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 sping along $\vec{v}$. This means we create the observable $M = \vec{v} \cdot \vec{\sigma}$ where $\vec{\sigma}$ is the general Pauli matrix equal to sum of the 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 the corresponding eigenvector of $O$. Let $\psi^{'}$ be the collapsed wavefunction after Alice's measurement.

  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. My whole simluation can be found at:


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.

Aside: If part of the answer is that $O$ should be $M \otimes M$ I emphatically consider that as 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.


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.


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