I've got a question regarding the position operator. As far as I learned it, the position operator simply gets defined by saying:

$\hat{x} \ \psi(x,t) = x \ \psi(x,t)$

And I also learned that the momentum representation of the position operator is defined as ($p,k$ used interchangeably):

$(\hat{x})_p = i \frac{\partial}{\partial k}$

I'm a bit confused about that. Why is this the "momentum" representation? Let's say I have a plane wave

$\psi(x,t)=e^{i(kx-\omega t)}$

If I wanted to define the position operator, I would just do it like that:

$\hat{x} e^{i(kx-\omega t)} = -i \frac{\partial}{\partial k} e^{i(kx-\omega t)} = x e^{i(kx-\omega t)}$

So why does this look like the momentum representation of the position operator (except for the sign)?

I was just wondering since I never learned the "explicit form" of the position operator. I just learned what it does, namely:

$\hat{x} \ \psi(x,t) = x \ \psi(x,t)$

But my question is, what does $\hat{x} = \cdots$ look like specifically? Does the explicit form of $\hat{x} = \cdots$ depend on the function I'm using it on?

  • 1
    $\begingroup$ Related : Hermiticity of Momentum Operator (matrix) Represented in Position Basis. $\endgroup$ – Frobenius Apr 27 at 14:46
  • $\begingroup$ This is not how position operator defined, but the eigenvalue equation for its eigenfunctions. Note that all these questions are dealt with in the first chapters of any QM textbook... $\endgroup$ – Roger Vadim Apr 27 at 14:51
  • $\begingroup$ Keep in mind the momentum representation of the position operator should be applied to the momentum wave function, not the position wave function. $\endgroup$ – R. Romero Apr 27 at 14:58
  • $\begingroup$ Much of your dilemma evaporates upon using the proper definition, $\hat x \equiv \int\!\!dx ~ |x\rangle x\langle x|= \int\!\!dp ~|p\rangle i\partial_p \langle p|$. $\endgroup$ – Cosmas Zachos Apr 27 at 19:51
  • $\begingroup$ @Vadim I think the OP is referring to $\langle x|\hat X|\psi\rangle=x\psi(x)$. It doesn't have to be about position eigenstates. Additionaly, I don't think Griffiths QM hits deeper formalism like this until chapter 3, and even then I don't recall it being entirely rigorous (it's rigorous enough for undergrad at least). $\endgroup$ – BioPhysicist Apr 27 at 20:06

I hope that this addresses your question(s). In the following we set $\hbar=1$.

Let us first clarify that things like $\hat{x}\,\psi(x)$ in fact mean

$$ (\hat{x}\psi)(x) \equiv \langle x| \hat{x} |\psi\rangle \quad .$$ We further define the wave function as $$ \psi(x) \equiv \langle x|\psi\rangle \quad. $$

Similar equations hold for the case of the momentum operator / basis.

The position representation of the position and momentum operator is (very nicely explained here):

\begin{align} \langle x|\hat{x} &= \langle x| \,x \\ \langle x| \hat{p} &= - i \,\frac{\mathrm{d}}{\mathrm{d}x} \langle x| \quad . \end{align} This should look familiar to you: Evaluated in the position basis, the position operator acts as a multiplicative operator while the momentum operator acts as a derivative. Using the conventions from the beginning we then find \begin{align} (\hat{x}\psi)(x) &= x\, \psi(x) \\ (\hat{p}\psi)(x) &= - i \,\frac{\mathrm{d}}{\mathrm{d}x} \psi(x) \quad , \end{align}

which means for example that the projection of the vector $\hat{x}|\psi\rangle$ onto a state of the position basis is the number $ x\, \psi(x)$.

However, this does not mean that the momentum operator is a derivative; in the same sense, the position operator is not an operator multiplying 'everything' it acts on with $x$. Indeed, evaluated in the momentum basis, we find that the momentum operator is multiplicative and the position operator acts as a derivative: \begin{align} \langle p | \hat{x} &= i\, \frac{\mathrm{d}}{\mathrm{d}p} \langle p|\\ \langle p | \hat{p } &= \langle p |\,p \quad . \end{align} See also this question and the links therein. This tells us that e.g. the projection of the vector $\hat{x}|\psi\rangle$ onto an element of the momentum basis is the number $i\, \frac{\mathrm{d}}{\mathrm{d}p} \psi(p)$.

In conclusion, we can express the action of the position operator $\hat{x}$ on a state $|\psi\rangle$ either as $$ \hat{x}|\psi\rangle = \int \mathrm{d}x\, x\, |x\rangle \langle x|\psi\rangle$$ or $$\hat{x}|\psi\rangle = i\,\int \mathrm{d}p \, |p\rangle \frac{\mathrm{d}}{\mathrm{d}p}\langle p|\psi\rangle \quad .$$

Regarding your example: We know that $\langle x |p\rangle = e^{ipx}$ (up to some constant factor which I'll omit) and let's say we want to calculate $ \langle x |\hat{x}|p\rangle$, which is I think you meant by $\hat{x}\, e^{ipx}$.

Using the position representation of the position operator, we easily find $$\langle x |\hat{x}|p\rangle = x \, \langle x|p\rangle = x \,e^{ipx} \quad . $$

Performing the calculation within the momentum representation of the position operator yields:

$$\langle x |\hat{x}|p\rangle = \left({\langle p |\hat{x}^\dagger|x\rangle }\right)^* = \left({\langle p |\hat{x}|x\rangle }\right)^* = \left(i \frac{\mathrm{d}}{\mathrm{d}p} e^{-ipx}\right)^* = x \, e^{ipx} \quad ,$$ as expected.


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