I am trying to understand Box 2.7 on page 113 of Quantum Computation and Quantum Information book by Nielsen and Chuang. They start out with following wave function:

\begin{equation} \psi = \frac{|01\rangle-|10\rangle}{\sqrt 2} \end{equation}

Now let $\vec{v} = k(2,3,5)$ where $k$ is just normalization factor (equal to $\frac{1}{\sqrt {38}}$) so that length of $\vec{v}$ is 1. Can someone now tell me what does it even mean that Alice will measure $\vec{v} \cdot \vec{\sigma}$ on her qubit?

And what happens when Bob measures $\vec{v} \cdot \vec{\sigma}$ on his qubit?

I am looking for a worked out answer with numbers in it rather than mathematical symbols.


1 Answer 1


The Physical System (Background)

The system in Box 2.7 is a spin singlet, which physically refers to two electrons coupled such that their spin angular momentum cancels out perfectly. In many regards, two electrons in this state behave as a single particle with zero spin angular momentum.

The angular momentum of a classical particle has a continuous spectrum of possible directions and magnitude. In stark contrast, electron spin has only two possible directions (up and down) and one possible magnitude ($\frac{1}{2} \hbar$). In a spin singlet state one electron has up-spin, and the other has down-spin.

When two electrons in a singlet state are spatially separated, they continue to behave as a single particle in certain ways. This is the type of the non-local behavior that made many physicists uneasy (notably the EPR authors).

The combination of spin and electric charge in an electron gives it a magnetic dipole moment. To determine the spin state of an electron, Stern-Gerlach measurements pass electrons through an inhomogeneous magnetic field such that electrons experience a force dependent on the orientation of their magnetic moment. The result is that electrons with up and down spin are deflected in two distinct directions.

Choice of Measurement Basis, $\vec v \cdot \vec \sigma$ (Answer)

When Alice and Bob perform a Stern-Gerlach measurement on their respective electrons, they must each choose an orientation of the magnetic field described above. In other words they must pick a $\vec v$, which is equivalent to picking a point on a unit sphere. Here, the Pauli matrices ($\vec \sigma$) can be thought of as a convenient basis for real 3-space.

In your example, Alice has chosen $$A = \vec v \cdot \vec \sigma = \frac{1}{\sqrt{38}} (2,3,5) \cdot \vec \sigma = \frac{1}{\sqrt{38}} \begin{bmatrix} 5 & 2-3i \\ 2+3i & -5 \end{bmatrix}.$$

This matrix is called a single particle spin operator and represents the orientation of Alice's magnetic field relative to some agreed coordinate reference frame. This matrix is Hermitian, which means that it is "real" (as opposed to complex or imaginary) in every way that matters. Bob's decision to choose the same orientation means that the joint spin measurement operator is given by $$A \otimes A = \frac{1}{38}\begin{bmatrix} 25 & 10-15i & 10-15i & -5-12i \\ 10+15i & -25 & 13 & -10+15i \\ 10+15i & 13 & -25 &-10+15i \\ -5+12i & -10-15i & -10-15i & 25 \end{bmatrix}.$$

The wavefunction of a singlet states is given by $$\vert \psi \rangle = \frac{\vert 01 \rangle- \vert 10 \rangle}{\sqrt 2}= \begin{bmatrix} 0 \\ \frac{1}{\sqrt{2}} \\ -\frac{1}{\sqrt{2}} \\ 0 \end{bmatrix}.$$

Now we have everything we need to evaluate the wavefunction over the joint spin measurement operator with nothing more than matrix multiplication. $$\langle \psi \vert A \otimes A \vert \psi \rangle = -1,$$ where the value $-1$ represents perfect anti-correlation (as claimed in Box 2.7).

It's worth noting that this result, and any others obtained in the same manner, is not sensitive to the coordinate reference frame agreed upon by Alice and Bob. As constructed any (non-relativistic) change in coordinate reference frame is effected by a unitary transformation, $A' = U^\dagger A U$, and unitary transformations preserve inner products.

If you want to go through the calculations, here's the Matlab script I used to double check it myself.

>> A=1/38^.5*[5,2-3i;2+3i,-5]

A =

0.8111 + 0.0000i   0.3244 - 0.4867i
0.3244 + 0.4867i  -0.8111 + 0.0000i

>> AoA=kron(A,A)

AoA =

0.6579 + 0.0000i   0.2632 - 0.3947i   0.2632 - 0.3947i  -0.1316 - 0.3158i
0.2632 + 0.3947i  -0.6579 + 0.0000i   0.3421 + 0.0000i  -0.2632 + 0.3947i
0.2632 + 0.3947i   0.3421 + 0.0000i  -0.6579 + 0.0000i  -0.2632 + 0.3947i
-0.1316 + 0.3158i  -0.2632 - 0.3947i  -0.2632 - 0.3947i   0.6579 + 0.0000i

>> psi=1/2^.5*[0;1;-1;0]

psi =


>> psi'*AoA*psi

ans =

  • $\begingroup$ Thanks for your answer @Jonathan. However I am not sure if this is how the EPR experiment is advertised in the books. The books say first Alice measures her qubit, then Bob. So I don't think you can take tensor product of A with A. It should be tensor product of A with I (the identity matrix) to reflect Alice's measurement of her qubit. And if you do that you will find the wave function does not collapse to a definite state post Alice's measurement and which contradicts the gist of EPR experiment as stated in the books. $\endgroup$
    – morpheus
    Dec 29, 2020 at 20:38
  • $\begingroup$ @morpheus Very welcome. If Alice and Bob are measuring two halves of a singlet state in the same measurement basis, their results will be perfectly anti-correlated (this is the gist Box 2.7). You're right that this is a classical correlation, which is expected when the angle between detectors is zero (see e.g. here). The quantum correlations that concern EPR and Bell/CHSH tests occur when Alice and Bob measure in different bases. $\endgroup$ Dec 30, 2020 at 0:07
  • $\begingroup$ Whether Alice measures first or second doesn't change the math, but it should be clear that when Alice and Bob choose different measurement bases the joint spin measurement operator becomes $A \otimes B$. If you're interested, here is a worked example of the equations when Alice and Bob measure in different bases. $\endgroup$ Dec 30, 2020 at 0:24
  • $\begingroup$ @morpheus Based on your earlier comment, I'm not sure what your question is. The math I laid out above is conventional, but seemingly not what you're looking for to understand the meaning of $\vec v \cdot \vec \sigma$. It's not clear what you're asking. $\endgroup$ Dec 31, 2020 at 19:19
  • $\begingroup$ I posted a follow up question here as we can't discuss this over comments: quantumcomputing.stackexchange.com/questions/15353/… $\endgroup$
    – morpheus
    Jan 1, 2021 at 1:08

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