Today, the first part of the lecture was on the quantum mechanics of spin-1/2 particles. We introduced the Pauli matrices σx, σy, and σz, and gave their commutation relations:
σx σy = −σy σx = i σzand
σy σz = −σz σy = i σx
σz σx = −σx σz = i σy
σx2 = σy2 = σz2 = 1The three Pauli matrices and the identity for a basis for the space of 2 × 2 matrices.
We reviewed von Neumann, or projective, measurements. What I'm calling a complete projective measurement corresponds to an orthonomal basis. If a quantum state is expressed in this basis as
α1 | e 1 〉 + α2 | e 2 〉 + ... + αn | e n 〉Then the measurement result | ek 〉 is observed with probability αk2.
We then explained how a Hermitian matrix corresponds to an observable. To find the expected value of an observable M on a state | ψ 〉 , we can compute
〈 ψ | M | ψ 〉which may be a lot simpler than finding all the probabilities of measurement outcomes and then summing the value of the observable on these to obtain the expectation.
We next showed how a spin in the v = (α,β,γ) direction corresponds to the Hermitian matrix
α σx + β σy + γ σzwhich we will call v⋅σ. We proved that this matrix has eigenvalues ±1. For a spin ½ particle, the observable ½ v⋅σ is the spin along the v axis. To obtain the unitary matrix giving a rotation around the v-axis by an angle θ, we need to exponentiate
exp(- ½ i θ v⋅σ)(Recall the imaginary exponential of a Hermitian matrix is a unitary matrix.)
In the second part of the lecture, we talked about joint state spaces and tensor products. I'm going to go over this in much more detail on Tuesday, so don't worry if you didn't understand it completely. Anyway, the space that is the tensor product of two linear spaces A and B is the linear closure of states | a 〉 | b 〉. The state | a 〉 | b 〉 is itself the tensor product of states | a 〉 ∈ A and | b 〉 ∈ B. Tensor products can be represented several different ways.
| a 〉 | b 〉 = | a b 〉 = | a 〉 ⊗ | b 〉This tensor product space contains vectors which are not themselves tensor products. These states are called entangled. One such state is
( | 00 〉 + | 11 〉 ).We calculated that this state was invariant under the change of basis
| v0 〉 = cos φ | 0 〉 + sin φ | 1 〉so
| v1 〉 = sin φ | 0 〉 − cos φ | 1 〉
1 / √2 ( | 00 〉 + | 11 〉 ) = ( | v0 〉 | v0 〉 + | v1 〉 | v1 〉 )On Tuesday, I will be taking this calculation and using it to show how this state gives rise to the EPR paradox, which was one of the earliest demonstrations of the weirdness of quantum mechanics.