- #1
Peeter
- 305
- 3
Homework Statement
For
[tex]\begin{align*}H = \frac{\mathbf{p}^2}{2m} + V(\mathbf{r})\end{align*} [/tex]
use the properties of the double commutator [itex]\left[{\left[{H},{e^{i \mathbf{k} \cdot \mathbf{r}}}\right]},{e^{-i \mathbf{k} \cdot \mathbf{r}}}\right][/itex] to obtain
[tex]\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*} [/tex]
Homework Equations
Above.
The Attempt at a Solution
For the commutators I get:
[tex]\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*} [/tex]
and
[tex]\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*} [/tex]
I have also reduced the energy expression as follows to something just involving the expectation of [itex]\mathbf{k} \cdot \mathbf{p}[/itex]
[tex]\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*} [/tex]
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
[tex]\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*} [/tex]
but if I take this further I just get
[tex]\begin{align*}-\frac{1}{{m}} (\hbar \mathbf{k})^2 = -\frac{1}{{m}} (\hbar \mathbf{k})^2 \end{align*} [/tex]
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?