To each observable in quantum mechanics there is an operator corresponding to it. I don't understand what's the meaning of the eigenvalues of the $\hat{x}$ operator. Since $\hat{x}$ is hermitian, eigenvalues correspond to real numbers, what's their physical meaning? If they describe a particle localized at a particular point, isn't it in opposition with the statistical nature of quantum mechanics?

Second: Since position and momentum operators do not commute, the eigenvectors of the Hamiltonian are usually different from the eigenvectors of both position and momentum operators. But I see in books applying the $\hat{x}$ to the $\psi_n$a representing the autostates of the Hamiltonian operator. How do I find the autostates for $\hat{x}$ if they are not the same obtained by solving Schrödinger equation?


2 Answers 2


Eigenvalues are the values that are measured in the experiment, i.e. eigenvalues of $\hat{x}$ are the values of the position obtained when measuring it. Every measurement will produce a different result, i.e. a different eigenvalue, unless the system was prepared in an eigenstate of the measured quantity - this is the statistical nature of quantum mechanics. Thus, after $N$ measurements we have a sample average $$ \bar{x} = \frac{1}{N}\sum_{i=1}^N x_i, $$ whose value approaches the one estimated in quantum mechanics, $\langle \hat{x}\rangle$. (All this looks trivial, but the problem of modern physics is that QM is often taught to students who has never taken a course in statistics.)

The state of the system does not have to be eigenstate of all the operators, which is pretty much the point of the uncertainty principle. Thus, an eigenstate of the Hamiltonian operator has well-defined energy, but uncertain position and often uncertain momentum. For example, the eigenstate of a harmonic oscillator gives a Gaussian distribution of position, i.e. the emasurements $x_i$ measured above will be distributed, as if they come from a Gaussian distribution, with variance $\sigma_x^2 = \langle \hat{x}^2\rangle - \langle \hat{x}\rangle^2$.

  • $\begingroup$ So when the state of the system is not an eigenstate for the $\hat{x}$ operator, how do you calculate the position? Is there an analogous equation to Schrödinger for $\hat{x}$?. And are these possible positions quantized (corresponding to the eigenvalues), in the sense there are only certain values for positions that you can measure? $\endgroup$ Sep 29, 2020 at 11:34
  • $\begingroup$ Let me first point out that all this is explained in QM textbooks. Since QM is statistical in nature, you do not calculate the position, but its expectation value: $\langle \hat{x}\rangle = \int dx \psi^*(x)\hat{x}\psi(x)$. The wave function depends on the state in which your system is, and its time evolution is governed by the Schrödinger equation - there is no need for another equation. $\endgroup$ Sep 29, 2020 at 11:38
  • $\begingroup$ Sure, but in this way the eigenvalues of $\hat{x}$ are undefined. $\endgroup$ Sep 29, 2020 at 11:44
  • $\begingroup$ Eigenvalues are defined - they depend only on the operator itself: $\hat{x}\phi_x(x) = x\phi_x(x)$, but the system is not in the eigenstate of this operator. Every time the measurement of $x$ is carried out, the system collapses into one of the eigenstates eigenstate of $\hat{x}$, with probability $\psi_(x)$. You have to note here the difference between the wave function of the system, $\psi(x)$, and the eigenfunctions of $\hat{x}$ - $\phi_x(x)$. It makes things even more confusing, because we are looking at a wave-functions in the position representations, i.e. as functions of $x$... $\endgroup$ Sep 29, 2020 at 11:52
  • $\begingroup$ How do you obtain the eigenfunctions of $\hat{x}, \phi_x{x}$? $\endgroup$ Sep 29, 2020 at 11:55

It is sometimes said that the delta distributions $\delta(x-x_0)$ are the eigenstates of the position operator $\hat x$. However, as its name may imposes, those are not classical functions and precisely not elements of the Hilbert space $x$ is defined on. The same is true for the momentum operator and the functions $e^{i\langle x,k\rangle}$. They are not square integrable and, thus, no elements of the Hilbert space. Clearly, an eigenvector has to be an element of the Hilbert space, so what's the matter?

The problem we're dealing with here is the infinite dimension of our Hilbert space. Things get more sophisticated then in the finite dimensional scenario where we can just compute eigenstates (and we know that these are elements of the Hilbert space). Still an operator $\hat A$ can have eigenstates defined by non-trivial solutions of $$(\hat A - \lambda \cdot\operatorname{id})| \psi\rangle = 0$$ for $\lambda\in\mathbb{C}$ and a $|\psi\rangle\in\mathcal{H}$ in the domain of $\hat A$. In infinite dimensions operators are not necessarily defined on the whole Hilbert space. Take for example $\hat x$, it can kick a function out of the Hilbert space by returning a non-square integrable function. The eigenvalues $\lambda$ obtained that way are called the point-spectrum of $\hat A$. However, as mentioned above, the delta distributions are not elements of the Hilbert space and, thus, they cannot contribute to the point-spectrum. Instead, they represent another kind of spectrum, the contineous spectrum. The position and momentum operators for free particles do not posess a point spectrum, but only a contineous one. The Hamilton operator of a Hydrogen atom is an example of an operator with point spectrum (bound states) as well as contineous spectrum ($E>0$, scattering states). For a nice introduction to infinite dimensional quantum systems and their peculiarities check out the lectures of F. Schuller.

For the point spectrum, we find a discrete set of eigenvectors (for example by solving the Schrödinger equation). However, the contineous spectrum is usually discribed by an operator valued measure $\mu$ on $\mathbb{R}$. In principle, it tells us the properbility of measuring a state with a measuring result in a subset $A\subseteq\mathbb{R}$. Now, finding these "eigenstates" of the contineous spectrum corresponds to finding the precise measure $\mu$.

I hope this can help you somehow. The explict construction of the measure can be done by using the resolvent operator for example. This is explained in this lecture video. Cheers!

  • $\begingroup$ @JoshuaTS I got your concern, but you are missing an important detail. Completeness of a metric space is by definition the convergence of all Cauchy sequences. However, there is no need for an arbitrary sequence to converge. Take for examle the sequence of functions $(n\mapsto\sin(nx))_{n\in\mathbb{N}}$. It does not converge in the space of square integrable functions although it is well known that the latter is complete. The sequences that converge to the delta distribution follow another notion of convergence, i.e. as a distribution. Hope this helps to clear the confusion! $\endgroup$ Sep 29, 2020 at 21:05
  • $\begingroup$ I don't see the connection of the eigenfunctions of $\hat{x}$ with the real physical system so with $\Psi(\mathbf{x},t)$. For example if you take $\hat{H}$ the eigenvalue equations are given by the Time-independent Schrödinger equation. In the case of $\hat{x}$ I don't see this connection. For example If we measured the electron’s position to be at r = 2.3 a.u., the corresponding eigenfunction would be δ(r − 2.3 a.u.). How do I set up the eigenfunctions if I've not made any measurement? $\endgroup$ Sep 30, 2020 at 9:56

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