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Say I have the singlet $|s\rangle = |0\rangle - |1\rangle$ and $m=0$ triplet $|t\rangle = |0\rangle + |1\rangle$ quantum spin states (say in the $S_x$ basis). If all I can do are spin measurements, how can I distinguish these two states? If all I do are measurements in the $S_x$ basis, these two states give identical results upon repeated measurement. I may think that these two states are identical forever, because nothing from the $S_x$ basis measurements tells me to try anything else. How would I know that I should actually try something different, like a measurement in the $S_z$ basis, in order to distinguish the two states?

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  • $\begingroup$ You can use the latex commands \rangle (or \langle) for bras (and kets) instead of the > < ascii characters. It looks neater. Thanks. $\endgroup$
    – joseph h
    May 6, 2023 at 5:07
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    $\begingroup$ (I made the change for you, heh) $\endgroup$
    – CR Drost
    May 6, 2023 at 5:09
  • $\begingroup$ Is the singlet state $\lvert s \rangle \equiv \frac{1}{\sqrt{2}} (\lvert 10 \rangle - \lvert 01 \rangle$? $\endgroup$ May 6, 2023 at 5:19

3 Answers 3

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We cannot conclude two states are identical because they generate the same probability distribution for only one observable. Provided an ensemble of identical states, quantum state tomography may be used to reconstruct the density matrix for that state. This amounts to measuring over an appropriate basis of observables for the Hilbert space many times.

For the spin states above, quantum state tomography is able to reconstruct the density matrix using measurements on $S_x$, $S_y$, and $S_z$.

$$\rho = \frac{1}{2}\left[\mathbb{I} + \langle \sigma_x\rangle \sigma_x + \langle \sigma_y\rangle \sigma_y + \langle \sigma_z\rangle \sigma_z\right]$$

where $\mathbb{I}$ is the identity matrix, and $\sigma_{x,y,z}$ the Pauli matrices. Measuring the expectation value of the spin in the $x$, $y$, and $z$ directions provides enough information to reconstruct the density matrix for the original state, and thus distinguish it from others. Measuring the spin in the $x$ direction alone is not enough.

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  • $\begingroup$ The proposed tomography expression is trivially $\rho=\frac12 I$ for both proposed states, as they are spherically symmetric with respect to the spin operators. $\endgroup$
    – CR Drost
    May 6, 2023 at 7:11
  • $\begingroup$ @CRDrost can you explain what you mean by spherically symmetric with respect to the spin operators? I did the calculation and found the states have distinct density matrices. $\endgroup$
    – Aiden
    May 6, 2023 at 15:50
  • $\begingroup$ Sure, just posted a quick answer if it helps. :) I mean that $\langle \mathbf S\rangle=\mathbf 0$ for both configurations and thus you cannot reconstruct the state from the average spin in each direction. $\endgroup$
    – CR Drost
    May 6, 2023 at 17:39
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The state that a system was in pre-measurement is not completely determined by what measurement value you get out with one operator. It is also not determined by what distribution of measurement values you get out with one operator (via measuring an infinite ensemble of identical states). In fact, you provide a counterexample. Hence, you should not expect that the aforementioned information allows you to, with veracity, distinguish between the two states.

(Edits made upon reading Aiden's answer).

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So, given a Hamiltonian with an $S^2$ term or so (total energy depends on total spin), a two-spin level splits into two energy levels, one with degeneracy 1 ($s_\text{total}=0$) and one with degeneracy 3 ($s_\text{total}=3$) known respectively as the spin singlet and spin triplet: $$ \newcommand{\bra}[1]{\langle{#1}|} \newcommand{\ket}[1]{|{#1}\rangle} \newcommand{\sqh}{\sqrt{\frac12}} \newcommand{\uu}{\ket{\uparrow\uparrow}} \newcommand{\ud}{\ket{\uparrow\downarrow}} \newcommand{\du}{\ket{\downarrow\uparrow}} \newcommand{\dd}{\ket{\downarrow\downarrow}} \begin{align} \ket{0,0}~~~&=\sqh\big(\ud-\du\big),\\ \ket{1,-1}&=\dd\\ \ket{1,0}~~~&=\sqh\big(\ud+\du\big),\\ \ket{1,+1}&=\uu.\end{align} $$ and one can work out this ambiguity that you mentioned that the average spin along all three axes is zero for both $\ket{0,0}$ and $\ket{1,0}$, $$ \bra{0,0}\mathbf S\ket{0,0} = \mathbf 0,\\ \bra{1,0}\mathbf S\ket{1,0} = \mathbf 0,$$ which I mentioned in my comment to Aiden as saying that these are both “spherically symmetric”, and this is probably an incorrect (or at least imprecise) way to generalize, as the phase of the wavefunction should be asymmetric.

The boring answer is, measure $S^2$ and $S_z$. This is boring because those are the two quantum numbers in $\ket{\ell, m},$ obviously if you measure $S^2$ you will tell the difference between the $\ell=1$ $\ket{1,0}$ state and the $\ell=0$ $\ket{0,0} state.

The more interesting answer is, $\ket{0,0}$ is an eigenvector of $S_x$ and $S_y$ in addition to being an eigenvector of $S_z$, but $\ket{1,0}$ is only an eigenvector of $S_z$ and not of $S_{x,y}.$ This means that once you have ruled out the possibilities of $\uu$ and $\dd$ and you are only trying to tell the difference between $\ket{0,0}$ and $\ket{1,0}$, you have the cute trick that $$ \begin{align} \ket{0,0} &=\sqh\big(\ket{\leftarrow\rightarrow}- \ket{\rightarrow\leftarrow}\big)\\ \ket{1,0}&=\sqh\big(\ket{\leftarrow\leftarrow}- \ket{\rightarrow\rightarrow}\big) \end{align} $$ (there's like a 50/50 chance there's a sign error there but it's immaterial if there is).

In other words if you measure $S_x$ on these two, that measurement will be destructive for $\ket{1,0}$ but will reveal either a spin of $\pm 1$ in the $x$-direction, never zero. Meanwhile, the measurement will be non-destructive on $\ket{0,0}$ and will always reveal a spin of zero.

So if you had a jet of spin states flying by and you didn't know if they were singlet or triplet or some mix, a single Stern-Gerlach apparatus could probably tell you, and two could definitely tell you.

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