega and m are classical quantities.

This problem can be used to illustrate some of the reasons why we study the different pictures (Heisenberg, Interaction and Schr\”{o}dinger). This is a problem well suited to all of these (FIXME: lookup an example of this with the interaction picture. The book covers H and S methods.

We attack this with a non-intuitive, but cool technique. Introduce the raising a^\dagger and lowering a operators:

\begin{aligned}a &= \sqrt{\frac{m \omega}{2 \hbar}} \left( X + i \frac{P}{m\omega} \right) \\ a^\dagger &= \sqrt{\frac{m \omega}{2 \hbar}} \left( X - i \frac{P}{m\omega} \right)\end{aligned} \hspace{\stretch{1}}(8.171)

\paragraph{Question:} are we using the dagger for more than Hermitian conjugation in this case.
\paragraph{Answer:} No, this is precisely the Hermitian conjugation operation.

Solving for X and P in terms of a and a^\dagger, we have

\begin{aligned}a + a^\dagger &= \sqrt{\frac{m \omega}{2 \hbar}} 2 X  \\ a - a^\dagger &= \sqrt{\frac{m \omega}{2 \hbar}} 2 i \frac{P }{m \omega}\end{aligned}


\begin{aligned}X &= \sqrt{\frac{\hbar}{2 m \omega}} (a^\dagger + a) \\ P &= i \sqrt{\frac{\hbar m \omega}{2}} (a^\dagger -a)\end{aligned} \hspace{\stretch{1}}(8.173)

Express H in terms of a and a^\dagger

\begin{aligned}H &= \frac{P^2}{2m} + \frac{1}{{2}} K X^2  \\ &= \frac{1}{2m} \left(i \sqrt{\frac{\hbar m \omega}{2}} (a^\dagger -a)\right)^2+ \frac{1}{{2}} m \omega^2\left(\sqrt{\frac{\hbar}{2 m \omega}} (a^\dagger + a) \right)^2 \\ &= \frac{-\hbar \omega}{4} \left(a^\dagger a^\dagger + a^2 - a a^\dagger - a^\dagger a\right)+ \frac{\hbar \omega}{4}\left(a^\dagger a^\dagger + a^2 + a a^\dagger + a^\dagger a\right) \\ \end{aligned}

\begin{aligned}H= \frac{\hbar \omega}{2} \left(a a^\dagger + a^\dagger a\right) = \frac{\hbar \omega}{2} \left(2 a^\dagger a + \left[{a},{a^\dagger}\right]\right) \end{aligned} \hspace{\stretch{1}}(8.175)

Since \left[{X},{P}\right] = i \hbar \mathbf{1} then we can show that \left[{a},{a^\dagger}\right] = \mathbf{1}. Solve for \left[{a},{a^\dagger}\right] as follows

\begin{aligned}i \hbar &=\left[{X},{P}\right] \\ &=\left[{\sqrt{\frac{\hbar}{2 m \omega}} (a^\dagger + a) },{i \sqrt{\frac{\hbar m \omega}{2}} (a^\dagger -a)}\right] \\ &=\sqrt{\frac{\hbar}{2 m \omega}} i \sqrt{\frac{\hbar m \omega}{2}} \left[{a^\dagger + a},{a^\dagger -a}\right] \\ &= \frac{i \hbar}{2}\left(\left[{a^\dagger},{a^\dagger}\right] -\left[{a^\dagger},{a}\right] +\left[{a},{a^\dagger}\right] -\left[{a},{a}\right] \right)  \\ &= \frac{i \hbar}{2}\left(0+2 \left[{a},{a^\dagger}\right] -0\right)\end{aligned}

Comparing LHS and RHS we have as stated

\begin{aligned}\left[{a},{a^\dagger}\right] = \mathbf{1}\end{aligned} \hspace{\stretch{1}}(8.176)

and thus from 8.175 we have

\begin{aligned}H = \hbar \omega \left( a^\dagger a + \frac{\mathbf{1}}{2} \right)\end{aligned} \hspace{\stretch{1}}(8.177)

Let {\lvert {n} \rangle} be the eigenstate of H so that H{\lvert {n} \rangle} = E_n {\lvert {n} \rangle}. From 8.177 we have

\begin{aligned}H {\lvert {n} \rangle} =\hbar \omega \left( a^\dagger a + \frac{\mathbf{1}}{2} \right) {\lvert {n} \rangle}\end{aligned} \hspace{\stretch{1}}(8.178)


\begin{aligned}a^\dagger a {\lvert {n} \rangle} + \frac{{\lvert {n} \rangle}}{2} = \frac{E_n}{\hbar \omega} {\lvert {n} \rangle}\end{aligned} \hspace{\stretch{1}}(8.179)

\begin{aligned}a^\dagger a {\lvert {n} \rangle} = \left( \frac{E_n}{\hbar \omega} - \frac{1}{{2}} \right) {\lvert {n} \rangle} = \lambda_n {\lvert {n} \rangle}\end{aligned} \hspace{\stretch{1}}(8.180)

We wish now to find the eigenstates of the “Number” operator a^\dagger a, which are simultaneously eigenstates of the Hamiltonian operator.

Observe that we have

\begin{aligned}a^\dagger a (a^\dagger {\lvert {n} \rangle} ) &= a^\dagger ( a a^\dagger {\lvert {n} \rangle} ) \\ &= a^\dagger ( \mathbf{1} + a^\dagger a ) {\lvert {n} \rangle}\end{aligned}

where we used \left[{a},{a^\dagger}\right] = a a^\dagger - a^\dagger a = \mathbf{1}.

\begin{aligned}a^\dagger a (a^\dagger {\lvert {n} \rangle} ) &= a^\dagger \left( \mathbf{1} + \frac{E_n}{\hbar\omega} - \frac{\mathbf{1}}{2} \right) {\lvert {n} \rangle} \\ &= a^\dagger \left( \frac{E_n}{\hbar\omega} + \frac{\mathbf{1}}{2} \right) {\lvert {n} \rangle},\end{aligned}


\begin{aligned}a^\dagger a (a^\dagger {\lvert {n} \rangle} ) = (\lambda_n + 1) (a^\dagger {\lvert {n} \rangle} )\end{aligned} \hspace{\stretch{1}}(8.181)

The new state a^\dagger {\lvert {n} \rangle} is presumed to lie in the same space, expressible as a linear combination of the basis states in this space. We can see the effect of the operator a a^\dagger on this new state, we find that the energy is changed, but the state is otherwise unchanged. Any state a^\dagger {\lvert {n} \rangle} is an eigenstate of a^\dagger a, and therefore also an eigenstate of the Hamiltonian.

Play the same game and win big by discovering that

\begin{aligned}a^\dagger a ( a {\lvert {n} \rangle} ) = (\lambda_n -1) (a {\lvert {n} \rangle} )\end{aligned} \hspace{\stretch{1}}(8.182)

There will be some state {\lvert {0} \rangle} such that

\begin{aligned}a {\lvert {0} \rangle} = 0 {\lvert {0} \rangle}\end{aligned} \hspace{\stretch{1}}(8.183)

which implies

\begin{aligned}a^\dagger (a {\lvert {0} \rangle}) = (a^\dagger a) {\lvert {0} \rangle} = 0\end{aligned} \hspace{\stretch{1}}(8.184)

so from 8.180 we have

\begin{aligned}\lambda_0 = 0\end{aligned} \hspace{\stretch{1}}(8.185)

Observe that we can identify \lambda_n = n for

\begin{aligned}\lambda_n = \left( \frac{E_n}{\hbar\omega} - \frac{1}{{2}} \right) = n,\end{aligned} \hspace{\stretch{1}}(8.186)


\begin{aligned}\frac{E_n}{\hbar\omega} = n + \frac{1}{{2}}\end{aligned} \hspace{\stretch{1}}(8.187)


\begin{aligned}E_n = \hbar \omega \left( n + \frac{1}{{2}} \right)\end{aligned} \hspace{\stretch{1}}(8.188)

where n = 0, 1, 2, \cdots.

We can write

\begin{aligned}\hbar \omega \left( a^\dagger a + \frac{1}{{2}} \mathbf{1} \right) {\lvert {n} \rangle} &= E_n {\lvert {n} \rangle} \\ a^\dagger a {\lvert {n} \rangle} + \frac{1}{{2}} {\lvert {n} \rangle} &= \frac{E_n}{\hbar \omega} {\lvert {n} \rangle} \\ \end{aligned}


\begin{aligned}a^\dagger a {\lvert {n} \rangle} = \left( \frac{E_n}{\hbar \omega} - \frac{1}{{2}} \right) {\lvert {n} \rangle} = \lambda_n {\lvert {n} \rangle} = n {\lvert {n} \rangle}\end{aligned} \hspace{\stretch{1}}(8.189)

We call this operator a^\dagger a = N, the number operator, so that

\begin{aligned}N {\lvert {n} \rangle} = n {\lvert {n} \rangle}\end{aligned} \hspace{\stretch{1}}(8.190)

Relating states.

Recall the calculation we performed for

\begin{aligned}L_{+} {\lvert {lm} \rangle} &= C_{+} {\lvert {l, m+1} \rangle} \\ L_{-} {\lvert {lm} \rangle} &= C_{+} {\lvert {l, m-1} \rangle}\end{aligned} \hspace{\stretch{1}}(9.191)

Where C_{+}, and C_{+} are constants. The next game we are going to play is to work out C_n for the lowering operation

\begin{aligned}a{\lvert {n} \rangle} = C_n {\lvert {n-1} \rangle}\end{aligned} \hspace{\stretch{1}}(9.193)

and the raising operation

\begin{aligned}a^\dagger {\lvert {n} \rangle} = B_n {\lvert {n+1} \rangle}.\end{aligned} \hspace{\stretch{1}}(9.194)

For the Hermitian conjugate of a {\lvert {n} \rangle} we have

\begin{aligned}(a {\lvert {n} \rangle})^\dagger = ( C_n {\lvert {n-1} \rangle} )^\dagger = C_n^{*} {\lvert {n-1} \rangle}\end{aligned} \hspace{\stretch{1}}(9.195)


\begin{aligned}({\langle {n} \rvert} a^\dagger) (a {\lvert {n} \rangle}) = C_n C_n^{*} \left\langle{{n-1}} \vert {{n-1}}\right\rangle = {\left\lvert{C_n}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(9.196)

Expanding the LHS we have

\begin{aligned}{\left\lvert{C_n}\right\rvert}^2 &={\langle {n} \rvert} a^\dagger a {\lvert {n} \rangle} \\ &={\langle {n} \rvert} n {\lvert {n} \rangle} \\ &=n \left\langle{{n}} \vert {{n}}\right\rangle \\ &=n \end{aligned}


\begin{aligned}C_n = \sqrt{n}\end{aligned} \hspace{\stretch{1}}(9.197)


\begin{aligned}({\langle {n} \rvert} a^\dagger) (a {\lvert {n} \rangle}) = B_n B_n^{*} \left\langle{{n+1}} \vert {{n+1}}\right\rangle = {\left\lvert{B_n}\right\rvert}^2\end{aligned} \hspace{\stretch{1}}(9.198)


\begin{aligned}{\left\lvert{B_n}\right\rvert}^2 &={\langle {n} \rvert} \underbrace{a a^\dagger}_{a a^\dagger - a^\dagger a = \mathbf{1}} {\lvert {n} \rangle} \\ &={\langle {n} \rvert} \left( \mathbf{1} + a^\dagger a \right) {\lvert {n} \rangle} \\ &=(1 + n) \left\langle{{n}} \vert {{n}}\right\rangle \\ &=1 + n \end{aligned}


\begin{aligned}B_n = \sqrt{n + 1}\end{aligned} \hspace{\stretch{1}}(9.199)

Heisenberg picture.

\paragraph{How does the lowering operator a evolve in time?}

\paragraph{A:} Recall that for a general operator A, we have for the time evolution of that operator

\begin{aligned}i \hbar \frac{d A}{dt} = \left[{ A },{H}\right]\end{aligned} \hspace{\stretch{1}}(10.200)

Let’s solve this one.

\begin{aligned}i \hbar \frac{d a}{dt} &= \left[{ a },{H}\right] \\ &= \left[{ a },{ \hbar \omega (a^\dagger a + \mathbf{1}/2) }\right] \\ &= \hbar\omega \left[{ a },{ (a^\dagger a + \mathbf{1}/2) }\right] \\ &= \hbar\omega \left[{ a },{ a^\dagger a }\right] \\ &= \hbar\omega \left( a a^\dagger a - a^\dagger a a \right) \\ &= \hbar\omega \left( (a a^\dagger) a - a^\dagger a a \right) \\ &= \hbar\omega \left( (a^\dagger a + \mathbf{1}) a - a^\dagger a a \right) \\ &= \hbar\omega a \end{aligned}

Even though a is an operator, it can undergo a time evolution and we can think of it as a function, and we can solve for a in the differential equation

\begin{aligned}\frac{d a}{dt} = -i \omega a \end{aligned} \hspace{\stretch{1}}(10.201)

This has the solution

\begin{aligned}a = a(0) e^{-i \omega t}\end{aligned} \hspace{\stretch{1}}(10.202)

here a(0) is an operator, the value of that operator at t = 0. The exponential here is just a scalar (not effected by the operator so we can put it on either side of the operator as desired).


\begin{aligned}a' = a(0) \frac{d}{dt} e^{-i \omega t} = a(0) (-i \omega) e^{-i \omega t} = -i \omega a\end{aligned} \hspace{\stretch{1}}(10.203)

A couple comments on the Schr\”{o}dinger picture.

We don’t do this in class, but it is very similar to the approach of the hydrogen atom. See the text for full details.

In the Schr\”{o}dinger picture,

\begin{aligned}-\frac{\hbar^2}{2m} \frac{d^2 u}{dx^2} + \frac{1}{{2}} m \omega^2 x^2 u = E u\end{aligned} \hspace{\stretch{1}}(11.204)

This does directly to the wave function representation, but we can relate these by noting that we get this as a consequence of the identification u = u(x) = \left\langle{{x}} \vert {{u}}\right\rangle.

In 11.204, we can switch to dimensionless quantities with

\begin{aligned}\xi = \text{``xi (z)''} = \alpha x\end{aligned} \hspace{\stretch{1}}(11.205)


\begin{aligned}\alpha = \sqrt{\frac{m \omega}{\hbar}}\end{aligned} \hspace{\stretch{1}}(11.206)

This gives, with \lambda = 2E/\hbar\omega,

\begin{aligned}\frac{d^2 u}{d\xi^2} + (\lambda - \xi^2) u = 0\end{aligned} \hspace{\stretch{1}}(11.207)

We can use polynomial series expansion methods to solve this, and find that we require a terminating expression, and write this in terms of the Hermite polynomials (courtesy of the clever French once again).

When all is said and done we will get the energy eigenvalues once again

\begin{aligned}E = E_n = \hbar \omega \left( n + \frac{1}{{2}} \right)\end{aligned} \hspace{\stretch{1}}(11.208)

Back to the Heisenberg picture.

Let us express

\begin{aligned}\left\langle{{x}} \vert {{n}}\right\rangle = u_n(x)\end{aligned} \hspace{\stretch{1}}(12.209)


\begin{aligned}a {\lvert {0} \rangle} = 0,\end{aligned} \hspace{\stretch{1}}(12.210)

we have

\begin{aligned}0  =\left( X + i \frac{P}{m \omega} \right) {\lvert {0} \rangle},\end{aligned} \hspace{\stretch{1}}(12.211)


\begin{aligned}0 &= {\langle {x} \rvert} \left( X + i \frac{P}{m \omega} \right) {\lvert {0} \rangle} \\ &= {\langle {x} \rvert} X {\lvert {0 } \rangle} + i \frac{1}{m \omega} {\langle {x} \rvert} P {\lvert {0} \rangle} \\ &= x \left\langle{{x}} \vert {{0}}\right\rangle + i \frac{1}{m \omega} {\langle {x} \rvert} P {\lvert {0} \rangle} \\ \end{aligned}

Recall that our matrix operator is

\begin{aligned}{\langle {x'} \rvert} P {\lvert {x} \rangle} = \delta(x - x') \left( -i \hbar \frac{d}{dx} \right)\end{aligned} \hspace{\stretch{1}}(12.212)

\begin{aligned}{\langle {x} \rvert} P {\lvert {0} \rangle} &={\langle {x} \rvert} P \underbrace{\int {\lvert {x'} \rangle} {\langle {x'} \rvert} dx' }_{= \mathbf{1}}{\lvert {0} \rangle} \\ &=\int {\langle {x} \rvert} P {\lvert {x'} \rangle} \left\langle{{x'}} \vert {{0}}\right\rangle dx' \\ &=\int \delta(x - x') \left( -i \hbar \frac{d}{dx} \right)\left\langle{{x'}} \vert {{0}}\right\rangle dx' \\ &=\left( -i \hbar \frac{d}{dx} \right)\left\langle{{x}} \vert {{0}}\right\rangle\end{aligned}

We have then

\begin{aligned}0 =x u_0(x) + \frac{\hbar}{m \omega} \frac{d u_0(x)}{dx}\end{aligned} \hspace{\stretch{1}}(12.213)

NOTE: picture of the solution to this LDE on slide…. but I didn’t look closely enough.

Posted in Math and Physics Learning. | Tagged: , , , , , , , , , , , | Leave a Comment »