From my professor's notes on statistical mechanics. $\left|\bf{k}\right\rangle$ is eigenstate of the hamiltonian of the free particle with periodic boundary conditions: $$ \left\langle{\bf r}|{\bf k}\right\rangle = \frac{1}{\sqrt{V}} e^{i\bf{k}\cdot\bf{r}}. $$

The states $\left|\bf{k}\right\rangle$ are eigenstates of the kinetic energy: $T_1|{\bf k}\rangle = (\hbar^2 k^2/2m)\left|\bf{k}\right\rangle$. Thus, the single-particle kinetic energy is diagonal in momentum space $$ \langle{\bf k'} | T_1 | {\bf k}\rangle = \delta_{{\bf k},{\bf k'}}\frac{\hbar^2 k^2}{2m}. $$

One gets \begin{align} \delta_{{\bf k},{\bf k'}}\frac{\hbar^2 k^2}{2m} &= \int d{\bf r'} \int d{\bf r} \langle{\bf k'}|{\bf r'}\rangle \langle{\bf r'} | T_1 | {\bf r}\rangle \langle{\bf r}|{\bf k}\rangle \\ &= \int d{\bf r'} \int d{\bf r} \frac{1}{\sqrt{V}} e^{-i\bf{k'}\cdot\bf{r'}} \langle{\bf r'} | T_1 | {\bf r}\rangle \frac{1}{\sqrt{V}} e^{i\bf{k}\cdot\bf{r}} \end{align} from which we have the well-known coordinate representation of the kinetic-energy operator $$ \langle{\bf r'} | T_1 | {\bf r}\rangle = \left(\frac{-\hbar^2 \nabla^2}{2m}\right)\delta({\bf r}-{\bf r'}). $$

What I don't understand is how it concludes the very last equation from the one preceeding it. I could verify it is right by substituting it back, but I'm interested in another explanation and more importantly its uniqueness. I think there is a trivial way to say that, but I couldn't find it. Any help is greatly appreciated.


2 Answers 2


I've relabeled $\bf r,\, r'$ from the initial integral to:

$$ \delta_{\bf k,k'} \frac{\hbar^2k^2}{2m} = \frac{1}{V}\int\mathrm{d}{\bf r'_1}\int\mathrm{d}{\bf r_1}e^{-i\bf k' \cdot r'_1}\langle{\bf r'_1}|T_1|{\bf r_1}\rangle e^{i\bf k \cdot r_1} $$

As ZachMcDargh mentions in his answer, the following identity is useful:

$$ \sum_{\bf k} e^{-i\bf k \cdot r} = V \delta(\bf r) $$

Taking two (discrete) inverse Fourier transforms on both sides:

\begin{align} \sum_{\bf k'}\sum_{\bf k}e^{i\bf k' \cdot r'}\delta_{\bf k,k'} \frac{\hbar^2k^2}{2m}e^{-i\bf k \cdot r} & = \frac{1}{V}\sum_{\bf k'}\sum_{\bf k}\int\mathrm{d}{\bf r'_1}\int\mathrm{d}{\bf r_1}\;e^{-i\bf k' \cdot (r'_1 - r')}\langle{\bf r'_1}|T_1|{\bf r_1}\rangle e^{i\bf k \cdot (r_1 - r)} \\ & = V\int\mathrm{d}{\bf r'_1}\int\mathrm{d}{\bf r_1}\;\delta({\bf r_1' - r'})\delta({\bf r_1 - r})\langle{\bf r'_1}|T_1|{\bf r_1}\rangle \\ \implies \sum_{\bf k} \frac{1}{V}\frac{\hbar^2k^2}{2m}e^{-i\bf k \cdot (r-r')} & = \langle{\bf r'}|T_1|{\bf r}\rangle \end{align}

Evaluate the sum on the LHS in Cartesian coordinates:

$$ \implies \langle{\bf r'}|T_1|{\bf r}\rangle = \frac{\hbar^2}{2mV}\sum_{\bf k} [k_x^2 + k_y^2 + k_z^2]\;e^{-i\bf k \cdot (r-r')}$$

Note that (via partial differentiation under the integral sign, as long as $\bf r$ is differentiable):

$$ \sum_{\bf k}\; k_m^2 \;e^{-i\bf k \cdot (r-r')} = -\frac{\partial^2}{\partial m^2}\sum_{\bf k}\;e^{-i\bf k \cdot (r-r')},\quad m = x,\,y,\,z$$


\begin{align} \langle{\bf r'}|T_1|{\bf r}\rangle & = -\frac{\hbar^2}{2mV}\sum_{\bf k} \; \left[\frac{\partial^2}{\partial x^2} + \frac{\partial^2}{\partial y^2} + \frac{\partial^2}{\partial x^2}\right]e^{-i\bf k \cdot (r-r')} \\ & = -\frac{\hbar^2 \nabla^2}{2mV}\sum_{\bf k}\;e^{-i\bf k \cdot (r-r')} = -\frac{\hbar^2 \nabla^2}{2m} \delta(\bf r - r') \end{align}

The same as the result noted by AccidentalFourierTransform (not so accidental in this answer, heh).

Unfortunately I don't know distribution theory, so I can't prove uniqueness yet.

  • $\begingroup$ Neither the plane wave nor the Dirac delta are square integrable functions, so the justification of uniqueness is wrong. One should instead use distribution theory. $\endgroup$
    – Adam
    Sep 20, 2018 at 20:03
  • $\begingroup$ Just a note, the $k$ form a discrete set ($k \propto n$ with $n \in Z^3$). Does this change much in your derivation? $\endgroup$
    – pp.ch.te
    Sep 26, 2018 at 10:15
  • $\begingroup$ I've updated the answer for the discrete case. As Adam noted, I was incorrect about the justification of uniqueness before, which is currently reflected in my answer. $\endgroup$ Sep 27, 2018 at 9:17

In order to derive the last equation from the one before it, multiply both sides by $e^{-i \boldsymbol{k}\cdot \boldsymbol{r}''}e^{i \boldsymbol{k}'\cdot \boldsymbol{r}'''}$, and sum over both $\boldsymbol{k}$ and $\boldsymbol{k}'$, i.e. perform inverse Fourier transforms. Using the identity $$ \sum_{\boldsymbol{k}} e^{i \boldsymbol{k}\cdot\boldsymbol{r}}= V\delta(\boldsymbol{r}) $$ you'll quickly see that $$ \langle \boldsymbol{r}' | T_1 | \boldsymbol{r}\rangle = \sum_{\boldsymbol{k}} \frac{1}{V} e^{-i \boldsymbol{k} \cdot(\boldsymbol{r} - \boldsymbol{r}')} \frac{\hbar^2 k^2}{2 m}. $$ Then we just identify the Fourier transform of $k^2$ as $-V\delta(\boldsymbol{r}-\boldsymbol{r}')\nabla^2$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.