Suppose $A$ and $B$ are compatible observables (i.e. $[A,B] = 0$). We take the eigenkets of $A$ to be $|a_1 \rangle \ldots |a_N \rangle$. Further, we suppose that the first $k$ eignekets of $A$ are degenerate while other eigenkets are non-degenerate. That is $A |a_i \rangle = a |a_i \rangle$ for $i = 1, \ldots, k$. But as is typically the case we suppose that the observable $B$ breaks the degeneracy and we may label the first $k$ eigenkets by their distinct $B$ eigenvalue. That is, we take $|a_i \rangle = | a, b_i\rangle$ for $i = 1,\ldots, k$ where $A | a, b_i\rangle = a | a, b_i\rangle$ and $B | a, b_i\rangle = b_i | a, b_i\rangle$.

Now let us say we have a state ket that is in a general superposition of the $A$ eigenkets: $$ |\psi \rangle = \sum_{i=1}^{k} c_i | a,b_i \rangle + \sum_{i=k+1}^N c_i|a_i \rangle . $$ We measure the observable $A$ and obtain the value $a$ (the degenerate eigenvalue). I have a few questions about this situation:

  1. It seems that this should happen with probability $|c_1|^2 + \ldots + |c_k|^2$. Is this correct?

  2. What state does the system jump into after this measurement? Sakurai (section 1.4) claims that the system jumps into a linear combination of the $|a,b_i \rangle$. What are the coefficients though? Are they the same as that in the original state but normalized to retain the overall normalization of the state ket? That is, do we have $$ |\psi \rangle \to \sum_{i=1}^{k} \frac{c_i}{|c_1|^2 + \ldots + |c_k|^2} | a,b_i \rangle ? $$

  3. (Assuming that the assertion in 2. is correct) Would there be any experimental difference if instead we postulated that the system jumps into a particular $|a,b_i \rangle$ for some $i$ with probability $|c_i|^2$? For example, if all we did was measure $B$ afterwards we would get all the values of $b_i$ with the same probabilities under both postulates.

  • $\begingroup$ The process of measuring $A$ doesn't know anything about $B$, so how could it know to collapse to one of its eigenkets. Instead of $B$ you could just as well contrive another compatible observable $C$, which has a completely different basis of eigenkets spanning the degenerate eigenspace of $A$. Then would measurements of $A$ project onto an eigenket of $B$ or $C$? $\endgroup$ May 4, 2015 at 3:03

1 Answer 1

  1. Is the probability $|c_1|^2 + \ldots + |c_k|^2$? Yes or no. It happens $\frac{|c_1|^2 + \ldots + |c_k|^2}{|c_1|^2 + \ldots + |c_N|^2}$ fraction of the time in the long run. (Assuming the states you listed were all normalized.)

  2. What state does the system jump into after this measurement? There is no experimental evidence of jumps or anything discontinuous. But it is called the projection postulate because the assumption is that the final state is the orthogonal projection onto the eigenspace in question. This fits with the idea that a repeated measurement should produce the same result.

This would be written $$ |\psi \rangle \to \sum_{i=1}^{k} \frac{c_i}{|c_1|^2 + \ldots + |c_k|^2} | a,b_i \rangle. $$ (Again assuming the states $| a,b_i \rangle$ are all normalized.)

  1. Would there be any experimental difference if instead we postulated that the system jumps into a particular $|a,b_i \rangle$ for some $i$ with probability $|c_i|^2$?

Yes, if you did that you will get answers that contradict the theroy and contradict measurements. Take a hydrogen atom (in an external magnetic field oriented in the x direction if you want to be 100% sure the following measurements can be experimentally performed in reality). As an example, assume that A is $L^2$, the total orbital angular momentum. Then there are multiple possible $B$ that commute with $L^2$ that lift the degeneracy, for instance $L_z$ and $L_x$. If we favored one of them, say $L_z$ and said that any measurement of $L^2$ also projects onto a common eigenspace of $L^2$ and $L_z$ then we have a problem with the fact that $L_x$ and $L^2$ commute.

Measure $L^2$ then $L_x$ then $L^2$ again then finally $L_x$ again. By your method, the two results of measuring $L_x$ can give different answers and with nonzero frequency will give different answers (because the second measurement of $L^2$ which should have done nothing at all instead shifted it to an eigenstate of $L_z$ and each eigenstate of $L_x$ is a nontrivial linear combination of $L_z$ eigenvectors). The projection postulate requires/enforces that the two measurements of $L_x$ above give the exact same answer.

Since $L^2$ and $L_z$ commute it could be possible in principle to measure both simultaneously, and then it would do what you describe. In fact measuring one then the other would do exactly that, so measure them in some order. Do it fast, and call that the measurement in question. But that is different than measuring just $L^2$, so you need to call it what it is, a joint measurement of $L^2$ and $L_z$.


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