# Position operator explicit form & definition

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?

• – Frobenius Apr 27 at 14:46
• 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... – Roger Vadim Apr 27 at 14:51
• Keep in mind the momentum representation of the position operator should be applied to the momentum wave function, not the position wave function. – R. Romero Apr 27 at 14:58
• Much of your dilemma evaporates upon using the proper definition, $\hat x \equiv \int\!\!dx ~ |x\rangle x\langle x|= \int\!\!dp ~|p\rangle i\partial_p \langle p|$. – Cosmas Zachos Apr 27 at 19:51
• @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). – 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.