Today, we first showed that if we have two particles A and B in the state

| 00 ⟩ + | 11 ⟩and we measure one of them in the basis

| vthen we find that the probability that we simultaneously measure them to be the same is cos_{0}⟩ = cos &phi | 0 ⟩ − sin φ | 1 ⟩

| v_{1}⟩ = sin φ | 0 ⟩ + cos φ | 1 ⟩

We now have two particles A and B, which can be separated, Suppose now that we
make a joint measurement. Quantum mechanics tells us a probability
distribution for measuring A at angle φ_{A} and B at angle
φ_{B};
each particle has a $frac12 probability of being measured as 0 and 1, and
they agree with probability $cos^{2} (φ_{A}
− φ_{B}). Common sense tells us that how we choose to measure
one does not affect the measurement process of the other particle.
Quantum mechanics, it turns out, disagrees with common sense.

Let us assume that each particle "knows" what its measurement results will be, and let us all assume that it knows deterministically. We can create a table such as the following:

This table corresponds to a set of repeated trials of the experiment. The first column is the result we would obtain if we measure particle A with angle φ=0, and so on. What does quantum mechanics say about the entries of this table? Well, since the outcome of the measurement on each particle is equally likely to be |v

A,φ=0 B,φ=π/6 A,φ=π/3 B,φ=π/2 1 1 0 0 0 0 0 1 0 0 1 1 1 0 0 0 0 1 1 1 1 0 0 0 1 1 1 0 0 0 1 1

Although I didn't explain this in class in detail, it is not hard to further reason that if, instead of a definite outcome given for each measurement on particles A and B, a probability distribution is given for the outcomes of each measurement, it is still impossible to construct such a table. One can replace each row with a random choice of definite outcomes according to the probability distributions, and the law of large numbers says that the proportion of entries in the rows still will approach the right value in the limit.

What does this mean? Philosphers of quantum mechanics can argue about it endlessly. One interpretation is that you cannot ask counterfactual questions. That is, because the two measurements of particle A are non-commuting, you are not allowed to write down a table that contains both of them, the way the above table does. People who believe in classical probability are compelled to say that the choice of measurement performed on particle B (together with its outcome) somehow alters the probability distribution of the outcomes on the measurement of particle A. I'm going to try to remain agnostic about what is really going on in this class.

We next talked about commuting observables and simultaneously diagonalizable Hermitian matrices. Suppose we have two observables, i.e., Hermitian matrices. If they commute, then they are simultaneously diagonalizable (theorem from linear algebra), so you can find a measurement basis which will reveal the values of both observables. On the other hand, if they do not commute, then there is no way to ascertain the value of both simultaneously. For an example relating to what we will be doing next lecture, let us define on two qubits

JWe showed J_{x}= σ_{x}⊗ id + id ⊗ σ_{x}

J_{y}= σ_{y}⊗ id + id ⊗ σ_{y}

J_{z}= σ_{z}⊗ id + id ⊗ σ_{z}

0These are the Bell states. Note that the Bell state 1/√2 ( | 01 ⟩ − | 10 ⟩ ) has 0 angular momentum in all directions. In physics, this state is called the "singlet state" while the other three are called the "triplet states."

J _{x}^{2}J _{y}^{2}J _{z}^{2}1/√2 ( | 00 ⟩ + | 11 ⟩ ) 1 0 1 1/√2 ( | 00 ⟩ − | 11 ⟩ ) 1 1 1/√2 ( | 01 ⟩ + | 10 ⟩ ) 1 1 0 1/√2 ( | 01 ⟩ − | 10 ⟩ ) 0 0 0