I start by showing how I tried to obtain the position operator analogously to how the momentum operator is obtained:

If we differentiate the wave function in one dimension $\Psi(x,t) = e^{i(px-Et)/\hbar}$, with respect to x:

$$ \frac{\partial}{\partial x}\Psi(x,t) = \frac{ip}{\hbar}~\Psi(x,t) \tag1 $$

from which we get the momentum operator: $\hat{p}=-i\hbar\frac{\partial}{\partial x}$

But suppose I differentiate $\Psi$ with respect to momentum:

$$ \frac{\partial}{\partial p}\Psi(x,t) = \frac{i}{\hbar}(x-\frac{p}{m}t)~\Psi(x,t) \tag2 $$

which gives $x-\frac{p}{m}t = -i\hbar\frac{\partial}{\partial p}$

Now, setting $t=0$, I suppose we would get the operator $$\hat x_0 = -i\hbar\frac{\partial}{\partial p}$$ Which would be something like an operator for initial position. But it doesn't look right, because I know that the position operator does not have a minus sign. And is there even such a thing as "initial position operator"?

So, what is wrong with this? The reason I'm asking is because I want to show that the expectation value for position satisfies the following relation (in accordance with the correspondance principle):

$$\left<x\right> = \left<x\right>_{t=t_0}+\frac{\left<p\right>}{m}(t-t_0)$$

and it was given as a hint to start as in $(2)$ and take it from there. I sort of know what to do, but the minus sign in the position operator confuses me.

The hint also suggests $(2)$ should lead to $x = i\hbar\frac{\partial}{\partial p} + \frac{p}{m}t$, but somehow the minus sign isn't there.

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    $\begingroup$ you are applying $d/dp$ to the position space wave function, which is not an eigenvalue of $\hat x$. $\endgroup$ – JEB Apr 20 at 22:58
  • 1
    $\begingroup$ if you want to express $\hat{x}$ in terms of $\frac{d}{dp}$ then you need to convert $\Psi (x)$ to $\Psi (p)$ first $\endgroup$ – Andrew Apr 20 at 23:21

It is a bit more subtle, and this subtlety is important here.

The definition of an operator is that by acting on a wave function the operator determines the expectation value of its corresponding physical quantity: $$\langle \hat{O}\rangle =\int dx \Psi(x)^*\hat{O}\Psi(x).$$ Therefore, just like the wave function, the operator has different forms in different representations.

Let us start with the position expectation in the position representation: $$\langle x\rangle = \int dx x|\Psi(x)|^2 = \int dx \Psi(x)^*x\Psi(x).$$ We easily read the position operator from this expression as $$\hat{x}=x.$$ The momentum operator in this representation is given by $\hat{p}=-i\hbar\partial_x$, as can be verified by considering its action on a state with a definite momentum: $$\psi_p(x)=\frac{1}{\sqrt{2\pi}}e^{i\frac{px}{\hbar}}.$$ Note that the minus sign in this operator is a matter of convention: if we defined plane waves as $\psi_p(x)=\frac{1}{\sqrt{2\pi}}e^{-i\frac{px}{\hbar}}$, we would have to choose $\hat{p}=i\hbar\partial_x$.

Let us now look at the momentum representation. The wave function in the momentum representation us given by $$\Phi(p)= \int dx \psi_p(x)^*\Psi(x).$$ Now $|\Phi(p)|^2$ is the probability density for momentum states and the momentum operator is simply $\hat{p}=p$, as follows from $$\langle p\rangle = \int dp p|\Phi(p)|^2 = \int dp \Phi(p)^*p\Phi(p).$$ For the position operator we have $$\langle x\rangle = \int dp \Phi(p)^*\hat{x}\Phi(p)= \int dp \int dx \Psi(x)^*\psi_p(x)\hat{x}\int dx'\psi_p(x')^*\Psi(x) = \int dx \Psi(x)^*x\Psi(x),$$ where the position operator has to be defined in such a way that $$\int dp \psi_p(x)\hat{x}\psi_p(x')^* = \frac{1}{2\pi}\int dp e^{i\frac{px}{\hbar}}\hat{x}e^{-i\frac{px}{\hbar}} x\delta(x-x').$$ Choosing $\hat{x} = i\hbar\partial_p$ we satisfy this condition. The sign of the position operator is different than the sign of the momentum operator. If, as I mentioned in the beginning, we defined the plane wave with momentum $p$ as $e^{-i\frac{px}{\hbar}}$, teh signs of both operators would be different.

To summarize:

  • In the position representation: $\hat{x}=x, \hat{p}=-i\hbar\partial_x$.
  • In the momentum representation: $\hat{x}=i\hbar\partial_p, \hat{p}=p$
  • The signs before the derivatives in the above expressions are always opposite, but depend on how we define the plane wave with definite momentum.
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The position and momentum distributions are connected by the Fourier transform: $\frac{1}{(2\pi)^{1/2}}\exp(ixp /\hbar)$ is the base vector in the position space and $\frac{1}{(2\pi)^{1/2}}\exp(-ixp /\hbar)$ is the base vector in the momentum space.

Note the following relations from Fourier analysis and quantum mechanics:

$$-i\frac{d}{d x}f(x)=\frac{1}{(2\pi)^{1/2}}\int_{\mathbb{R^3}}kg(k)\exp(ixk)dk$$ and $$p=\hbar k.$$

Now you can do the usual expectation value integral for momentum in momentum space and translate it into the position space.

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Probably the clearest way to check the result is to write the operator explicitly in ket notation in terms of the momentum basis (with $\hbar=1)$

$$ X = \int d^3p |p\rangle i\frac \partial{\partial p} \langle p|$$ and apply this to a position state $$ \begin{align} X|x\rangle &= \int d^3p |p\rangle i\frac \partial{\partial p} \langle p|x\rangle \\ & = \frac 1{(2\pi)^{3/2}}\int d^3p |p\rangle i\frac \partial{\partial p} e^{-ip\cdot x}\\ & = \frac 1{(2\pi)^{3/2}}\int d^3p |p\rangle x e^{-ip\cdot x}\\ & = x \int d^3p |p\rangle \langle p|x\rangle\\ & = x |x\rangle\\ \end{align}$$ where the resolution of unity $$ 1 = \int d^3 p|p\rangle \langle p| $$ has been used. It is seen that the minus sign comes from the conjugate, $\langle p|x\rangle$ rather than $\langle x|p\rangle$, as we would have for the momentum operator. More generally, the position operator can be written $$ X = \int d^3x |x\rangle x \langle x|$$ Then $$ \begin{align} X &= \int d^3 p \int d^3 q \int d^3x |p\rangle \langle p|x\rangle x \langle x |q\rangle \langle q|\\ &= \frac 1{(2\pi)^{3}}\int d^3 p \int d^3 q\int d^3x |p\rangle e^{-ip\cdot x}x e^{iq\cdot x} \langle q|\\ &= \frac 1{(2\pi)^{3}}\int d^3 p \int d^3 q\int d^3x |p\rangle i \frac\partial{\partial p} e^{i(q-p)\cdot x} \langle q|\\ &= \int d^3 p \int d^3 q |p\rangle i \frac\partial{\partial p} \delta (q-p) \langle q|\\ &= \int d^3 p |p\rangle i \frac\partial{\partial p} \langle p|\\ \end{align}$$

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