I'm trying to wrap my head around QM, so forgive me (and please correct me) if I my question or any way in which I explain the concepts as I understand them are wrong.

So this goes my understanding so far:

For a Hermitian operator $\mathcal{\hat A}$ representing the observable $\mathsf{A}$, the eigenfunctions of $\mathcal{\hat A}$, (let's call them $\psi_i$), are the different 'quantum states' (which are all the possible functions for the wave function $\Psi$ to collapse to, I think? sidenote: if that's the case would that mean that $\psi_i$ are not actually wave functions?) for which if you measure the system for the observable $\mathsf{A}$ you will definitely get $\psi_i$ for some particular $i$. And in fact the actual measurement you make is the eigenvalue of $\psi_i$, call it $a_i$. For two different eigenfunctions $\psi_1$ and $\psi_2$ if they have eigenvalues $a_1$ and $a_2$ such that $a_1 \not= a_2$ then $\psi_1$ and $\psi_2$ are orthogonal.

So my question is, if you have two different eigenfunctions $\psi_{n_1}$ and $\psi_{n_2}$ which have the same eigenvalue, $a_n$, then when you measure for $\mathsf{A}$, can you get different quantum states (i.e either $\psi_{n_1}$ or $\psi_{n_2}$)?

And if the actual measurement you record is $a_n$, then how do you know which $\psi_{n_i}$ it corresponds to.

I guess I'm still not clear on what $\psi_i$ is REALLY representing, if after all it is the eigenvalue $a_i$ that we measure wouldn't the quantum state of the system be defined by $a_i$?

Thanks, and apologies in advance is my question is long winded or unclear.


1 Answer 1


Suppose you have an eigenbasis for the operator $\mathcal{A}$. Let's say its eigenvalues $\{a_i\}_{i=0}^n$ are all non-degenerate (i.e. they are associated with only one eigenstate $\psi_i$) except for $a_0$, which is degenerate (i.e. it's associated with multiple nontrivial eigenstates $\{\phi_j\}_{j=1}^m$). Then, since any linear combination of the $\phi_j$ is also an eigenstate of $\mathcal{A}$ with the same eigenvalue $a_0$, there is a whole $m$-dimensional space of possible eigenstates.

If you measure $\mathsf{A}$ and get $a_0$ as a result, the "collapse" of the wavefunction is only partial. What's really happening is that the wavefunction you're measuring is projected into the $m$-dimensional eigenspace, so whatever $\phi_j$ components it had before measurement are all that's left.

For example, if we let $n=3$ and $m=2$, we write our wavefunction in this basis as


If you measure $\mathsf{A}$ and get $a_1$, then the wavefunction collapses into $\psi_1$, and likewise for $a_2$ and $\psi_2$, and $a_3$ and $\psi_3$. The coefficients $c_1,c_2,c_3$ determine the probability of measurement of $a_1,a_2,a_3$ (namely, the probability of measuring $a_1$ is $|c_1|^2$, etc.). If you measure $a_0$, then the wavefunction collapses into $k_1\phi_1+k_2\phi_2$ after measurement. Your probability of measuring $a_0$ is $|k_1|^2+|k_2|^2$.

  • $\begingroup$ Thank you! That helped with my understanding. What is $\psi_k$ if $a_k$ is the quantity we measure? And what conditions determine the degree to which the wavefunction collapses? $\endgroup$
    – Premez
    Jan 27, 2018 at 19:30
  • 1
    $\begingroup$ $\psi_k$ is the eigenstate of $\mathcal{A}$ associated with the eigenvalue $a_k$. It's uniquely determined by the diagonalization of $\mathcal{A}$. The degree to which the wavefunction collapses is determined by the geometric multiplicity of the eigenvalues of $\mathcal{A}$ in whatever state space $\mathcal{A}$ operates on. $\endgroup$ Jan 27, 2018 at 19:33
  • $\begingroup$ I understand that $\psi_k$ is the eigenstate of $\mathcal{A}$ mathematically, I meant what does it mean physically? $\endgroup$
    – Premez
    Jan 27, 2018 at 19:47
  • $\begingroup$ The eigenstate $\psi_k$ is a quantum state. Whether a quantum state is a physical object or merely a representation of non-determinism depends on the interpretation of quantum mechanics that you're using (and in some of those interpretations, "wavefunction collapse" isn't a thing). $\endgroup$ Jan 27, 2018 at 19:53
  • 1
    $\begingroup$ Again, that depends on the interpretation you're using. $\endgroup$ Jan 27, 2018 at 20:33

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