homework-and-exercisesmomentumoperatorsquantum mechanics
I know that
$$\tag{1}\hat{p}~=~-i\hbar \frac{\partial}{\partial x}~.$$
How can I get
$$\tag{2}\hat{x}~=~i\hbar \frac{\partial}{\partial p}~?$$
I think this simple and I'm just over thinking it, but knowing the momentum operator (1), how can I get the position operator (2)?
1) Notice that by inserting a complete set of position states we can write $$ \hat p \psi(x) = \langle x|\hat p|\psi\rangle = \int dx'\langle x|\hat p|x'\rangle\langle x'|\psi\rangle =\int dx'\langle x|\hat p|x'\rangle \psi(x') $$ so if we set $$ \langle x|\hat p|x'\rangle = -i\hbar \frac{\partial}{\partial x}\delta(x-x') =i\hbar \frac{\partial}{\partial x'}\delta(x-x') $$ then we can use integration by parts to obtain $$ \hat p \psi(x) =i\hbar \int dx'\frac{\partial}{\partial x'}\delta(x-x') \psi(x') = -i\hbar \int dx'\delta(x-x') \frac{d \psi}{dx'}(x') = -i\hbar \frac{d\psi}{dx}(x) $$ So your expression is correct. The derivative of a delta function is essentially defined by the integration by parts manipulation that I just performed; in fact derivatives of distributions in general are defined in an analogous way. See this lecture for example.
Hope that helps; let me know of any typos!
Cheers!
There is a heuristic way to look at this.
The Dirac delta function corresponds to a spike when its argument is zero. You can view it as the limit of a sequence of Gaussian functions whose areas are all one but whose width goes to zero. The derivative of a Gaussian function looks like this:
So in the limit, the derivative of the Dirac function is something like an up spike infinitesimally to the left of the origin followed by a down spike infinitesimally to the right. So the matrix elements you're looking at aren't actually diagonal, they're infinitesimally off-diagonal.
These kinds of heuristics can be useful, but they can also be dangerous, so don't take what I say too literally.
Update: Another way to look at this is to approach the derivative Dirac delta via a discretisation. If the wavefunction is represented by a vector of equally spaced samples, the derivative can be represented by central differences. Assuming periodic boundary conditions we get a matrix like:
$\frac{1}{2}\pmatrix{ 0 & 1 & 0 & 0 & \ldots & -1 \\ -1 & 0 & 1 & 0 & \ldots & 0 \\ 0 & -1 & 0 & 1 & \ldots & 0 \\ 0 & 0 & -1 & 0 & \ldots & 0 \\ & & \vdots \\ 1 & 0 & 0 & 0 & \ldots & 0 \\ }$
We have 1's just above the diagonal and -1's just below. As the discretisation gets finer we get a matrix where they entries are more and more concentrated near the diagonal even though all of the non-zero terms are actually off-diagonal. In the limit you can again imagine something that is infinitesimally off-diagonal.
Best Answer
Here's how you do it.
First, notice that for any state $|\psi\rangle$, we have \begin{align} \langle p|[\hat x, \hat p]|\psi\rangle &= \langle p|\hat x\hat p-\hat p\hat x|\psi\rangle \\ &= \langle p|\hat x\hat p|\psi\rangle - \langle p|\hat p \hat x|\psi\rangle \\ &= \langle p|\hat x \hat p|\psi\rangle - p\langle p|\hat x|\psi\rangle \end{align} Now, recall the canonical commutation relation between $\hat x$ and $\hat p$; \begin{align} [\hat x, \hat p] &= i\hbar \hat I \end{align} where $\hat I$ is the identity operator. Using this fact on the left hand side of the manipulation we just performed, and doing a little rearranging, we find that \begin{align} p\langle p|\hat x|\psi\rangle = \langle p|\hat x \hat p|\psi\rangle -i\hbar\langle p|\psi\rangle. \tag{1} \end{align} Now, focus on the first term on the right. we have \begin{align} \langle p|\hat x \hat p|\psi\rangle &= \int dx\,\langle p|\hat x|x\rangle\langle x|\hat p|\psi\rangle \\ &= \int dx\, x e^{ipx/\hbar}\left[-i\hbar \frac{d}{dx}\psi(x)\right] \\ &= i\hbar \int dx\, \frac{d}{dx}(xe^{ipx/\hbar})\psi(x) \\ &= i\hbar \int dx\, e^{ipx/\hbar}\psi(x) + i\hbar\int dx\,x\left(\frac{ip}{\hbar}\right)e^{ipx/\hbar}\psi(x) \\ &= i\hbar\psi(p) -p\int dx\,xe^{ipx/\hbar}\psi(x)\\ &= i\hbar\psi(p)+i\hbar p\frac{d}{dp}\int dx\,e^{ipx/\hbar}\psi(x) \\ &= i\hbar\psi(p)+i\hbar p \frac{d}{dp}\psi(p) \tag{2} \end{align} where we have made liberal use of integration by parts, and the following identities: \begin{align} \int dx\, |x\rangle\langle x| = \hat I, \qquad \psi(x) \overset{\mathrm{def}}{=} \langle x|\psi\rangle, \qquad \psi(p) \overset{\mathrm{def}}{=} \langle p|\psi\rangle,\qquad \langle x|\hat p|\psi\rangle = -i\hbar\frac{d}{dx}\psi(x). \end{align} The last fact is precisely the statement that the position space representation of the momentum operator is $-i\hbar d/dx$. Now combine $(1)$ and $(2)$ to obtain \begin{align} p\langle p|\hat x|\psi\rangle = i\hbar p\frac{d}{dp}\psi(p) \tag{3} \end{align} Next, we simply note that if we denote the momentum space representation of the position operator as $D^{(p)}(\hat x)$, then by definition \begin{align} D^{(p)}(\hat x)\psi(p) = \langle p|\hat x|\psi\rangle, \end{align} Combining this with $(3)$, and noting that the resulting equation should hold for all $p$, and dividing both sides by $p$, we obtain the desired result; \begin{align} D^{(p)}(\hat x) = i\hbar \frac{d}{dp} \end{align}