Lecture 4

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
| v0 ⟩ = cos &phi | 0 ⟩ − sin φ | 1 ⟩
| v1 ⟩ = sin φ | 0 ⟩ + cos φ | 1 ⟩
then we find that the probability that we simultaneously measure them to be the same is cos2 φ (meaning this is the probability we obtain | b ⟩ and | vb ⟩ for b either 0 or 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 φA and B at angle φB; each particle has a $frac12 probability of being measured as 0 and 1, and they agree with probability $cos2 (φA − φB). 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:

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
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 |v0⟩ or |v1⟩, as the table grows bigger, the proportion of 0s in each column must approach ½. Similarly, the proportion of changes between the first column and the second should approach ¼. However, we also know that the first column and the last column must be entirely different, since we are measuring them at an angle of φ=0 and φ=π/2. It is easy to see that these two conditions are incompatible. If every row changes at least once between its first entry and its last entry, then each pair of adjacent columns on average must contain at least 1/3 differences. Thus, quantum mechanics tells us that we cannot write down such a table.

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

Jx = σx ⊗ id + id ⊗ σx
Jy = σy ⊗ id + id ⊗ σy
Jz = σz ⊗ id + id ⊗ σz
We showed Jx2, Jy2, and Jz2 all commute, so by the theorem above they have simultaneous eigenvectors. These eigenvectors, together with their eigenvalues, are
0
  Jx2 Jy2 Jz2
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
These 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."