1
$\begingroup$

Consider a quantum system in an eigenstate $|x\rangle$ of the position operator $\hat{x}$, which means that $\hat{x}|x\rangle=x|x\rangle$. I hope that the expected value of $\hat{x}$ will be $x$, since the state $|x\rangle$ is the one in which the system is located in the position $x$. To prove it we do the following: \begin{equation} \langle \hat{x} \rangle = \langle x|\hat{x}|x\rangle = \langle x|x|x\rangle = x\langle x|x\rangle \end{equation} Now, $\langle x'|x\rangle=\delta(x-x')$, so I think that $\langle x|x\rangle$ would be $+\infty$?, and then: \begin{equation} \langle \hat{x} \rangle = \pm\infty?\quad (\text{depending on the sign of $x$}) \end{equation} Which is clearly wrong. How do you calculate this mean value?

$\endgroup$
  • $\begingroup$ @JohnForkosh Sorry, but you're wrong. The inner product involves integral of the wavefunction squared. $\endgroup$ – OON Sep 4 '17 at 3:03
5
$\begingroup$

The formula for the expectation value $\langle A\rangle=\langle\psi|\hat{A}|\psi\rangle$ is given for the normalized states $\langle\psi|\psi\rangle=1$. You can generalize it as \begin{equation} \langle A\rangle=\frac{\langle\psi|\hat{A}|\psi\rangle}{\langle\psi|\psi\rangle} \end{equation} Of course this expression would still be ill-defined for $|x\rangle$ as it's not a proper quantum state, just as $\delta(x-x')$ is actually not a function but rather a distribution.

$\endgroup$
4
$\begingroup$

I would like to add something to @OON's answer.

The spectrum of the position operator is purely continuous, and that means that there is no eigenvector for it in $L^2(\mathbb{R})$. Its eigenvectors are in fact in the space of tempered distributions $\mathscr{S}'(\mathbb{R})\supset L^2(\mathbb{R})$. Nonetheless, there are "almost eigenvectors" of $\hat{x}$ in $L^2$:

Given any $\epsilon>0$ and $x_0\in \mathbb{R}$, there exists at least one (actually many) vector $\psi_{x_0,\epsilon}\in L^2$ such that: $$\int_{\mathbb{R}}x\; \bar{\psi}_{x_0,\epsilon}(x)\psi_{x_0,\epsilon}(x)\;\mathrm{d}x=x_0\; ,\; \int_{\mathbb{R}} \bar{\psi}_{x_0,\epsilon}(x)\psi_{x_0,\epsilon}(x)\;\mathrm{d}x =1\; ,$$ and $\mathrm{supp}(\psi_{x_0,\epsilon})\subseteq [x_0-\epsilon,x_0+\epsilon]$.

Such function is extremely localized around $x_0$ (so it is almost an eigenfunction of $\hat{x}$ with eigenvalue $x_0$), it is normalized, and has the "right" expectation (for it to be an eigenfunction). Finally, the limit $\epsilon\to 0$ of $\psi_{x_0,\epsilon}$ (in the sense of distributions) yields the "correct" (generalized) eigenfunction $\delta(x-x_0)$ (that is however not normalizable since not in $L^2$).

Such construction of "approximated" eigenfunctions is always possible for observables with continuous spectrum, and it is an easy but useful consequence of the so-called spectral theorem for self-adjoint operators.

$\endgroup$

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.