Hermitian observables.

So much for quantum states. Next I look at measurement. Generally, a measurement of something in quantum mechanics (spin, energy, momentum ...) is associated with a Hermitian operator, say $A$, on the Hilbert space of states. In finite dimensional cases, possible results of the measurement must be eigenvalues of $A$. If $v$ is an eigenvector of $A$ with eigenvalue $\lambda$, then measuring a system with state vector $v$ will always give you the result $\lambda$.

What happens if the state vector of the system is not an eigenvector? Here we skate perilously close to philosophy! To make things simple⁴, suppose $(v_1,...,v_n)$ is an orthonormal basis for $H$ consisting of eigenvectors for $A$, say $Av_i=\lambda_i v_i$. Suppose the system has state vector $w=a_1 v_1 +\ldots+a_n v_n$, and assume $w$ is normalized (i.e., $\vert w\vert=1$). The so-called ``collapse'' interpretation of quantum mechanics says: (i) The measurement process forces the system to jump randomly to a state with state vector $v_1$, or $v_2$, or ... or $v_n$. (ii) The probability that the system jumps to a state with state vector $v_i$ is $\vert a_i\vert^2 = \vert\langle w, v_i\rangle\vert^2$. (iii) If the system ends up with state vector $v_i$, then the measurment yields result $\lambda_i$. (Note that you'll get real-valued measurement results because $A$ is Hermitian.)

Nearly everyone agrees that the collapse interpretation will correctly predict the results of experiments. Whether it is what's ``really going on'' is the subject of endless debates. What we have here are rules for calculating probabilities. At least four different philosophies have been draped around the rules.

How does measurement look for the electron? Say we want to measure the component of spin along the z-axis. Since the electron is charged, it acts like a little magnet, with north and south poles along the axis of rotation. (Circulating charge causes a magnetic field. Think of an electromagnet-- a coil of wire with an electric current flowing around in it.) A physicist would say that the electron has a magnetic moment.

We can use the magnetic moment to measure the spin. Stern and Gerlach got a Nobel prize for doing just that. They sent a beam of electrically neutral silver atoms through a magnetic field. It turns out that the magnetic moments of the electrons in a silver atom cancel out in pairs except for one electron, so we can pretend (so far as the spin is concerned) that we're looking at a beam of electrons passing through a magnetic field.

The magnetic field was designed to produce a force on the electrons. An electron with spin pointing up would look like an magnet, and would experience an upward force; an electron with spin pointing down would look like an magnet, and would experience a downward force. Classically, you would expect an electron with spin at angle $\alpha$ to the vertical to experience an upward force proportional to $\cos\alpha$.⁵ So the electron beam should be spread out into a vertical smear, according to classical mechanics.

In fact, the beam splits into two beams, one up, one down. In other words, if we measure the component of the spin along the vertical axis, we always find that the spin is entirely up or entirely down. This is the most basic sense in which the ``spinning ball'' analogy is wrong. The same two-valued behavior holds for any measurement axis.

Classically this is inexplicable. How can the electron have spin up and spin sideways at the same time? Answer: it doesn't. After you've measured the spin along the z-axis, the electron has vertical spin (say spin up). If you take your vertically spinning electron and measure its spin along the x-axis, you have a 50-50 chance at getting spin left or spin right. If you now repeat the spin measurement along the z-axis, you have a 50-50 chance of getting spin up or spin down. The x-axis measurement has destroyed the information obtained from the z-axis measurement.

Let $A$ be the Hermitian operator corresponding to ``measure the spin along the z-axis''. The eigenvalues (i.e., possible results) will be 1 and $-1$, if we choose our units right. Pick a basis of two eigenvectors; then the matrix for $A$ in this basis is just $\left[ \begin{array}{cc} 1 & 0\\ 0 & -1\\ \end{array}\right]$, i.e., the Pauli matrix $\sigma_z$. (Common notation for the eigenvectors is $\vert{\rm up}\rangle$ and $\vert{\rm down}\rangle$, although $\vert{\rm dead}\rangle$ and $\vert{\rm alive}\rangle$ are popular for that other famous two-state system, Schrödinger's Cat.)

If $\sigma_z$ is here, can $\sigma_x$ and $\sigma_y$ be far behind? In fact, these are the matrices for measuring the x-component (respectively y-component) of spin, provided we continue to use the same basis $\vert{\rm up}\rangle$ and $\vert{\rm down}\rangle$.

It turns out that $(1:0)$ represents spin up along the z-axis, $(1:1)$ represents spin along the x-axis, and $(i:1)$ represents spin along the y-axis. In a different notation, $\vert{\rm up}\rangle +\vert{\rm down}\rangle$ and $i\vert{\rm up}\rangle +\vert{\rm down}\rangle$ are the state vectors for these two spin directions. The x-axis and y-axis state vectors are not eigenvectors of $\sigma_z$. The rules for calculating probabilities (clothed in any philosophy you like) yield the 50-50 chances mentioned earlier.

As an exercise, you may like to chew on these remarks: if two measurements can be done simultaneously, then the associated Hermitian operators must have the same set of eigenvectors, and so the operators must commute. But the $\sigma$ matrices don't commute. This accounts mathematically for the non-intuitive (or at least non-classical) results of the Stern-Gerlach experiment. The Heisenberg uncertainty principle stems from the same sort of considerations.

Now a general comment. Any linear operator on a Hilbert space $H$ induces a mapping on the space of states, since if $x$ and $c x$ are two state vectors for the same state, then $Ax$ and $Acx=cAx$ will represent the same state. Can I dispense with the Hilbert space entirely and just work with the space of states and the induced mappings? The answer is yes, but it would be inconvenient. If $A$ is an observable with eigenvector $v$, say $Av=\lambda v$, then the eigenvalue $\lambda$ has physical significance. But when we look at the action of $A$ on the space of states, all we notice (at first) is that $A$ leaves the state corresponding to $v$ fixed.

Nonetheless, the results of measurement are encoded in the action of $A$ on the space of states. $A$ and $cA$, $c\neq 0$, induce the same mapping on the states, and the converse holds for the cases of interest to us (if $A$ and $B$ induce the same state mapping, then $A=cB$ for some scalar $c\neq 0$). This scalar $c$ will be real for Hermitian $A$ and $B$. If $c\neq 1$, then $A$ and $B$ really represent the same measurement, but expressed in different units (e.g., foot-pounds vs. ergs.)

Example: suppose $Av=\lambda v$ and $Aw=\mu w$, and $Bv=\lambda'v$, $Bw=\mu w$, with $\lambda\neq\lambda'$. $A$ and $B$ do the same thing to the quantum states determined by $v$ and $w$ - namely, the states are left fixed. However, $A$ and $B$ send the state determined by $v+w$ to different states.

© 2001 Michael Weiss

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