Expectations and uncertainties

Expectations and averages

Two ways we will denote averages of repeated measurements, say of $S_z$, are: $$ \nonumber \overline S_z \text{ or } \langle S_z \rangle.$$ This time we'll use the angled brackets, because of its visual resemblance to one way to calculate it, which you're about to see...

Consider repeated measurements of $S_z$ made, each time, on a state that has been previously prepared as a superposition $\ket \psi = a\ket + + b\ket-$. The average value, or expectation value is... [This is quite different from *successive* measurements...] $$\langle S_z \rangle=+\frac \hbar 2\P_+ -\frac\hbar 2\P_-.$$

Using projection operators, McIntyre shows in Eqs (2.75) and (2.76) that this can be written as: $$\langle S_z\rangle =\bra\psi \hat S_z \ket \psi.$$ To actually calculate using this expression:

Group this as $\bra \psi \big(\hat S_z\ket \psi\big)$.
Evaluate the expression in (...), which will give you a column vector,
Then evaluate the inner product of $\bra \psi$ with that column vector.

Calculate $\langle S_z\rangle$ if $\ket \psi=\ket -$. $\bra - \hat S_z \ket - = -\hbar/2$.
Calculate $\langle S_z\rangle$ if $\ket \psi=\ket +_x$. ${}_x\bra + \hat S_z \ket +_x = 0$.

For the second case, were any of the individual measurement of $S_z$ equal to $\langle S_z\rangle$? No! Any measurement of $S_z$ will only ever result in $+\hbar/2$ or $-\hbar/2$. In this case, $\P_+=1/2$ and $\P_-=1/2$ so the expectation value is $+1/2\times\hbar/2 -1/2\times \hbar/2=0$.

Uncertainty

We saw that $\sigma$, the spread in measurement values of a random variable $w$ was: $$\sigma^2 \equiv \overline{ (\Delta w)^2 } = \overline{ (w - \overline{ w })^2 }=\overline{w^2}-\overline{w}^2 .$$ where the last expression is the easiest way to calculate it.

So, for an operator $\hat A$ that represents some observable, we can define the spread in measurements as:

Measurement uncertainty $\Delta A$: $$\Delta A \equiv \sqrt{ \langle \hat A^2\rangle -\langle \hat A \rangle^2}.$$

The operator $\hat A^2$ means "apply the operator twice", that is $$\hat A^2\ket \psi = \hat A \big(\hat A\ket \psi \big).$$
This is the underlying measurement spread solely due to the probabilistic nature of quantum measurement, before taking into account any measurement apparatus uncertainty.

Make sure that you understand the circumstances and calculations on P53-54. Can you write down a plausible sample of 10 measurements for each of these cases?

Measuring $S_z$ on a beam prepared in the $\ket +$ state: $$ \langle \hat S_z \rangle = +\hbar/2\ \ \text{ and }\ \ \Delta S_z = 0.$$
Measuring $S_z$ on a beam prepared in the $\ket +_x$ state: $$ \langle \hat S_z \rangle = 0\ \ \text{ and }\ \ \Delta S_z = \hbar / 2.$$

Compatibility and Commutators

Consider spin experiment 3:

Swap the $X$ analyzer and the second $Z$ analyzer. Same connections but in the order $Z$ $Z$ $X$. Out of 10,000 counts, roughly what would expect in the various counters?

Now hook up the experiment on the Spins Experiment simulator, and try it!

Apparently, the order of measurement matters! In this case: $$\hat Z \hat X \ket + \neq \hat X \hat Z \ket +.$$ Or... $$\begineq \Rightarrow 0 & \neq \big(\hat Z \hat X-\hat X\hat Z\big)\ket +\\ 0&\neq [\hat Z, \hat X]\ket +\endeq$$ The [] is a new operator, defined as:

The commutator of two operators is defined as: $$[\hat A, \hat B]\equiv \hat A\hat B-\hat B\hat A$$

If $[\hat A,\hat B]\neq 0$, we say that the two operators are incompatible. It is impossible to prepare a state $\ket \psi$ which is both an eigenstate of $\hat A$ and an eigenstate of $\hat B$.

On the other hand, if $[\hat A, \hat B]=0$, then the order of measurement does not matter. The two operators (and the measured observables) are "compatible", and an eigenstate of one operator must be an eigenstate of the other.

Calculate the matrix representation of $[S_z,S_x]$ to show that these two are incompatible.

Uncertainty Principle $$\Delta\hat A\Delta \hat B \geq \frac 12\Big|\ \langle[\hat A, \hat B]\rangle \ \Big|$$