What is the significance of the double commutator in energy relations?

[Peeter]

Homework Statement

For
\begin{align*}H = \frac{\mathbf{p}^2}{2m} + V(\mathbf{r})\end{align*}

use the properties of the double commutator \left[{\left[{H},{e^{i \mathbf{k} \cdot \mathbf{r}}}\right]},{e^{-i \mathbf{k} \cdot \mathbf{r}}}\right] to obtain

\begin{align*}\sum_n (E_n - E_s) {\left\lvert{{\langle {n} \rvert} e^{i\mathbf{k} \cdot \mathbf{r}} {\lvert {s} \rangle}}\right\rvert}^2\end{align*}

Homework Equations

Above.

The Attempt at a Solution

For the commutators I get:

\begin{align*}\left[{H},{ e^{i \mathbf{k} \cdot \mathbf{r}}}\right]&= \frac{1}{{2m}}e^{i \mathbf{k}\cdot \mathbf{r}} \left((\hbar\mathbf{k})^2 + 2 (\hbar \mathbf{k}) \cdot \mathbf{p} \right)\end{align*}

and
\begin{align*}\left[{\left[{H},{e^{i \mathbf{k} \cdot \mathbf{r}}}\right]},{e^{-i \mathbf{k} \cdot \mathbf{r}}}\right]&=- \frac{1}{{m}} (\hbar \mathbf{k})^2 \end{align*}

I have also reduced the energy expression as follows to something just involving the expectation of \mathbf{k} \cdot \mathbf{p}

\begin{align*}\sum_n (E_n - E_s) {\left\lvert{{\langle {n} \rvert} e^{i\mathbf{k} \cdot \mathbf{r}} {\lvert {s} \rangle}}\right\rvert}^2&=\sum_n (E_n - E_s) {\langle {s} \rvert} e^{-i\mathbf{k} \cdot \mathbf{r}} {\lvert {n} \rangle} {\langle {n} \rvert} e^{i\mathbf{k} \cdot \mathbf{r}} {\lvert {s} \rangle} \\ &=\sum_n {\langle {s} \rvert} e^{-i\mathbf{k} \cdot \mathbf{r}} H {\lvert {n} \rangle} {\langle {n} \rvert} e^{i\mathbf{k} \cdot \mathbf{r}} {\lvert {s} \rangle} -{\langle {s} \rvert} e^{-i\mathbf{k} \cdot \mathbf{r}} {\lvert {n} \rangle} {\langle {n} \rvert} e^{i\mathbf{k} \cdot \mathbf{r}} H {\lvert {s} \rangle} \\ &={\langle {s} \rvert} e^{-i\mathbf{k} \cdot \mathbf{r}} H e^{i\mathbf{k} \cdot \mathbf{r}} {\lvert {s} \rangle} -{\langle {s} \rvert} e^{-i\mathbf{k} \cdot \mathbf{r}} e^{i\mathbf{k} \cdot \mathbf{r}} H {\lvert {s} \rangle} \\ &={\langle {s} \rvert} e^{-i\mathbf{k} \cdot \mathbf{r}} \left[{H},{e^{i\mathbf{k} \cdot \mathbf{r}}}\right] {\lvert {s} \rangle} \\ &=\frac{1}{{2m}} {\langle {s} \rvert} e^{-i\mathbf{k} \cdot \mathbf{r}} e^{i \mathbf{k}\cdot \mathbf{r}} \left((\hbar\mathbf{k})^2 + 2 (\hbar \mathbf{k}) \cdot \mathbf{p} \right){\lvert {s} \rangle} \\ &=\frac{1}{{2m}} {\langle {s} \rvert} (\hbar\mathbf{k})^2 + 2 (\hbar \mathbf{k}) \cdot \mathbf{p} {\lvert {s} \rangle} \\ &=\frac{(\hbar\mathbf{k})^2}{2m} + \frac{1}{{m}} {\langle {s} \rvert} (\hbar \mathbf{k}) \cdot \mathbf{p} {\lvert {s} \rangle} \\ \end{align*}

I figure there is some trick to evaluating that last expectation value related to the double commutator, so expanding the expectation of that seems appropriate

\begin{align*}-\frac{1}{{m}} (\hbar \mathbf{k})^2 &={\langle {s} \rvert} \left[{\left[{H},{e^{i \mathbf{k} \cdot \mathbf{r}}}\right]},{e^{-i \mathbf{k} \cdot \mathbf{r}}}\right] {\lvert {s} \rangle} \\ &={\langle {s} \rvert} \left[{H},{e^{i \mathbf{k} \cdot \mathbf{r}}}\right] e^{-i \mathbf{k} \cdot \mathbf{r}}-e^{-i \mathbf{k} \cdot \mathbf{r}}\left[{H},{e^{i \mathbf{k} \cdot \mathbf{r}}}\right] {\lvert {s} \rangle} \\ &=\frac{1}{{2m }} {\langle {s} \rvert} e^{ i \mathbf{k} \cdot \mathbf{r}} ( (\hbar \mathbf{k})^2 + 2 \hbar \mathbf{k} \cdot \mathbf{p}) e^{-i \mathbf{k} \cdot \mathbf{r}}-e^{-i \mathbf{k} \cdot \mathbf{r}} e^{ i \mathbf{k} \cdot \mathbf{r}} ( (\hbar \mathbf{k})^2 + 2 \hbar \mathbf{k} \cdot \mathbf{p}) {\lvert {s} \rangle} \\ &=\frac{1}{{m}} {\langle {s} \rvert} e^{ i \mathbf{k} \cdot \mathbf{r}} (\hbar \mathbf{k} \cdot \mathbf{p}) e^{-i \mathbf{k} \cdot \mathbf{r}}-\hbar \mathbf{k} \cdot \mathbf{p} {\lvert {s} \rangle} \end{align*}

but if I take this further I just get

\begin{align*}-\frac{1}{{m}} (\hbar \mathbf{k})^2 = -\frac{1}{{m}} (\hbar \mathbf{k})^2 \end{align*}

which isn't very helpful. I don't actually like the approach I've used, where I took the magic expression and blundered through it attempting to get the desired answer.

Can anybody supply a tip that uses a more fundamental principle, where after a natural sequence of steps would arrive and the desired end result (instead of fluking upon it by lucky algebraic manipulation). I'd also settle for the trick as a last resort, or a hint of what it could be.

Also, does this energy relationship have a name?

 

[betel]

I assume your bold letters represent vectors and not neccessarily operators? i.e. k is not an operator?

 

[Peeter]

Yes:

\mathbf{k} \cdot \mathbf{r} = k_x x + k_y y + k_z z

and all the k_n's are just numbers, not operators.

 

[betel]

In the third line you already have the double commutator.
Try writing the double commutator explicitly to see what you need to get. The last term in your line is almost correct if you combine the exponentials. In the first you only need some justification to swap the minus to the right (I can give another hint there if you want)

 

[Peeter]

Using

\hbar \mathbf{k} \cdot \mathbf{p} e^{-i \mathbf{k} \cdot \mathbf{r}} = e^{-i \mathbf{k} \cdot \mathbf{r}} \left( -(\hbar \mathbf{k})^2 + \mathbf{k} \cdot \mathbf{p} \right)

I can try your suggestion (as I interpretted it) of swapping the exponentials in the first term of the last line:
\begin{align*}\frac{1}{{m}} {\langle {s} \rvert} e^{ i \mathbf{k} \cdot \mathbf{r}} (\hbar \mathbf{k} \cdot \mathbf{p}) e^{-i \mathbf{k} \cdot \mathbf{r}}-\hbar \mathbf{k} \cdot \mathbf{p}{\lvert {s} \rangle} \\ &=\frac{1}{{m}} {\langle {s} \rvert} (-(\hbar \mathbf{k})^2 + \hbar \mathbf{k} \cdot \mathbf{p})-\hbar \mathbf{k} \cdot \mathbf{p}{\lvert {s} \rangle} \\ &=\frac{1}{{m}} {\langle {s} \rvert} -(\hbar \mathbf{k})^2 {\lvert {s} \rangle}\end{align*}

But, like I said, this just takes me full circle.

 

[betel]

The last part of your calculation is just going full circle. You don't have to use your "trick". You can transform your following line to get the double comm.
<br /> \begin{align*}\sum_n (E_n - E_s) {\left\lvert{{\langle {n} \rvert} e^{i\mathbf{k} \cdot \mathbf{r}} {\lvert {s} \rangle}}\right\rvert}^2&amp;={\langle {s} \rvert} e^{-i\mathbf{k} \cdot \mathbf{r}} H e^{i\mathbf{k} \cdot \mathbf{r}} {\lvert {s} \rangle} -{\langle {s} \rvert} e^{-i\mathbf{k} \cdot \mathbf{r}} e^{i\mathbf{k} \cdot \mathbf{r}} H {\lvert {s} \rangle} \end{align*} <br />
Now you should write the double commutator with operators, i.e. without using any commutation relations and compare with this line. Then you should see what you need to transform.

 

[Peeter]

Thanks a lot. I've got it now. My problem was not knowing how the two sign variants of these Hamiltonian exponential sandwiches compared:

\begin{align*}e^{-i\mathbf{k} \cdot\mathbf{r}} &amp;H e^{i\mathbf{k} \cdot\mathbf{r}} \\ e^{i\mathbf{k} \cdot\mathbf{r}} &amp;H e^{-i\mathbf{k} \cdot\mathbf{r}} \\ \end{align*}

Expanding them out (the dumb and simple way) with \cos + i\sin I see that they have the same real part. Since they are both also Hermitian, the expectation value of each is real and only these (equal) real parts can contribute to the respective expectation values. That's enough to equate (up to a constant factor of two) the expectation of the double commutator and this (unnamed?) weighted energy difference sum expression.