Degeneracy and measurement · Quantum Mechanics

Intuition

When several independent states share the same eigenvalue, finding that value does not say which of them the system is in. The measurement then keeps the part of the state inside that eigenvalue’s whole subspace, superposition and all, and throws away the rest. A second, compatible observable can then split the subspace further.

Sorting mail by city keeps each city’s letters in one pile without ordering them by street. The pile is everything with that city on it; sorting by street is a second, finer question.

Measuring a degenerate eigenvalue

Let $\hat{P}_{a}$ be the projector onto all eigenvectors of $\hat{A}$ with eigenvalue $a$ . The probability of the result $a$ is the squared length of the part of the state in that subspace, and the state after the measurement is that part, normalised. This refinement of the Born rule is called the Lüders rule.

P(a)=\bra{\psi}\hat{P}_{a}\ket{\psi},\qquad \ket{\psi}\longrightarrow\frac{\hat{P}_{a}\ket{\psi}}{\sqrt{P(a)}}

How it works

With an orthonormal basis $\ket{a,k}$ of the eigenspace, $P(a)=\sum_{k}|\braket{a,k|\psi}|^{2}$ : the probabilities of the states sharing the value add.

Probabilities over an eigenspace add

\bra{\psi}\hat{P}_{a}\ket{\psi}=\sum_{k}\bigl|\braket{a,k|\psi}\bigr|^{2}

Write the projector onto the eigenspace as the sum of the ket-bras of an orthonormal basis of it, and sandwich it in the state. Each term becomes the modulus squared of one overlap, so the probability of the degenerate result is the sum of the probabilities of the states that share it — the same number whichever basis of the eigenspace is used.

Proof steps

\hat{P}_{a}=\sum_{k}\ket{a,k}\bra{a,k}

The projector onto a subspace is the sum over an orthonormal basis of it.

\bra{\psi}\hat{P}_{a}\ket{\psi}=\sum_{k}\braket{\psi|a,k}\braket{a,k|\psi}

Put it between the bra and the ket of the state.

=\sum_{k}\bigl|\braket{a,k|\psi}\bigr|^{2}

Each term is an overlap times its conjugate.

\hat{P}_{a}\ \text{does not depend on the choice of}\ \ket{a,k}

The projector is fixed by the subspace alone, so the probability is too.

Applications

Practice

Add Over the Eigenspace

The probability of a degenerate result is the sum of the probabilities of all the states that share it.

P(a)=\sum_{k}\bigl|\braket{a,k|\psi}\bigr|^{2}

Try it

$\hat{A}=\operatorname{diag}(1,\ 1,\ 2)$ in the basis $\ket{1},\ket{2},\ket{3}$ and $\ket{\psi}=\tfrac12\ket{1}+\tfrac12\ket{2}+\tfrac{1}{\sqrt2}\ket{3}$ . What is the probability of the result 1?