I'm determining the density operator in momentum space and by working out the Fourier transform from the coordinate representation I get to:

$$\hat n_q=\sum_{kk'ss'}\left<k,s|e^{-iq\hat r}|k's'\right>a^\dagger_{ks}a_{k's'}$$

How does the exponential defined above shifts the momenta according to: $$\hat n_q=\sum_{kk'ss'}\delta_{ss'}\left<k-q|k'\right>a^\dagger_{ks}a_{k's'}$$ How did the exponential act on the bra and shifted the momenta?

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    $\begingroup$ Hint : If $\hat{p}$ is the generator of translations for eigenstates of $\hat{r}$, then what do you think $\hat{r}$ does to eigenstates of $\hat{p}$? $\endgroup$
    – insomniac
    Mar 29, 2020 at 17:25

2 Answers 2


Momentum generates translations in position, which means that $e^{i \hat{q} r}$ shifts position by $r$. And this is completely symmetric: position generates translations in momentum, which means that $e^{i q \hat{r}}$ shifts momentum by $-q$.

To see this explicitly, note that $$\hat{r} = i \frac{\partial}{\partial p}$$ and so, expanding in a Taylor series, $$e^{i q \hat{r}} \tilde{\psi}(k) = \sum_{n=0}^\infty \frac{(iq \hat{r})^n}{n!} \tilde{\psi}(k) = \sum_{n=0}^\infty \frac{(-q)^n}{n!} \tilde{\psi}^{(n)}(k) = \tilde{\psi}(k-q)$$ where the last step follows from the definition of Taylor series. So for momentum eigenstates, $$e^{i q \hat{r}} |k \rangle = |k-q \rangle.$$


One way to approach it is by direct calculation (in the position representation): $$\langle k |e^{-iqx}|k'\rangle = \int_{-\infty}^{+\infty}dx \frac{1}{\sqrt{2\pi}}e^{-ikx} e^{-iqx} \frac{1}{\sqrt{2\pi}}e^{ik'x} = \delta (k+q-k') = \int_{-\infty}^{+\infty}dx \frac{1}{\sqrt{2\pi}}e^{-i(k+q)x} \frac{1}{\sqrt{2\pi}}e^{ik'x} = \langle k+q |k'\rangle.$$ (Note that I get a different sign - this may have to do with how the momentum eigenstates are defined in your book.)

A more elegant and convincing approach is to think of it in the momentum representation, where the position operator corresponds to differentiating in respect to momentum: $$\hat{x} = i\hbar\frac{d}{dp} = i\frac{d}{dk},$$ ($p = \hbar k$) that is $$e^{-iq\hat{x}} = e^{q\frac{d}{dk}},$$ which is a shift operator for $k$-dependent functions: $$ e^{q\frac{d}{dk}}f(k) = \sum_{n=0}^{n=+\infty}\frac{1}{n!}\frac{d^n}{dk^n}f(k) = f(k+q).$$


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