Quantumania 12 - Unexpected Values

We’ve encountered two kinds of operator, unitary (including rotations, reflections and phase factors), and hermitian (the kind that scales in orthogonal directions.)

In QM the role of hermitian operators is to represent observables. This sounds more abstract than it actually is: we’ve seen how if you have a qubit, its state vector can be pointing in any direction in the state space (which corresponds to a direction in physical space), but to find anything out about it you have to choose an alignment in physical space, and the only thing you can discover is a single bit (in information theory terms): which way along your chosen alignment it points (as of now, since you measured it.) Each possible answer corresponds to a different post-measurement state vector, and for one observable, all the possible post-measurement state vectors are mutually orthogonal.

Every observable has a hermitian operator that scales along its post-measurement alignments. So the $\vert u \rangle$ and $\vert d \rangle$ vectors of the Z basis get scaled differently by the operator for “measure along Z”. If we represent the operator as a matrix in the Z basis, it’s a diagonal matrix, and the values on the diagonal are the eigenvalues. For $\vert u \rangle$, the value is $+1$ and for $\vert d \rangle$, it’s $-1$.

Now, we know that the state vector’s coordinates in a measurement basis are each directly related to the probability of getting the measurement outcome represented by that coordinate’s basis vector. The modulus squared of the coordinate is the probability, which we get by multiplying the coordinate by its complex conjugate.

And we just learned that we can do this for all the coordinates, sum them, and get $1$, expressed by this notation, a bra-ket combination:

\[\langle \psi \vert \psi \rangle\]

which in terms of matrices is just matrix multiplication between the column vector on the right and a row vector on the left, the latter containing the complex conjugates of the coordinates on the right. This literally just means we’re finding the modulus squared of all the coordinates and summing them. An exhaustive set of mutually exclusive probabilities has to sum to $1$, but also this is the inner product of a unit vector with itself, which also means it has the value $1$.

So what happens if we insert our hermitian operator for our observable in the middle?

\[\langle \psi \vert H \vert \psi \rangle\]

The symmetry of the notation is indicative of the associative nature of matrix multiplication: it doesn’t matter which side we multiply first. But the extra ingredient provided by the central operator is the measured values: the eigenvalues associated with each measurement outcome. Because it’s a diagonal matrix (in its own basis), there’s no “crosstalk” between the dimensions. Each coordinate is being scaled by its measurable value, as well as being converted into a probability, so we’re in effect multiplying measurable values by the probability of obtaining them, and finally summing all those numbers.

This is the probability-weighted average of the measurable values, often called the expectation value by physicists, and unfortunately called the “expected” value in many other fields. Why unfortunate? Because it may well be a value that it is impossible to measure in any given experiment, so it hardly seems right to call it expected. Then again, “expectation” isn’t super clear about that either. It’s really the average result of making the same measurement many times on identically prepared systems.

The notation for it is also a bit confusing in this context:

\[\langle H \rangle\]

It looks like we’ve erased the state vectors from Dirac’s notation, but this is just a notational overlap. The angle brackets are used for the expectation value of a variable in statistics, never mind QM.

Here’s how it works when our observable is “measure along the Z axis”, so our basis vectors are $\vert u \rangle$ and $\vert d \rangle$ and their eigenvalues are $+1$ and $-1$ respectively, meaning that $Z = \operatorname{diag}(1, -1)$. We’re going to work out:

\[\langle Z \rangle = \langle \psi \vert Z \vert \psi \rangle\]

Expressing our state vector as a linear combination of the basis vectors (I’ve once again chosen numbers that will come out neat when squared):

\[\vert \psi \rangle = 0.8367 \vert u \rangle + 0.5477 \vert d \rangle\]

Apply $Z$ to it (it just flips the sign of the down coordinate):

\[Z \vert \psi \rangle = 0.8367 \vert u \rangle - 0.5477 \vert d \rangle\]

And finally applying $\langle \psi \vert$:

\[\langle Z \rangle = 0.7 - 0.3 = 0.4\]

The only values we can possibly measure are $+1$ and $-1$, but the expectation value comes out as $0.4$, because I set up $\vert \psi \rangle$ to be weighted more toward up than down. If I’d made both coordinates $1/\sqrt{2}$ then $\langle Z \rangle$ would have come out as zero, and if $\vert \psi \rangle = \vert u \rangle$ then it would simply be $1$.