I fail to understand the EPR experiment. From Wikipedia:

Alice now measures the spin along the z-axis. She can obtain one of two possible outcomes: +z or −z. Suppose she gets +z. Informally speaking, the quantum state of the system collapses into state I. The quantum state determines the probable outcomes of any measurement performed on the system. In this case, if Bob subsequently measures spin along the z-axis, there is 100% probability that he will obtain −z. Similarly, if Alice gets −z, Bob will get +z.

I emphatically do not reproduce above claim. As example I consider measurement of spin along an arbitrary axis. Specially I consider measuring spin along $\vec{v} = \frac{1}{\sqrt{38}}(2,3,5)$ and with input state $\psi = \frac{1}{\sqrt 2}(|01\rangle - |10\rangle)$. Here are my steps:

  1. Alice wants to measure spin along $\vec{v}$. This means we create the observable $M = \vec{v} \cdot \vec{\sigma}$ where $\vec{\sigma}$ is the vector of individual Pauli matrices i.e., $\vec{\sigma} = (X , Y , Z)$.

  2. $M$ is a $2\times2$ matrix and $\psi$ is a $4\times1$ vector. To make a measurement corresponding to $M$ on the first qubit we need to create a tensor product of $M$ with the $2\times2$ identity matrix $I$. So let $O = M \otimes I$.

  3. Now to perform the measurement we will do a eigen decomposition of $O$. Note that gives me 4 eigenvalues not 2. Its true that there are 2 unique eigenvalues $\{+1, -1\}$ and I understand these to mean spin "up" or spin "down".

  4. I get following probabilities of the 4 eigenvalues:

Alice's Eigenvalues (i.e., possible measurement outcomes)
[-1.00000000000000, -1.00000000000000, 1.00000000000000, 1.00000000000000]
Alice's probabilities
[0.452776776413453, 0.0472232235865468, 0.0472232235865468, 0.452776776413453]
  1. Now suppose Alice measures the first eigenvalue $-1$ (spin "down").

  2. We then collapse the wavefunction $\psi$ to one of the corresponding eigenvectors of $O$. Let $\psi^{'}$ be the collapsed wavefunction after Alice's measurement, which is one of the two eigenvectors of $O$ with eigenvalue $-1$.

  3. Now Bob will measure spin along the $\vec{v}$ axis on the second qubit. The observable corresponding to this is $O^{'} = I \otimes M$ and input to this observable is $\psi^{'}$.

  4. And now the kick: when I simulate Bob's measurement I do not get a spin "up" outcome with 100% probability. I find that there are 2 unique eigenvalues associated with non-zero probabilities.

Bob's Eigenvalues
[-1.00000000000000, -1.00000000000000, 1.00000000000000, 1.00000000000000]
Bob's probabilities
[0.00892013138361998, 0.0855263157894737, 0.0855263157894737, 0.820027237037433]

And this is my dilemma.

Can someone explain me what it is that I have done wrong? I would like a worked out answer with numbers in it rather than mathematical symbols. An answer showing what needs to be corrected to my code would be even better.

Aside: If part of the answer is that $O$ should be $M \otimes M$, I emphatically consider that to be cheating. The experiment states:

Alice now measures the spin along the z-axis... the quantum state of the system collapses into state I... Bob subsequently measures spin along the z-axis, there is 100% probability that he will obtain −z. Similarly, if Alice gets −z, Bob will get +z.


2 Answers 2


I believe your problem is that you're splitting up your measurement observable too much. Yes, $M\otimes I$ has four eigenvalues, $-1,-1,1,1$. BUT, when you make the measurement, you do not get 4 difference answers. You only get 2.

To see why what you've done cannot work, note that the choice of eigenvectors is not unique. So how can you sub-divide the results you get?

The correct way to do this is to split things up into two subspaces corresponding to the $\pm 1$ eigenvalues: $$ P_{\pm}=(I\pm M)\otimes I/2 $$ The two outcomes $\pm 1$ correspond to getting one of these measurement results, which occur with probabilities $$ p_{\pm}=\langle\psi|P_{\pm}|\psi\rangle, $$ and if you get a given result $\pm1$, the overall state of the system after the measurement is $$ |\psi_{\pm}\rangle=\frac{P_{\pm}|\psi\rangle}{\sqrt{p_{\pm}}}. $$

If you like, one way of thinking about it that closely parallels your attempt is that, as I already mentioned, the eigenvectors are not unique. There are two eigenvectors in the +1 eigenspace. So, you could pick one that is orthogonal to $|\psi\rangle$, and hence always has 0 probability. Similarly for the -1 eigenstates. You're then left with two eigenvectors that you could keep going with, and you'd get the same results as I've claimed.

  • $\begingroup$ re: the comment "You're then left with two eigenvectors that you could keep going with", the eigenvectors are orthonormal and thus distinguishable. refer Nielsen & Chuang p. 86 section 2.2.4 Distinguishing quantum states. So the wave function will collapse to a definite eigenvector and is inline with what the books say. $\endgroup$
    – morpheus
    Commented Jun 14, 2021 at 20:45

I suspect that there is a bug in the code. Here is a short analysis following the steps of your scenario. First, diagonalize $M$

$$ M = |a\rangle\langle a| - |b\rangle\langle b| = P_a - P_b $$

where $|a\rangle = \alpha_0|0\rangle + \alpha_1|1\rangle$, $|b\rangle = \beta_0|0\rangle + \beta_1|1\rangle$ and $P_a = |a\rangle\langle a|$ and $P_b = |b\rangle\langle b|$ are the projectors onto the eigenspaces of the two eigenvalues of $M$: $+1$ and $-1$, respectively.

Probability that Alice measures $-1$ is

$$ \begin{align} \langle\psi | P_b \otimes I |\psi\rangle &= \frac{1}{2}(\langle 01|-\langle 10|) P_b \otimes I (|01\rangle - |10\rangle) \\ &= \frac{1}{2}\left(\langle 01|P_b \otimes I|01\rangle -\langle 01| P_b \otimes I |10\rangle \\ - \langle 10|P_b \otimes I|01\rangle + \langle 10|P_b \otimes I|10\rangle\right) \\ &= \frac{1}{2} \left(|\beta_0|^2 - 0 - 0 + |\beta_1|^2\right) \\ &= \frac{1}{2} \end{align} $$

and similarly for $+1$.

Now, as in your scenario let us compute the collapsed state following Alice measuring $-1$

$$ |\psi'\rangle = \frac{P_b \otimes I |\psi\rangle}{\sqrt{\langle\psi | P_b \otimes I |\psi\rangle}} = P_b \otimes I (|01\rangle - |10\rangle) = \beta_0^*|b1\rangle - \beta_1^*|b0\rangle. $$

Finally, let us find the probability that Bob measures $-1$

$$ \begin{align} \langle\psi' | I \otimes P_b |\psi'\rangle &= (\beta_0\langle b1| - \beta_1\langle b0|) I \otimes P_b (\beta_0^*|b1\rangle - \beta_1^*|b0\rangle) \\ &= |\beta_0|^2\langle b1|I \otimes P_b|b1\rangle - \beta_0\beta_1^*\langle b1|I \otimes P_b|b0\rangle\\ &- \beta_0^*\beta_1\langle b0|I \otimes P_b|b1\rangle + |\beta_1|^2\langle b0|I \otimes P_b|b0\rangle\\ &= |\beta_0|^2|\beta_1|^2 - |\beta_0|^2|\beta_1|^2 - |\beta_0|^2|\beta_1|^2 + |\beta_0|^2|\beta_1|^2 \\ &= 0 \end{align} $$

and the probability that he measures $+1$

$$ \begin{align} \langle\psi' | I \otimes P_a |\psi'\rangle &= (\beta_0\langle b1| - \beta_1\langle b0|) I \otimes P_a (\beta_0^*|b1\rangle - \beta_1^*|b0\rangle) \\ &= |\beta_0|^2\langle b1|I \otimes P_a|b1\rangle - \beta_0\beta_1^*\langle b1|I \otimes P_a|b0\rangle\\ &- \beta_0^*\beta_1\langle b0|I \otimes P_a|b1\rangle + |\beta_1|^2\langle b0|I \otimes P_a|b0\rangle\\ &= |\beta_0|^2|\alpha_1|^2 - \beta_0\beta_1^*\alpha_0^*\alpha_1 - \beta_0^*\beta_1\alpha_0\alpha_1^* + |\beta_1|^2|\alpha_0|^2 \\ &= (\alpha_1\beta_0 - \alpha_0\beta_1)(\alpha_1^*\beta_0^* - \alpha_0^*\beta_1^*) \\ &= \left|\det \begin{pmatrix} \alpha_1 & \alpha_0 \\ \beta_1 & \beta_0 \end{pmatrix}\right|^2 \\ &= 1. \end{align} $$

Thus, Bob's measurement of $M$ on the collapsed state has deterministic outcome, as expected.

Remark on "cheating": The calculations above hint at the fact that we can skip the computation of the collapsed state $|\psi'\rangle$ and instead go directly for the Bob's measurement outcome probabilities, for example

$$ \langle\psi' | I \otimes P_a |\psi'\rangle = \frac{\langle\psi| (P_b \otimes I) \circ (I \otimes P_a) \circ (P_b \otimes I) |\psi\rangle}{\langle\psi | P_b \otimes I |\psi\rangle} = 2 \langle\psi| P_b \otimes P_a |\psi\rangle. $$

Note the appearance of an eigenspace projector for the joint observable $M\otimes M$. This suggests that using the joint observable $M \otimes M$ is not "cheating". On the contrary, calculations with the joint observable $M \otimes M$ and the step-by-step calculations that first use $M\otimes I$ and then use $I \otimes M$ are mathematically equivalent.

In general, the joint probability distribution over Alice's and Bob's measurement outcomes as well as the post-measurements state are the same whether you

  • first measure $M \otimes I$ and then $I \otimes M$ or
  • first measure $I \otimes M$ and then $M \otimes I$ or
  • measure $M \otimes M$.

The three ways of doing calculations are mathematically equivalent because $M\otimes I$ and $I \otimes M$ commute and therefore their eigenspace projectors like $P_a \otimes I$ and $I \otimes P_b$ also commute.

Note that this equivalence follows from linear algebra and does not involve relativity. In particular, it holds regardless of whether Alice's and Bob's measurements are timelike, null or spacelike separated. The order independence property is reassuring though since it means that quantum mechanics and special relativity do not come into conflict over the fact that if Alice's and Bob's measurements are spacelike separated then the order in which the measurements take place is observer-dependent due to relativity of simultaneity.

  • $\begingroup$ Step two is incorrect; if you are measuring on the first qubit, the correct operator is $I\otimes M$. That is, the first factor in the tensor product has to correspond to the last qubit(2) $\endgroup$
    – Anna Naden
    Commented Dec 21, 2021 at 23:46
  • $\begingroup$ @Anna do you have a reference for that? It seems that you mean that the order in which the basis of a tensor product vector space is defined, is different from the order in which the tensor product of matrices defines the matrix elements... Why? $\endgroup$ Commented Feb 27 at 20:25

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