1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ 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 Aug 4 '16 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 Aug 4 '16 at 17:44
  • $\begingroup$ Yes, but that's not represented by the equation you've written $\endgroup$ – zeldredge Aug 4 '16 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 Aug 4 '16 at 17:54
  • $\begingroup$ Echoing @zeldredge' comment, OP's eq. (3) is not correct. $\endgroup$ – Qmechanic Aug 4 '16 at 18:01
4
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ 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 Aug 4 '16 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 Aug 4 '16 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 Aug 5 '16 at 11:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.