Compatible and incompatible measurements

Compatible measurements

We say that two measurments, corresponding to two operators \(A, B\), are compatible if we can specify the values of both in an experiment. That is, it makes sense to say that a state has an eigenvalue \(a\) of \(A\) and an eigenvalue \(b\) of \(B\). Thus we must be able to diagonalize both in the same basis. As we showed, iin Chapter 3, this is possible if and only of \([A,B] = 0\).

In many cases the operators we study have degenerate spectra. In this case, the measurement of a given operator does not lead to a unique stateafter the wavefunction collapses. In this case we try to search for a complete set of commuting observables (CSCO); that is, operators \(A_1,\ldots,A_k\) such that \([A_i,A_j] = 0\), and such that if we specify eigenvalues \(A_1,\ldots,a_k\) of each of these, we uniquely specify the state.

In some cases this could be a single operator (eg, an operator without degeneracy). An example in which we need several is the nonrelativistic hydrogen atom. The energy levels are highly degenerate; we split them by considering eigenvlaues of the angular momentum operators \({\vec L}^2\), \(L_z\), \({\vec S}^2\), \(S_z\).

Incompatible measurements and the uncertainty principle

Recall that we defined the variance of an observable in a given state by:

(258)\[(\Delta A)^2 = \bra{\psi} (A - \vev{A})^2\ket{\psi}\]

where \(\vev{A} = \bra{\psi}A\ket{\psi}\) is the expectation value of the observable in taht state. \(\vev{A}\) represents the average of the outcomes of measuring \(A\) in a given state (where we prepare this idential state and measure \(A\) in it many times). \((\Delta A)^2\) is an estimate of the spread or variancce in measurements of \(A\); a cartoon of this is shown here:

When is the variance zero? For a Hermitian operator t requires that

(259)\[\begin{split}\begin{align} \bra{\psi} (A - \vev{A})^2\ket{\psi} & = \bra{\psi} (A - \vev{A})^{\dagger}(A - \vev{A})\ket{\psi} \\ & = ||(A - \vev{A})\ket{\psi}||^2 \end{align}\end{split}\]

But the norm of a vector is zero if and only if the vector is zero, which means that \(A\ket{\psi} = \vev{A}\ket{\psi}\), that is, \(\ket{\psi}\) is an eigenvactor and \(\vev{A}\) the associated eigenvalue.

Now if we have two observables \(A,B\), and they commute, then we can specify them both perfectly and it is possible for both of their variances to be zero. If they do not commute it is a different story.

Let us consider the general case \([A,B] = i C\) for \(A,B,C\) Hermitian operators. The Uncertainty Principle in this general case is state in the following theorem:

(260)\[\Delta A \Delta B \geq \half \vev{C}\]

Proof. If the system is in state \(\ket{\psi}\), we can define

(261)\[\begin{split}\begin{align} \ket{f_A} & = (A - \vev{A})\ket{\psi}\\ \ket{f_B} & = (B - \vev{B})\ket{\psi} \end{align}\end{split}\]

Now, by the Schwarz inequality,

\[\begin{split}\begin{align} & |\brket{f_B}{f_A}| \leq ||\ket{f_A}||\ ||\ket{f_B}||\\ & \Rightarrow |\brket{f_B}{f_A}|^2 \leq \brket{f_A}{f_A}\brket{f_B}{f_B}\\ & \Rightarrow |\brket{f_B}{f_A}|^2 \leq (\Delta A)^2 (\Delta B)^2 \end{align}\end{split}\]

Now we need to show that the left hand side of this last inequality is an upper bound on \(\frac{1}{4}\vev{C}^2\).

\[\begin{split}\begin{align} |\brket{f_A}{f_B}|^2 & = (\text{Im} \brket{f_A}{f_B})^2 + (\text{Re} \brket{f_A}{f_B})^2 \\ & = (\text{Im}(\bra{\psi}(A - \vev{A})(B - \vev{B})\ket{\psi}))^2 + (\text{Re}(\bra{\psi}(A - \vev{A})(B - \vev{B})\ket{\psi}))^2\\ & = (\frac{1}{2i}(\bra{\psi}(A - \vev{A})(B - \vev{B})\ket{\psi} - \bra{\psi}(B - \vev{B})(A - \vev{A})\ket{\psi}))^2\\ & \ \ + (\half((\bra{\psi}(A - \vev{A})(B - \vev{B})\ket{\psi} + \bra{\psi}(B - \vev{B})(A - \vev{A})\ket{\psi})))^2\\ & = \left(\frac{1}{2i}\bra{\psi}[A,B]\ket{\psi}\right)^2 + \left(\half\bra{\psi}\{A,B\}\ket{\psi} - \vev{A}\vev{B}\right)^2\\ & \geq \frac{1}{4} \vev{C}^2 \end{align}\end{split}\]

When is this inequality saturated? Looking closely at the proof above, his would require

\[|\brket{f_B}{f_A}|^2 = \brket{f_A}{f_A}\brket{f_B}{f_B} = \frac{1}{4} \vev{C}^2 + (\text{Re}\brket{f_B}{f_A})^2 = \frac{1}{4}\vev{C}^2\]

By the Schwarz inequality, this means that \(\ket{f_A} = \beta \ket{f_B}\). If we then demand that \((\text{Re}\brket{f_B}{f_A})^2 = 0\) as required above, this means that \(\beta\) is pure imaginary, \(\beta = i \lambda\) for \(\lambda\) real.

Example: spin-half

As a simple example, consider measuring \(S_x, S_y\) in a spin-\(\half\) system. Here \([S_x,S_y] = i \hbar S_z\). Now if the state is \(\ket{+,x}\) you can show that \((\Delta S_x)^2 = 0\), \((\Delta S_y)^2 = \frac{\hbar^2}{4}\), \(\vev{S_z} = 0\). On the other hand, if we choose \(\ket{\psi} = \ket{+,z}\), \((\Delta S_x)^2 = (\Delta S_y)^2 = \frac{\hbar^2}{4}\), and \(\vev{Sz} = \frac{\hbar}{2}\). Here the inequality is in fact saturated, which it does not in general have to be.

Example: particle on a line

The classic example is the psotion-momentum uncertainty principle. As we have sh own, \([{\hat x},{\hat p}] = i\hbar\). The Heisenberg Uncertainty Principle is:

\[(\Delta x)(\Delta p) \geq \frac{\hbar}{2}\]

in other words, tehre is no circumstance in which we can sepcify both \(x, p\) with infinte precision. Of we specify \(x\) with complete precision, then \(\Delta p\) must be infinite. (And thus the kinetci energy would have to be infinite, which is why one could never actually prepare such a state!). Similarly, if we psecify \(p\) with infinite precision, \(x\) has infinite variance, as is make clear from the associated wavefunction

\[\brket{x}{p} = \frac{1}{\sqrt{2\pi \hbar}} e^{i p x/\hbar}\]

An interesting question is, what state saturates the uncertainty? We need

\[({\hat p} - \vev{p})\ket{\psi} = i\lambda({\hat x} - \vev{x})\ket{\psi}\]

wheer \(\ket{\psi}\) is specified by the wavefunction \(\psi(x)\). Given what we know about the action of \({\hat p}, {\hat x}\) on this wavefunction, we can write this as a differential equation

\[\frac{\hbar}{i} \frac{\del}{\del x} \psi - \vev{p}\psi = i \lambda(x - \vev{x})\psi\]

This is a differential equation we can solve; the solution is

\[\psi(x) = A e^{i\vev{p}x/\hbar} e^{-\frac{\lambda}{2\hbar}(x - \vev{x})^2}\]

Thsi is a Gaussian wavepacket modulated by a sinusoid. The width of the wavpacket is proportional to \(\sqrt{\frac{\hbar}{\lambda}}\).