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My QM text defines the position operator as follows:

The position operator $X= (X_1,X_2,X_3)$ is such that for $j=1,2,3: \ X_j \psi(x,y,z)= x_j \psi(x,y,z)$.

To me this can mean two things.

1) $X$ is a vector and acts as $X \psi(x,y,z)= (x \psi(x,y,z), y \psi(x,y,z), z \psi(x,y,z))$. But this doesn't make sense as $X$ is an observable/operator and so must send vectors to vectors (here functions).

2)There are three position operators $X_1, X_2, X_3$ and each act as defined.

How does the postion operator act on a state? Could anyone help me out here? Thanks!

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    $\begingroup$ Are your two options really incompatible? After all, if (2) is true then a reasonable shorthand for the three operators $X_1,X_2,X_3$ would be $X=(X_1,X_2,X_3)$ and what you've written for (1) would then be a reasonable notation for the action of each component operator. $\endgroup$ – jacob1729 Jun 1 at 16:12
  • $\begingroup$ I can see your point, but it's a bit confusing that they call $X$ the position operator. $\endgroup$ – user Jun 1 at 16:22
  • $\begingroup$ Let's be methodical. Are you comfortable with the def in the one-dim case? Is your problem about the 3-dim case? $\endgroup$ – A. Bordg Jun 1 at 16:24
  • $\begingroup$ @A. Bordg Yes, the 1D case is no problem for me. It's the 3D case that bothers me and the notation used here. $\endgroup$ – user Jun 1 at 16:25
  • $\begingroup$ I guess my text tried to define two notions at once and used the same word for them. $\endgroup$ – user Jun 1 at 16:36
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This is quite an odd way to introduce the position operator, I have to admit. Both definitions you have used are correct, they're just used in different ways in quantum mechanics.

In the the first one, $X$ is what is technically called a vector operators, in this case it's a little like a vector but the components are matrices (or operators). Sometimes it's useful to do this and we can do a sort of dot-product with other vector operators (which you will probably come across soon in QM).

$X$ is composed of the three operators you've defined in 2), and when we want to think about the position operator in three dimensions, definition 1) does actually work. It's a little bit odd, but like I said, $X$ isn't an operator, $X$ is a vector operator, and so the technical issue of mapping the wavefunction to a vector isn't actually an issue. If it's still puzzling, then you can try to think about it as each component of the vector $X\psi$ as being an individual function, and then noticing that this is really just a way of putting three separate scalar equations into one vector equation.

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    $\begingroup$ Also I'd recommend doing a bit of googling that introduces you to the position operator in one dimension, although it is entirely mathematically the same, it's probably a bit easier to build up understanding that way. $\endgroup$ – MC2k Jun 1 at 16:12
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    $\begingroup$ The first definition is a vector operator, not a pseudo vector, which is a different distinction. $\endgroup$ – Javier Jun 1 at 16:14
  • $\begingroup$ Thanks! This is very helpful. Just one question. Did you mean to write "incorrect" in your second line? $\endgroup$ – user Jun 1 at 16:25
  • $\begingroup$ Just to make sure I understand this completely. There are 3 position operators X, Y and Z acting on $\psi$ by $X \psi =x \psi(x,y,z)$ etc.. Further there is the seperate notion of a pseudo-vector which combines these three operators. Is this correct? $\endgroup$ – user Jun 1 at 16:40
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    $\begingroup$ @user Ah sorry, I meant to write correct. And yes, the second comment is true, although as javier pointed out, I made a mistake with the terminology earlier, it's a vector operator (like $\nabla$ from vector calculus), not a pseudo-vector $\endgroup$ – MC2k Jun 1 at 17:12
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The Hilbert space can be seen as a direct product of three independent Hilbert spaces. $$|x,y,z\rangle=|x\rangle\otimes|y\rangle\otimes|z\rangle$$ When expressed as such, the position operator is also seen as a direct product of three operators each acting on their corresponding spaces.

So technically they’re three eigenvalue equations combined into one.

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  • $\begingroup$ Thanks! I am unfamiliar with your notation. Which Hilbert space are you using? $\endgroup$ – user Jun 1 at 17:27
  • $\begingroup$ One that is composed of three independent ones. Think of it like complete set of committing observable. $\endgroup$ – Superfast Jellyfish Jun 1 at 22:07
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The position vector operator, $X=(X_1,X_2,X_3)$, is usually just used as shorthand for writing things like $P\cdot X$, which is defined to be $X_1P_1+X_2P_2+X_3P_3$. You are right that it is not an operator that acts on the Hilbert space of wave functions.

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