Assume we have a quantum system and its wavefunction $\psi$, but the wavefunction is not an eigenfunction of some operator (e.g. the z-component of the angular momentum $\hat{L_z}$).

In this case, how are we going to measure $L_z$?

Is it always the case that we can write $\psi$ as a sum of eigenfunctions of $\hat{L_z}$, and from that we extract an $L_z$ measurement?

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    $\begingroup$ Why do you think that your system needs to be in a definite state of some observable in order to measure that observable? $\endgroup$ – BioPhysicist Nov 4 '19 at 20:28
  • $\begingroup$ So if it is in a superposition how that is going to change the situation? $\endgroup$ – AHMED KRS Nov 4 '19 at 20:42
  • $\begingroup$ The superposition determines the probability of measuring a certain value of the observable. Your QM text/class should cover this. It's a pretty important part of QM. $\endgroup$ – BioPhysicist Nov 4 '19 at 20:47
  • $\begingroup$ we've covered this, but is the superposition necessarily an eigenfunction of the operator? $\endgroup$ – AHMED KRS Nov 4 '19 at 20:51
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    $\begingroup$ @AHMEDKRS This is sort-of the paradox with qm. If the wavefunction is not an eigenstate of an operator, then the corresponding property is not well-defined at that time. However, you can always write the wavefunction as a linear combination of eigenstates (because observable operators are hermitian), so it will instead be a superposition of those eigenstates (until a measurement changes it to a well-defined eigenstate). $\endgroup$ – Maximal Ideal Nov 4 '19 at 21:35

There seems to be some sort of misunderstanding here. Making a measurement of observable $A$ of a system in the state $|\psi\rangle$ does not mean we need to get a number from the calculation $A|\psi\rangle$. The issue here is that $A|\psi\rangle$ is still a vector. If you are expecting the measurement to give a value of $a$ for $A|\psi\rangle=a|\psi\rangle$ then this is still incorrect, as in general $|\psi\rangle$ will not be an eigenvector of $A$.

So how does the operator $A$ relate to the measurement of the observable associated with $A$? Well, all you have to do is express $|\psi\rangle$ in the eigenbasis of $A$ $$|\psi\rangle=\sum_nc_n|a_n\rangle$$ Quantum theory tells us that if we were to measure $A$ of our system that all can cen determine is the probability of measuring some value $a_n$. This probability is equal to $|c_n|^2=|\langle a_n|\psi\rangle|^2$.

So, from the operator we can determine two things:

  1. Its eigenvalues (possible measurement outcomes)
  2. Its eigenvectors (what we can use as basis vectors).

And from these two things we can then determine the probability of our system to have a value of $a_n$ when we measure $A$.

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  • $\begingroup$ Yeah, so you are saying that $\psi$ is generally is not an eigenvector of the operator and that when we want to get a measurement of some observable we can expand the wavefunction as a sum of the eigenbasis of that observable? we then use that to get the measurements. $\endgroup$ – AHMED KRS Nov 4 '19 at 21:57
  • $\begingroup$ @AHMEDKRS Kind of. We cannot "get a measurement" in QM. All we can get from $|\psi\rangle$ is the probabilities associated with certain measurements. There is no way to say what the result of a measurement will actually be though. $\endgroup$ – BioPhysicist Nov 4 '19 at 21:58
  • $\begingroup$ Yeah, I mean the probability of the measurement. $\endgroup$ – AHMED KRS Nov 4 '19 at 22:00
  • $\begingroup$ @AHMEDKRS Yes. $|\psi\rangle$ tells us the probability of measuring an eigenvalue of the operator associated with the observable in question. $\endgroup$ – BioPhysicist Nov 4 '19 at 22:02

We can always move to the eigenbasis given a hermitian operator like $L_z, H, L^2, \cdots.$ There is a theorem called Spectral theorem, which states that there exists an eigenbasis given hermitian operator, and this applies to finite dimensional, infinite dimensional, including Hilbert spaces.

Thus, since we have an eigenbasis, we can expand $|\psi\rangle=a_i|i\rangle$ in that basis with coefficients given by $\langle i | \psi \rangle$ where $| i \rangle$ is the given eigenbasis of the given hermitian operator $A, A|i\rangle=A_i|i\rangle$.

Then we are just left with writing down the eigenvalue problem $$\langle A \rangle = \langle \psi | A \psi \rangle = \langle i|a_i^* A a_j|j \rangle=\langle i|a_i^* a_jA_j|j\rangle=A_ja_i^*a_j\langle i|j\rangle = A_ja_i^*a_j \delta(i,j)=A_i|a_i| ^2. $$

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  • $\begingroup$ Is a sum implied on repeated indices? $\endgroup$ – Cham Nov 4 '19 at 20:28
  • $\begingroup$ Yes, Einstein summation notation is assumed. $\endgroup$ – Nugi Nov 4 '19 at 20:31
  • $\begingroup$ Then this may be misleading for the eigenvalue equation $A |i \rangle = A_i | i \rangle$. $\endgroup$ – Cham Nov 4 '19 at 20:33
  • $\begingroup$ Yes, but I thought it one could find it out that it is not the case there, and I thought it was better than putting summation symbols everywhere else. Also, it wouldn't make sense to sum over $i$ only on the right hand side and leave the left hand side with $i$. $\endgroup$ – Nugi Nov 4 '19 at 20:35
  • $\begingroup$ I found this helpful. Thank you! $\endgroup$ – AHMED KRS Nov 4 '19 at 22:04

It may not be an eigenstate of $\hat L_z$ but, if the system is in a pure state, it will be an eigenstate of $\hat L_{\hat n}=\hat n\cdot \vec L$, i.e. it will be an eigenvector of the projection of angular momentum in some direction $\hat n$.

It may be tricky to find this direction but one way might be to make a beam that travels through weak magnetic field and reorient the field gradient until one gets a single spot on a screen, i.e. until all particles are deflected in the same direction by the same amount.

Since $L_z$, together with $\vec L\cdot\vec L$ are a set of commuting hermitian operators, we are guaranteed their eigenvectors form a complete set, i.e. that any state can be expanded in these eigenvectors.

In our particular case, it’s not hard to write the eigenvectors of $\hat L_{\hat n}$ as a linear combination of eigenvectors of $\hat L_z$: just find the rotation ${\cal R}$ that takes $\hat z$ to $\hat n$ and rotate the eigenstates of $\hat L_z$ accordingly.

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  • $\begingroup$ I think I got the point. Thanks for your help. $\endgroup$ – AHMED KRS Nov 4 '19 at 22:09

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