A particle has spin $\hbar/2$. A measurement is made of the sum of x and z components of its spin angular momentum.

What are the possible results of the measurement?

  • $\begingroup$ By the postulates of quantum mechanics what are the possible values for an observable? $\endgroup$ – user1620696 Feb 24 '18 at 3:21
  • $\begingroup$ It's customary to say the particle has spin $s=\frac 1 2$, which means there a two possible eigenstates of angular momentum projection, $\pm \hbar/2$. The actual spin angular momentum of the particle is $\hbar\sqrt{s(s+1)}=\hbar\sqrt 3/2$. $\endgroup$ – JEB Feb 24 '18 at 4:50

Strictly speaking there is not enough information in your question to provide a definite answer.

In general, the possible outcomes of measuring spin in any directions must be the same since there is nothing preferential about one the other direction. In other words, suppose the possible outcomes of measuring spin along $\boldsymbol{\hat z}$ are $\pm \hbar/2$. I can relabel the $\boldsymbol{\hat z}$ axis as the $\boldsymbol{\hat x}$ axis, or for that matter as any unit vector $\boldsymbol{\hat n}$ and of course none would be the wiser about this labelling unless there is something to break the spherical symmetry of the problem.

What changes with the direction are the probabilities of each outcome. The probabilities depend on two things. One is the operator, and the other is the state. If $\vert\pm \hat n\rangle$ are the eigenstates of the spin operator $\sigma_{\hat n}$ in the direction $\boldsymbol{\hat n}$, and given a fixed initial state $\vert\psi\rangle$, then these probabilities are $\vert \langle \pm \hat n\vert \psi\rangle\vert^2$.

Note that an exception to the statement that the possible outcomes are the same for any direction occurs if one of $\vert \langle \pm \hat n\vert \psi\rangle\vert^2= 0$. The associated outcome has probability $0$ and is thus excluded as a possible outcome.

In particular, you can verify that the eigenvalues of $\sigma_x$ are those of $\sigma_z$ and those of $\sigma_y$, and those eigenvalues, by the postulates of quantum mechanics, are the possible outcomes.

Indeed you can verify that, if $U$ is unitary, then $U\sigma_zU^{-1}$ has the same eigenvalues (and thus same possible outcomes) as $\sigma_z$. The matrix $\sigma_x$ for instance, is given by $$ \sigma_x=U\sigma_z U^{-1}\, ,\qquad U=\left( \begin{array}{cc} \frac{1}{\sqrt{2}} & -\frac{1}{\sqrt{2}} \\ \frac{1}{\sqrt{2}} & \frac{1}{\sqrt{2}} \\ \end{array} \right)\, . $$

You can also verify for instance that, given the state $$ \vert \psi\rangle =\frac{1}{\sqrt{2}}\vert \hat z\rangle -\frac{1}{\sqrt{2}}\vert -\hat z\rangle $$ the possible outcomes of measuring $\sigma_z$ are $\pm\hbar/2$ since both $+\hbar/2$ and $-\hbar/2$ occur with non-zero probability. The state $\vert\psi\rangle$ is an eigenstate of $\sigma_x$, so there is one one possible outcome for $\sigma_z$ for this state.

This illustrates that probabilities depend on the operator and the state, but that possible outcomes (except when the associated probabilities are $0$), do not.


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