I have encountered a question:

What are the possible outcomes in measuring $L_{x}$ and what are the corresponding probability of the state: $\Psi(r,0)=1/2(\Psi_{200}+\Psi_{310}+\Psi_{311}+\Psi_{31-1})?$

All the functions above are for hydrogen.

I know how to calculate the expectation value of $L_{x}$ by using raising and lowering operators, but I don't know how to get the corresponding outcomes with corresponding probabilities.

  • $\begingroup$ Welcome to Physics Stack Exchange. What have you tried? Do you understand how to compute possible outcomes of a measurement at all? What is your exact question? $\endgroup$ – DanielSank Dec 10 '15 at 10:11
  • $\begingroup$ @DanielSank Oh, I think I know it. If we want to calculate the outcome with probability, I just use every eigenvector of the observable, then I get the coefficient of the linear combination. The square of coefficient should be the probability. But I don't know how to get eigenvector (or eigenfunction) for $L_{x}$ in this problem. If it is spin of 1/2-particle, I can get it. Then I think I want to know how to get the eigenfunctions. $\endgroup$ – Peter Liu Dec 10 '15 at 12:53

When you measure with $L_x$ you obtain an eigenstate of $L_x$. The state that you have prepared is $$|\psi\rangle = \frac 12 \left( |2\,0\,0\rangle + |3\,1\,0\rangle + |3\,1\,1\rangle + |3\,1\,-1\rangle \right) \,.$$ The normalization is correct, so expectation values are simply the overlap with some other state you expect. You need a set of eigenstates of $L_x$, let us call them $|nlm'\rangle$ where $m'$ is the $L_x$ component. Once you have those you can compute the expectation value with $$ \langle nlm' | nlm \rangle \,.$$ The left side is one of the $L_x$ eigenstates, the right side has one of the $L_z$ eigenstates. They are all eigenstates of $H$ and $L^2$ as well which gives you the quantum numbers $n$ and $l$ of course.

The hard part is the construction of the $L_x$ eigenstates. They need a different basis which corresponds to a rotation. I would suggest to write out the eigenstates as spherical harmonics $Y_{lm}$ and rotate them from the $z$-axis to the $x$-axis. Them you can use the orthogonality relations to get the overlaps. Perhaps there is an easier way, I hope this is something to get you started.

  • $\begingroup$ I have gotten an idea to construct the matrix of raising and lowering operator first, then use it to represent the matrix form of $L_{x}$. Is this right, because I just construct a matrix of 4*4 of these four wavefunctions. $\endgroup$ – Peter Liu Dec 10 '15 at 13:07
  • $\begingroup$ The states with $l = 0$ and $l = 1$ “live” in two different representations of the underlying so(3) algebra. The $l = 0$ case corresponds to a 1-dim representation whereas $l = 1$ has a 3-dim representation. You can then form $1 \oplus 3 = 4$ and have a 4-dim represenation which is block-diagonal. No of the $L_i$ operators can change $l$, they can only change $m = l_z$. Your operators should be block diagonal as well as they can never mix $l = 0$ and $l = 1$. I am not quite sure what you mean by ladder operator matrix, though. $\endgroup$ – Martin Ueding Dec 10 '15 at 14:09
  • $\begingroup$ Oh, I mean that I form a 4 dimensional vector space just as you mentioned: $(a b c d)^{T}$, corresponding to |2 0 0>, |3 1 0>, |3 1 1>, |3 1 -1>. Then the element of the matrix element will be $<\Psi_{m},|L_{+}|\Psi_{n}>$. Similarly, I find the lowering operator in matrix form. Then $L_{x}$ should be $(L_{+}+L_{-})/2$. I think it make sense to me. $\endgroup$ – Peter Liu Dec 10 '15 at 14:22

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