Collapse of Quantum State and Coefficients

Question

I have found the following exercise in one of my problem sheets:

Suppose we have an observable $Q$ and its corresponding operator $\hat{Q}$ has three eigenfunctions $\varphi_1, \varphi_2, \varphi_3$ with eigenvalues $2, 2,$ and $0$, respectively. Let $\psi$ be the following superposition state: $$\psi(t=0) = \varphi_1 + \frac{1}{\sqrt{2}}\varphi_2 + i\varphi_3$$ If a precise measurement of $Q$ yields the value $2$, what will be the wavefunction immediately after the measurement?

My guess was that $\psi$ must now be a superposition of the eigenstates $\varphi_1$ and $\varphi_2$, since these are the only ones with eigenvalue 2 for the observable $Q$. That is, immediately after the measurement, $\psi = a\varphi_1 + b\varphi_2$, for some $a, b \in \mathbb{C}$ (up to normalisation). However, the solution turns out to be that $\psi = \varphi_1 + \frac{1}{\sqrt{2}}\varphi_2$ up to a normalising constant. I do not understand why the coefficients $1$ and $\frac{1}{\sqrt{2}}$ are "preserved". Wouldn't any arbitrary linear combination of $\varphi_1$ and $\varphi_2$ be possible? Doesn't the state $\psi$ collapse into the eigenspace spanned by $\varphi_1$ and $\varphi_2$?

Answer 1

Your observation make sense. This is a known point with wavefunction collapse on degenerate eigenvalues.

Let's say you measure observable $A$, obtain eigenvalue $a$ and the orthogonal projector onto the eigenspace of $a$ is $\Pi$.

If the eigenvalue is non degenerate, $\Pi =|\phi\rangle\langle\phi|$. Now, immediately after the measurement the wavefunction must be $|\phi\rangle$ because if we repeated the measurement we would get $a$ with certainty.

However if $a$ is degenerate this argument doesn't work, and in principle one could answer as you do.

It is a separate axiom of quantum mechanics (verified in experiments) that in this case after the measurement the wavefunction is

$$ \frac{\Pi |\psi\rangle}{ \Vert \Pi |\psi\rangle \Vert}, $$

where $|\psi\rangle$ is the wavefunction before the measurement.

This procedure gives the result you saw in the book. You can interpret this extra axiom as a sort of Jaynes - maximum entropy principle.

Answer 2

I'm going to use bra ket notation for convenience. There are two relative states compatible with the eigenvalue 2: $|\varphi_1\rangle,|\varphi_2\rangle$. A measurement result that remains unchanged as a result of further measurements can be represented by a projector

https://arxiv.org/abs/quant-ph/0408125

The projector onto the subspace spanned by $|\varphi_1\rangle,|\varphi_2\rangle$ is $$P=|\varphi_1\rangle\langle\varphi_2|+|\varphi_2\rangle\langle\varphi_2|.$$ Changing the basis doesn't change this projection operator.

So if you get the measurement result 2 the relative state of the system changes to $$\frac{P|\psi\rangle}{\Vert P|\psi\rangle\Vert},$$ which is proportional to $$|\varphi_1\rangle+\tfrac{1}{\sqrt{2}}|\varphi_2\rangle.$$

Answer 3

There are two distinct cases:

1. If this measurement has three distinct outcomes but the same value is associated with two of them (e.g. precision is lost somewhere and two different values end up rounding to $2$), then the wave function after the measurement is one of the three eigenstates (here and later, I always mean "up to a nonzero scalar"). If the eigenstates of the two outcomes reported as $2$ are $\{φ_1,φ_2\}$ then the wave function after $2$ is measured will be $φ_1$ or $φ_2$, but you don't know which. If those eigenstates aren't $\{φ_1,φ_2\}$, the wave function after $2$ is measured would be something else of the form $aφ_1+bφ_2$.

2. If this measurement is a partial measurement which genuinely doesn't obtain any information about direction in the $\{φ_1,φ_2\}$ plane, then the state after $2$ is measured is the projection onto that plane of the state before the measurement.

There is no way to distinguish these cases in the operator formalism. By (useful) convention, degenerate eigenvalues in a measurement operator are associated with partial measurements (case 2). That means there is a discontinuity in the physical effect of an operator with eigenvalues $\{0,2,2{+}\epsilon\}$ at $\epsilon=0$, and it means that doing the measurement represented by $\hat O^2$ is not the same in general as doing the measurement represented by $\hat O$ and squaring the result. These are side effects of the convention and not really statements about physical reality.

Assuming relativistic locality and the Born rule for total measurements, if a particle may be in three places A, B, C, and you fail to find it at A, then you can't know in the short term whether it's at B or C. Therefore partial measurements are possible. Furthermore, if B and C are closer to each other than they are to A, and the result of the measurement at A were anything other than projection onto the $\{φ_B,φ_C\}$ plane, then you could detect it by measurements at B and C and build a FTL communication device. So I don't think the behavior of partial measurements is really a separate physical assumption, though it does show up in the axioms of QM.