We define the position operator $\hat{X}$ by $$\hat{X}|\psi\rangle := \bigg(\int dx |x \rangle x \langle x | \bigg) | \psi \rangle \tag{1}$$ for some state vector $| \psi \rangle \in \mathcal{H}$. We also have the state vector $| \psi \rangle$ in the position basis given by $\psi(y) = \langle y | \psi \rangle$. Then \begin{align} (\hat{X} \psi)(y) & := \langle y | \hat{X}| \psi \rangle = \int dx \langle y | x \rangle x \langle x | \psi \rangle = \int dx \delta(y-x)x \psi(x) \\ & = y \psi(y) \int dx \delta(y-x) = y \psi(y).\tag{2} \end{align} As I understand, one postulate of QM states that for any observable $\hat{A}$ we have that $$\hat{A} | \psi \rangle = a_n | \psi_n\rangle\tag{3}$$ where $a_n$ is some eigenvalue and $| \psi_n \rangle$ the corresponding eigenvector. For the position operator we thus have $$\hat{X} | \psi \rangle = x |x \rangle\tag{4}$$ where $x$ is some eigenvalue and $| x \rangle$ is the eigenvector of the position operator, therefore $$\langle y | \hat{X} | \psi \rangle = x \langle y | x \rangle = y \delta(y -x).\tag{5}$$ Therefore $$y\psi(y) = y\delta(y-x).\tag{6}$$ But then this implies that the state which we started with $| \psi \rangle$ is $$\psi(y) = \delta(y-x),\tag{7}$$ but this should be the state only after the position measurement. Why do these definitions of the position operator not agree?

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    $\begingroup$ That postulate is wrong. It should be summed over all $n$ (the entire eigenbasis), which corresponds to keeping the integral present in your formula. As a rule, you should become concerned if you have a delta function in a final result that won't be integrated over later. $\endgroup$
    – zeldredge
    Commented Aug 4, 2016 at 17:41
  • $\begingroup$ @zeldredge After measurement, does the wave function $| \psi \rangle$ not collapse to some eigenvector $| x \rangle$ corresponding to the measured eigenvalue value $x$? $\endgroup$
    – Alex
    Commented Aug 4, 2016 at 17:44
  • $\begingroup$ Yes, but that's not represented by the equation you've written $\endgroup$
    – zeldredge
    Commented Aug 4, 2016 at 17:51
  • $\begingroup$ @zeldredge How is it not? I start with state $| \psi \rangle$, take measurement $\hat{X} | \psi \rangle$ which leaves the state in eigenstate of position operator $| x \rangle$ with measured eigenvalue $x$. i.e. $\hat{X} | \psi \rangle = x |x \rangle$. $\endgroup$
    – Alex
    Commented Aug 4, 2016 at 17:54
  • $\begingroup$ Echoing @zeldredge' comment, OP's eq. (3) is not correct. $\endgroup$
    – Qmechanic
    Commented Aug 4, 2016 at 18:01

1 Answer 1


It is not the case that $$ \newcommand{\ket}[1]{\left| #1 \right\rangle} \hat{A} \ket{\psi} = a_n \ket{a_n} $$ where $\hat{A}$ is an observable, $a_n$ is an eigenvalue, and $\ket{a_n}$ is the corresponding eigenvector. You are getting confused with collapsing the wavefunction, which says that a measurement of the observable $\hat{A}$ results in the wavefunction collapsing to a particular eigenstate $\ket{a_n}$ if the value $a_n$ is measured, which is not the statement you have written. (The statement you've written is an equality--it doesn't tell you anything about a process occurring.)

What is true is that: $$ \hat{A} \ket{a_n} = a_n \ket{a_n} $$ which is the definition of an eigenvector. Therefore, since we can write $\ket{\psi}$ in the $\hat{A}$ basis: $$ \hat{A} \ket{\psi} = \int \mathrm{d} {n} \ \hat{A} \psi_n \ket{a_n} = \int \mathrm{d}n \ a_n \psi_n \ket{a_n} $$ Or, for $\hat{X}$: $$ \left\langle{y} \hat{X} \right\rangle{\psi} = \int \mathrm{d} x \ x \psi(x) \left\langle{y} | x \right\rangle = \int \mathrm{d} x \ x \psi(x) \delta(y - x) = y \psi (y) $$

Now, to address the incorrect postulate you claimed. We do not use the application of the observable $\hat{X}$ to represent the collapse of the wavefunction. After all, consider $\ket{\psi} = \ket{x_1} + \ket{x_2}$. Applying $\hat{X}$ to this sum of eigenvectors yields...another sum of eigenvectors, by linearity. You have to introduce the measurement result somehow--else how do you know which $x$ you get?

If you are given a wavefunction prior to a measurement and then given a measurement outcome, you should instead use the corresponding projection operator to project the wavefunction onto the measured eigenvector: $$ \ket{\psi} \to P \ket{\psi} = \left( \ket{x_1} \left\langle x_1 \right| \right) \ket{\psi} $$ assuming that $x_1$ was measured. This can be more complicated if the eigenvalues are degenerate, in which case you will collapse instead to the subspace of eigenvectors associated with that eigenvalue.

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    $\begingroup$ Good answer, I will add that the original confusion might have stemmed from the fact that often, people think that measurement has something to do with operators acting on a state vector. (I think this is the source of OP's incorrect equation.) Actually, in QM, you very rarely care about the actual action of an operator. $\endgroup$
    – knzhou
    Commented Aug 4, 2016 at 18:08
  • $\begingroup$ @knzhou I think it's a pretty common confusion. What it is the state vector, operators, etc represent is often pretty hard to grasp when you start on the subject -- I remember feeling pretty clearly that I had no intuition for any of the symbols until after a semester or so. $\endgroup$
    – zeldredge
    Commented Aug 4, 2016 at 18:20
  • $\begingroup$ @zeldredge Thanks for your answer. I have been mislead by a book I'm using which states the incorrect equation regarding the above postulate. But I see it doesn't make sense to fix the value of an operator in the way in which I did. $\endgroup$
    – Alex
    Commented Aug 5, 2016 at 11:46

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