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.

  • 3
    $\begingroup$ you are applying $d/dp$ to the position space wave function, which is not an eigenvalue of $\hat x$. $\endgroup$
    – JEB
    Apr 20, 2020 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, 2020 at 23:21

3 Answers 3


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 $e^{-i\frac{px}{\hbar}}$, the 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.

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.


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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