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Notes and problems for Desai chapter III.

Posted by peeterjoot on October 9, 2010

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Chapter III notes and problems for [1].

Some puzzling stuff in the interaction section and superposition of time-dependent states sections. Work through those here.


Problem 1. Virial Theorem.


With the assumption that \left\langle{{\mathbf{r} \cdot \mathbf{p}}}\right\rangle is independent of time, and

\begin{aligned}H = \frac{\mathbf{p}^2}{2m} + V(\mathbf{r}) = T + V\end{aligned} \hspace{\stretch{1}}(2.1)

show that

\begin{aligned}2 \left\langle{{T}}\right\rangle = \left\langle{{ \mathbf{r} \cdot \boldsymbol{\nabla} V}}\right\rangle.\end{aligned} \hspace{\stretch{1}}(2.2)


I floundered with this a bit, but found the required hint in physicsforums. We can start with the Hamiltonian time derivative relation

\begin{aligned}i\hbar \frac{d A_H}{dt} = \left[{A_H},{H}\right]\end{aligned} \hspace{\stretch{1}}(2.3)

So, with the assumption that \left\langle{{\mathbf{r} \cdot \mathbf{p}}}\right\rangle is independent of time, and the use of a stationary state {\lvert {\psi} \rangle} for the expectation calculation we have

\begin{aligned}0 &=\frac{d}{dt} \left\langle{{\mathbf{r} \cdot \mathbf{p}}}\right\rangle  \\ &=\frac{d}{dt} {\langle {\psi} \rvert} \mathbf{r} \cdot \mathbf{p} {\lvert {\psi} \rangle} \\ &={\langle {\psi} \rvert} \frac{d}{dt} ( \mathbf{r} \cdot \mathbf{p} ) {\lvert {\psi} \rangle} \\ &= \frac{1}{{i\hbar}} \left\langle{{ \left[{ \mathbf{r} \cdot \mathbf{p} },{H}\right] }}\right\rangle \\ &= -\left\langle{{ \left[{ \mathbf{r} \cdot \boldsymbol{\nabla} },{\frac{\mathbf{p}^2}{2m}}\right] }}\right\rangle -\left\langle{{ \left[{ \mathbf{r} \cdot \boldsymbol{\nabla} },{V(\mathbf{r})}\right] }}\right\rangle.\end{aligned}

The exercise now becomes one of evaluating the remaining commutators. For the Laplacian commutator we have

\begin{aligned}\left[{ \mathbf{r} \cdot \boldsymbol{\nabla} },{\boldsymbol{\nabla}^2}\right] \psi&=x_m \partial_m \partial_n \partial_n \psi - \partial_n \partial_n x_m \partial_m \psi \\ &=x_m \partial_m \partial_n \partial_n \psi - \partial_n \partial_n \psi - \partial_n x_m \partial_n \partial_m \psi \\ &=x_m \partial_m \partial_n \partial_n \psi - \partial_n \partial_n \psi - \partial_n \partial_n \psi - x_m \partial_n \partial_n \partial_m \psi \\ &=- 2 \boldsymbol{\nabla}^2 \psi\end{aligned}

For the potential commutator we have

\begin{aligned}\left[{ \mathbf{r} \cdot \boldsymbol{\nabla} },{V(\mathbf{r})}\right] \psi&=x_m \partial_m V \psi -V x_m \partial_m \psi  \\ &=x_m (\partial_m V) \psi x_m V \partial_m \psi -V x_m \partial_m \psi  \\ &=\Bigl( \mathbf{r} \cdot (\boldsymbol{\nabla} V) \Bigr) \psi\end{aligned}

Putting all the \hbar factors back in, we get

\begin{aligned}2 \left\langle{{ \frac{\mathbf{p}^2}{2m} }}\right\rangle = \left\langle{{ \mathbf{r} \cdot (\boldsymbol{\nabla} V) }}\right\rangle,\end{aligned} \hspace{\stretch{1}}(2.4)

which is the desired result.

Followup: why assume \left\langle{{\mathbf{r} \cdot \mathbf{p}}}\right\rangle is independent of time?

Problem 2. Application of virial theorem.

Calculate \left\langle{{T}}\right\rangle with V = \lambda \ln(r/a).

\begin{aligned}\mathbf{r} \cdot \boldsymbol{\nabla} V &= r \hat{\mathbf{r}} \cdot \hat{\mathbf{r}} \lambda \frac{\partial {\ln(r/a)}}{\partial {r}} \\ &= \lambda r \frac{1}{{a}} \frac{a}{r} \\ &= \lambda  \\ \implies \\ \left\langle{{T}}\right\rangle &= \lambda/2\end{aligned}

Problem 3. Heisenberg Position operator representation.

Part I.

Express x as an operator x_H for H = \mathbf{p}^2/2m.


\begin{aligned}{\langle {\psi} \rvert} x {\lvert {\psi} \rangle} = {\langle {\psi_0} \rvert} U^\dagger x U {\lvert {\psi_0} \rangle}\end{aligned}

We want to expand

\begin{aligned}x_H &= U^\dagger x U \\ &= e^{i H t/\hbar} x e^{-iH t/\hbar} \\ &= \sum_{k,l = 0}^\infty \frac{1}{{k!}} \frac{1}{{l!}} \left(\frac{i H t}{\hbar}\right)^k x \left(\frac{-i H t}{\hbar}\right)^l .\end{aligned}

We to evaluate H^k x H^l to proceed. Using p^n x = -i \hbar n p^{n-1} + x p^n, we have

\begin{aligned}H^k x &= \frac{1}{{(2m)^k}} p^2k x \\ &= \frac{1}{{(2m)^k}} \left( -i \hbar (2k) p^{2k -1} + x p^2k \right) \\ &= x H^k + \frac{1}{{2m}} (-i \hbar) (2k) p p^{2(k-1)}/(2m)^{k-1} \\ &= x H^k - \frac{i \hbar k}{m} p H^{k-1}.\end{aligned}

This gives us

\begin{aligned}x_H &= x - \frac{i \hbar p }{m} \sum_{k,l=0}^\infty \frac{k}{k!} \frac{1}{{l!}}\left(\frac{i t}{\hbar}\right)^k H^{k-1 + l}\left(\frac{-i t}{\hbar}\right)^l  \\ &= x - \frac{i \hbar p i t }{m \hbar} \end{aligned}


\begin{aligned}x_H &= x + \frac{p t }{m} \end{aligned} \hspace{\stretch{1}}(2.5)

Part II.

Express x as an operator x_H for H = \mathbf{p}^2/2m + V with V = \lambda x^m.

In retrospect, for the first part of this problem, it would have been better to use the series expansion for this exponential sandwich

Or, in explicit form

\begin{aligned}e^A B e^{-A}&=B + \frac{1}{{1!}} \left[{A},{B}\right]+ \frac{1}{{2!}} \left[{A},{\left[{A},{B}\right]}\right]+ \cdots\end{aligned} \hspace{\stretch{1}}(2.6)

Doing so, we’d find for the first commutator

\begin{aligned}\frac{i t}{2m \hbar} \left[{\mathbf{p}^2},{x}\right] = \frac{t p}{m},\end{aligned} \hspace{\stretch{1}}(2.7)

so that the series has only the first two terms, and we’d obtain the same result. That seems like a logical approach to try here too. For the first commutator, we get the same tp/m result since \left[{V},{x}\right] = 0.


\begin{aligned}x^n p = i \hbar n x^{n-1} + p x^n,\end{aligned} \hspace{\stretch{1}}(2.8)

I find

\begin{aligned}\left( \frac{i t}{\hbar} \right)^2 \left[{H},{\left[{H},{x}\right]}\right] &= \frac{i \lambda t^2}{\hbar m } \left[{x^n},{p}\right]  \\ &= - \frac{n t^2 \lambda}{m} x^{n-1} \\ &= - \frac{n t^2 V}{m x} \\ \end{aligned}

The triple commutator gets no prettier, and I get

\begin{aligned}\left( \frac{i t}{\hbar} \right)^3 \left[{H},{ \left[{H},{ \left[H, x\right] }\right] }\right]&= \frac{it}{\hbar} \left[{ \frac{\mathbf{p}^2}{2m} + \lambda x^n},{ - \frac{n t^2 V}{m x} }\right] \\ &= -\frac{it}{\hbar} \frac{n t^2 }{m } \frac{\lambda}{2m} \left[{\mathbf{p}^2},{ x^{n-1}}\right] \\ &= \cdots \\ &= \frac{n(n-1)t^3 V}{ 2 m^2 x^3 } (i \hbar n + 2 p x).\end{aligned}

Putting all the pieces together this gives

\begin{aligned}x_H =e^{iH t/\hbar} x e^{-iH t/\hbar}  &= x + \frac{tp}{m} - \frac{n t^2 V}{ 2 m x} + \frac{n(n-1)t^3 V}{ 12 m^2 x^3 } (i \hbar n + 2 p x) + \cdots\end{aligned} \hspace{\stretch{1}}(2.9)

If there is a closed form for this it isn’t obvious to me. Would a fixed lower degree potential function shed any more light on this. How about the Harmonic oscillator Hamiltonian

\begin{aligned}H = \frac{p^2}{2m} + \frac{m \omega^2 }{2} x^2\end{aligned} \hspace{\stretch{1}}(2.10)

… this one works out nicely since there’s an even-odd alternation.


\begin{aligned}x_H = x \cos (\omega^2 t^2 /2) + \frac{ p t }{m} \frac{\sin( \omega^2 t^2/2)}{ \omega^2 t^2/2 }\end{aligned} \hspace{\stretch{1}}(2.11)

I’d not expect such a tidy result for an arbitrary V(x) = \lambda x^n potential.

Problem 4. Feynman-Hellman relation.

For continuously parametrized eigenstate, eigenvalue and Hamiltonian {\lvert {\psi(\lambda)} \rangle}, E(\lambda) and H(\lambda) respectively, we can relate the derivatives

\begin{aligned}\frac{\partial {}}{\partial {\lambda}} ( H {\lvert {\psi} \rangle} ) &= \frac{\partial {}}{\partial {\lambda}} ( E {\lvert {\psi} \rangle} ) \\ \implies \\ \frac{\partial {H}}{\partial {\lambda}} {\lvert {\psi} \rangle} +H \frac{\partial {{\lvert {\psi} \rangle}}}{\partial {\lambda}} &= \frac{\partial {E}}{\partial {\lambda}} {\lvert {\psi} \rangle} + E \frac{\partial {{\lvert {\psi} \rangle} }}{\partial {\lambda}} \end{aligned}

Left multiplication by {\langle {\psi} \rvert} gives

\begin{aligned}{\langle {\psi} \rvert}\frac{\partial {H}}{\partial {\lambda}} {\lvert {\psi} \rangle} +{\langle {\psi} \rvert}H \frac{\partial {{\lvert {\psi} \rangle}}}{\partial {\lambda}} &= {\langle {\psi} \rvert}\frac{\partial {E}}{\partial {\lambda}} {\lvert {\psi} \rangle} +  E {\langle {\psi} \rvert}\frac{\partial {{\lvert {\psi} \rangle} }}{\partial {\lambda}} \\ \implies \\ {\langle {\psi} \rvert}\frac{\partial {H}}{\partial {\lambda}} {\lvert {\psi} \rangle} +({\langle {\psi} \rvert}E) \frac{\partial {{\lvert {\psi} \rangle}}}{\partial {\lambda}} &= {\langle {\psi} \rvert}\frac{\partial {E}}{\partial {\lambda}} {\lvert {\psi} \rangle} +  E {\langle {\psi} \rvert}\frac{\partial {{\lvert {\psi} \rangle} }}{\partial {\lambda}} \\ \implies \\ {\langle {\psi} \rvert}\frac{\partial {H}}{\partial {\lambda}} {\lvert {\psi} \rangle} &= \frac{\partial {E}}{\partial {\lambda}} \left\langle{{\psi}} \vert {{\psi}}\right\rangle,\end{aligned}

which provides the desired identity

\begin{aligned}\frac{\partial {E}}{\partial {\lambda}} = {\langle {\psi(\lambda)} \rvert}\frac{\partial {H}}{\partial {\lambda}} {\lvert {\psi(\lambda)} \rangle}\end{aligned} \hspace{\stretch{1}}(2.12)

Problem 5.


With eigenstates {\lvert {\phi_1} \rangle} and {\lvert {\phi_2} \rangle}, of H with eigenvalues E_1 and E_2, respectively, and

\begin{aligned}{\lvert {\chi_1} \rangle} &= \frac{1}{{\sqrt{2}}}( {\lvert {\phi_1} \rangle} +{\lvert {\phi_2} \rangle}) \\ {\lvert {\chi_2} \rangle} &= \frac{1}{{\sqrt{2}}}( {\lvert {\phi_1} \rangle} -{\lvert {\phi_2} \rangle})\end{aligned}

and {\lvert {\psi(0)} \rangle} = {\lvert {\chi_1} \rangle}, determine {\lvert {\psi(t)} \rangle} in terms of {\lvert {\phi_1} \rangle} and {\lvert {\phi_2} \rangle}.


\begin{aligned}{\lvert {\psi(t)} \rangle}&= e^{-i H t /\hbar} {\lvert {\psi(0)} \rangle} \\ &= e^{-i H t /\hbar} {\lvert {\chi_1} \rangle} \\ &= \frac{1}{{\sqrt{2}}} e^{-i H t /\hbar} ( {\lvert {\phi_1} \rangle} -{\lvert {\phi_2} \rangle}) \\ &= \frac{1}{{\sqrt{2}}} (e^{-i E_1 t /\hbar} {\lvert {\phi_1} \rangle} -e^{-i E_2 t /\hbar} {\lvert {\phi_2} \rangle} )\qquad\square\end{aligned}

Problem 6.


Consider a Coulomb like potential -\lambda/r with angular momentum l=0. If the eigenfunction is

\begin{aligned}u(r) = u_0 e^{-\beta r}\end{aligned} \hspace{\stretch{1}}(2.13)

determine u_0, \beta, and the energy eigenvalue E in terms of \lambda, and m.


We can start with the normalization constant u_0 by integrating

\begin{aligned}1 &= u_0^2 \int_0^\infty dr e^{-\beta r} e^{-\beta r}  \\ &=u_0^2 \left. \frac{e^{-2 \beta r}}{-2 \beta} \right\vert_{0^\infty} \\ &= u_0^2 \frac{1}{{2\beta}} \\ \end{aligned}

\begin{aligned}u_0 &= \sqrt{2\beta}\end{aligned} \hspace{\stretch{1}}(2.14)

To go further, we need the Hamiltonian. Note that we can write the Laplacian with the angular momentum operator factored out using

\begin{aligned}\boldsymbol{\nabla}^2 &= \frac{1}{{\mathbf{x}^2}} \left( (\mathbf{x} \cdot \boldsymbol{\nabla})^2 + \mathbf{x} \cdot \boldsymbol{\nabla} + (\mathbf{x} \times \boldsymbol{\nabla})^2 \right)\end{aligned} \hspace{\stretch{1}}(2.15)

With zero for the angular momentum operator \mathbf{x} \times \boldsymbol{\nabla}, and switching to spherical coordinates, we have

\begin{aligned}\boldsymbol{\nabla}^2 &= \frac{1}{{r}} \partial_r + \frac{1}{{r}} \partial_r r \partial_r \\ &= \frac{1}{{r}} \partial_r + \frac{1}{{r}} \partial_r+ \frac{1}{{r}} r \partial_{rr} \\ &= \frac{2}{r} \partial_r + \partial_{rr} \\ \end{aligned}

We can now write the Hamiltonian for the zero angular momentum case

\begin{aligned}H&= -\frac{\hbar^2}{2m} \left( \frac{2}{r} \partial_r + \partial_{rr} \right) - \frac{\lambda}{r}\end{aligned} \hspace{\stretch{1}}(2.16)

With application of this Hamiltonian to the eigenfunction we have

\begin{aligned}E u_0 e^{-\beta r} &=\left( -\frac{\hbar^2}{2m} \left( \frac{2}{r} \partial_r + \partial_{rr} \right) - \frac{\lambda}{r} \right) u_0 e^{-\beta r}  \\ &=\left( -\frac{\hbar^2}{2m} \left( \frac{2}{r} (-\beta) + \beta^2 \right) - \frac{\lambda}{r} \right) u_0 e^{-\beta r} .\end{aligned}

In particular for r = \infty we have

\begin{aligned}-\frac{\hbar^2 \beta^2 }{2m} &= E\end{aligned} \hspace{\stretch{1}}(2.17)

\begin{aligned}-\frac{\hbar^2 \beta^2 }{2m} &= \left( -\frac{\hbar^2}{2m} \left( \frac{2}{r} (-\beta) + \beta^2 \right) - \frac{\lambda}{r} \right)  \\ \implies \\ \frac{\hbar^2}{2m} \frac{2}{r} \beta &= \frac{\lambda}{r} \end{aligned}

Collecting all the results we have

\begin{aligned}\beta &= \frac{\lambda m}{\hbar^2} \\ E &= -\frac{\lambda^2 m}{2 \hbar^2} \\ u_0 &= \frac{\sqrt{2 \lambda m}}{\hbar}\end{aligned} \hspace{\stretch{1}}(2.18)

Problem 7.


A particle in a uniform field \mathbf{E}_0. Show that the expectation value of the position operator \left\langle{\mathbf{r}}\right\rangle satisfies

\begin{aligned}m \frac{d^2 \left\langle{\mathbf{r}}\right\rangle }{dt^2} = e \mathbf{E}_0.\end{aligned} \hspace{\stretch{1}}(2.21)


This follows from Ehrehfest’s theorem once we formulate the force e \mathbf{E}_0 = -\boldsymbol{\nabla} \phi, in terms of a potential \phi. That potential is

\begin{aligned}\phi = - e \mathbf{E}_0 \cdot (x,y,z)\end{aligned} \hspace{\stretch{1}}(2.22)

The Hamiltonian is therefore

\begin{aligned}H = \frac{\mathbf{p}^2}{2m} - e \mathbf{E}_0 \cdot (x,y,z).\end{aligned} \hspace{\stretch{1}}(2.23)

Ehrehfest’s theorem gives us

\begin{aligned}\frac{d}{dt} \left\langle{{x_k}}\right\rangle &= \frac{1}{{m}} \left\langle{{p_k}}\right\rangle \\ \frac{d}{dt} \left\langle{{p_k}}\right\rangle &= -\left\langle{{ \frac{\partial {V}}{\partial {x_k}} }}\right\rangle,\end{aligned}


\begin{aligned}\frac{d^2}{dt^2} \left\langle{{x_k}}\right\rangle &= -\frac{1}{{m}} \left\langle{{ \frac{\partial {V}}{\partial {x_k}} }}\right\rangle.\end{aligned} \hspace{\stretch{1}}(2.24)

\begin{aligned}\frac{\partial {V}}{\partial {x_k}} &= - e (\mathbf{E}_0)_k\end{aligned}

Putting all the last bits together, and summing over the directions \mathbf{e}_k we have

\begin{aligned}m \frac{d^2}{dt^2} \mathbf{e}_k \left\langle{{x_k}}\right\rangle = \mathbf{e}_k \left\langle{{ e (\mathbf{E}_0)_k }}\right\rangle= e \mathbf{E}_0\qquad\square\end{aligned}

Problem 8.


For Hamiltonian eigenstates {\lvert {E_n} \rangle}, C = AB, A = \left[{B},{H}\right], obtain the matrix element {\langle {E_m} \rvert} C {\lvert {E_n} \rangle} in terms of the matrix element of A.


I was able to get most of what was asked for here, with a small exception. I started with the matrix element for A, which is

\begin{aligned}{\langle {E_m} \rvert} A {\lvert {E_n} \rangle}={\langle {E_m} \rvert} BH - HB {\lvert {E_n} \rangle} =(E_n - E_m){\langle {E_m} \rvert} B {\lvert {E_n} \rangle} \end{aligned} \hspace{\stretch{1}}(2.25)

Next, computing the matrix element for C we have

\begin{aligned}{\langle {E_m} \rvert} C {\lvert {E_n} \rangle}&={\langle {E_m} \rvert} BHB - HB^2 {\lvert {E_n} \rangle} \\ &=\sum_a {\langle {E_m} \rvert} BH {\lvert {E_a} \rangle}{\langle {E_a} \rvert} B {\lvert {E_n} \rangle} - E_m {\langle {E_m} \rvert} B {\lvert {E_a} \rangle} {\langle {E_a} \rvert} B {\lvert {E_n} \rangle} \\ &=\sum_a E_a {\langle {E_m} \rvert} B {\lvert {E_a} \rangle}{\langle {E_a} \rvert} B {\lvert {E_n} \rangle} -E_m {\langle {E_m} \rvert} B {\lvert {E_a} \rangle} {\langle {E_a} \rvert} B {\lvert {E_n} \rangle} \\ &=\sum_a (E_a - E_m){\langle {E_m} \rvert} B {\lvert {E_a} \rangle}{\langle {E_a} \rvert} B {\lvert {E_n} \rangle} \\ &=\sum_a {\langle {E_m} \rvert} A {\lvert {E_a} \rangle} {\langle {E_a} \rvert} B {\lvert {E_n} \rangle} \\ &={\langle {E_m} \rvert} A {\lvert {E_n} \rangle} {\langle {E_n} \rvert} B {\lvert {E_n} \rangle} +\sum_{a \ne n} {\langle {E_m} \rvert} A {\lvert {E_a} \rangle} {\langle {E_a} \rvert} B {\lvert {E_n} \rangle} \\ &={\langle {E_m} \rvert} A {\lvert {E_n} \rangle} {\langle {E_n} \rvert} B {\lvert {E_n} \rangle} +\sum_{a \ne n} {\langle {E_m} \rvert} A {\lvert {E_a} \rangle} \frac{{\langle {E_a} \rvert} A {\lvert {E_n} \rangle}}{E_n - E_a}\end{aligned}

Except for the {\langle {E_n} \rvert} B {\lvert {E_n} \rangle} part of this expression, the problem as stated is complete. The relationship 2.25 is no help for with n = m, so I see no choice but to leave that small part of the expansion in terms of B.

Problem 9.


Operator A has eigenstates {\lvert {a_i} \rangle}, with a unitary change of basis operation U {\lvert {a_i} \rangle} = {\lvert {b_i} \rangle}. Determine in terms of U, and A the operator B and its eigenvalues for which {\lvert {b_i} \rangle} are eigenstates.


Consider for motivation the matrix element of A in terms of {\lvert {b_i} \rangle}. We will also let A {\lvert {a_i} \rangle} = \alpha_i {\lvert {a_i} \rangle}. We then have

\begin{aligned}{\langle {a_i} \rvert} A {\lvert {a_j} \rangle}&={\langle {b_i} \rvert} U A U^\dagger {\lvert {b_j} \rangle} \\ \end{aligned}

We also have

\begin{aligned}{\langle {a_i} \rvert} A {\lvert {a_j} \rangle}&=a_j {\langle {a_i} \rvert} {\lvert {a_j} \rangle} \\ &=a_j \delta_{ij}\end{aligned}

So it appears that the operator U A U^\dagger has the orthonormality relation required. In terms of action on the basis \{{\lvert {b_i} \rangle}\}, let’s see how it behaves. We have

\begin{aligned}U A U^\dagger {\lvert {b_i} \rangle}&= U A {\lvert {a_i} \rangle} \\ &= U \alpha_i {\lvert {a_i} \rangle} \\ &= \alpha_i {\lvert {b_i} \rangle} \\ \end{aligned}

So we see that the operators A and B = U A U^\dagger have common eigenvalues.

Problem 10.


With H {\lvert {n} \rangle} = E_n {\lvert {n} \rangle}, A = \left[{H},{F}\right] and {\langle {0} \rvert} F {\lvert {0} \rangle} = 0, show that

\begin{aligned}\sum_{n\ne 0} \frac{{\langle {0} \rvert} A {\lvert {n} \rangle} {\langle {n} \rvert} A {\lvert {0} \rangle} }{E_n - E_0} = {\langle {0} \rvert} AF {\lvert {0} \rangle}\end{aligned} \hspace{\stretch{1}}(2.26)


\begin{aligned}{\langle {0} \rvert} AF {\lvert {0} \rangle}&={\langle {0} \rvert} HF F - FH F{\lvert {0} \rangle} \\ &=\sum_n E_0 {\langle {0} \rvert} F {\lvert {n} \rangle}{\langle {n} \rvert} F {\lvert {0} \rangle} - E_n {\langle {0} \rvert} F {\lvert {n} \rangle} {\langle {n} \rvert} F{\lvert {0} \rangle} \\ &=\sum_n (E_0 -E_n) {\langle {0} \rvert} F {\lvert {n} \rangle}{\langle {n} \rvert} F {\lvert {0} \rangle}  \\ &=\sum_{n\ne0} (E_0 -E_n) {\langle {0} \rvert} F {\lvert {n} \rangle}{\langle {n} \rvert} F {\lvert {0} \rangle}  \\ \end{aligned}

We also have

\begin{aligned}{\langle {0} \rvert} A {\lvert {n} \rangle} {\langle {n} \rvert} A {\lvert {0} \rangle}&={\langle {0} \rvert} HF -F H {\lvert {n} \rangle} {\langle {n} \rvert} A {\lvert {0} \rangle} \\ &=(E_0 - E_n) {\langle {0} \rvert} F {\lvert {n} \rangle} {\langle {n} \rvert} HF - FH {\lvert {0} \rangle} \\ &=-(E_0 - E_n)^2 {\langle {0} \rvert} F {\lvert {n} \rangle} {\langle {n} \rvert} F {\lvert {0} \rangle} \\ \end{aligned}

Or, for n \ne 0,

\begin{aligned}{\langle {0} \rvert} F {\lvert {n} \rangle} {\langle {n} \rvert} F {\lvert {0} \rangle} &=-\frac{{\langle {0} \rvert} A {\lvert {n} \rangle} {\langle {n} \rvert} A {\lvert {0} \rangle}}{(E_0 - E_n)^2 }.\end{aligned}

This gives

\begin{aligned}{\langle {0} \rvert} AF {\lvert {0} \rangle}&=-\sum_{n\ne0} (E_0 -E_n) \frac{{\langle {0} \rvert} A {\lvert {n} \rangle} {\langle {n} \rvert} A {\lvert {0} \rangle}}{(E_0 - E_n)^2 } \\ &=\sum_{n\ne0} \frac{{\langle {0} \rvert} A {\lvert {n} \rangle} {\langle {n} \rvert} A {\lvert {0} \rangle}}{E_n - E_0 } \qquad\square\end{aligned}

Problem 11. commutator of angular momentum with Hamiltonian.

Show that \left[{\mathbf{L}},{H}\right] = 0, where H = \mathbf{p}^2/2m + V(r).

This follows by considering \left[{\mathbf{L}},{\mathbf{p}^2}\right], and \left[{\mathbf{L}},{V(r)}\right]. Let

\begin{aligned}L_{jk} = x_j p_k - x_k p_j,\end{aligned} \hspace{\stretch{1}}(2.27)

so that

\begin{aligned}\mathbf{L} = \mathbf{e}_i \epsilon_{ijk} L_{jk}.\end{aligned} \hspace{\stretch{1}}(2.28)

We now need to consider the commutators of the operators L_{jk} with \mathbf{p}^2 and V(r).

Let’s start with p^2. In particular

\begin{aligned}\mathbf{p}^2 x_m p_n&=p_k p_k x_m p_n \\ &=p_k (p_k x_m) p_n \\ &=p_k (-i\hbar \delta_{km} + x_m p_k) p_n \\ &=-i\hbar p_m p_n + (p_k x_m) p_k p_n \\ &=-i\hbar p_m p_n + (-i \hbar \delta_{km} + x_m p_k ) p_k p_n \\ &=-2 i\hbar p_m p_n + x_m p_n \mathbf{p}^2.\end{aligned}

So our commutator with \mathbf{p}^2 is

\begin{aligned}\left[{L_{jk}},{\mathbf{p}^2}\right]&=(x_j p_k - x_j p_k) \mathbf{p}^2 -( -2 i\hbar p_j p_k + x_j p_k \mathbf{p}^2 +2 i\hbar p_k p_j - x_k p_j \mathbf{p}^2 ).\end{aligned}

Since p_j p_k = p_k p_j, all terms cancel out, and the problem is reduced to showing that

\begin{aligned}\left[{\mathbf{L}},{H}\right] &= \left[{\mathbf{L}},{V(r)}\right] = 0.\end{aligned}

Now assume that V(r) has a series representation

\begin{aligned}V(r) &= \sum_j a_j r^j = \sum_j a_j (x_k x_k)^{j/2}\end{aligned}

We’d like to consider the action of x_m p_n on this function

\begin{aligned}x_m p_n V(r) \Psi&= -i \hbar x_m \sum_j a_j \partial_n (x_k x_k)^{j/2} \Psi \\ &= -i \hbar x_m \sum_j a_j (j x_n (x_k x_k)^{j/2-1} + r^j \partial_n \Psi) \\ &= -\frac{i \hbar x_m x_n}{r^2} \sum_j a_j j r^j + x_m V(r) p_n \Psi\end{aligned}

\begin{aligned}L_{mn} V(r) &=(x_m p_n - x_n p_m) V(r) \\ &= -\frac{i \hbar x_m x_n}{r^2} \sum_j a_j j r^j+\frac{i \hbar x_n x_m}{r^2} \sum_j a_j j r^j + V(r) (x_m p_n - x_n p_m )\\ &= V(r) L_{mn}\end{aligned}

Thus \left[{L_{mn}},{V(r)}\right] = 0 as expected, implying \left[{\mathbf{L}},{H}\right] = 0.


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


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