Quantization

Schrödinger titled his first great paper on quantum mechanics ``Quantization as an Eigenvalue Problem''. I have alluded to the meaning of this several times already. A state specified by $v$ has quantum number $q$ for operator $Q$ if $v$ satisfies the eigenvalue equation $Qv=qv$. Borrowing the imagery of the old quantum theory, we say that $Q$ has the value $q$ in the state specified by $v$.

Other jargon says that $Q$ is ``sharp'' or ``definite'' in the state $v$. According to the quantum theory of measurement, if we prepare an ensemble of systems, all in the state $v$, and measure $Q$ in each one, we will always get the value $q$. But if $v$ is not an eigenvector of $Q$, then we will various eigenvalues of $Q$ with different probabilities. There is statistical spread, and the value of $Q$ for $v$ is not ``sharp''.

What about ``stationary''? Here we must bring in another concept I've mentioned before: the energy operator governs the time evolution of a quantum system. A state $v$ is absolutely stationary (does not change at all) if and only if it is an eigenstate of the energy operator $E$. To be a touch more precise: suppose $Ev=\omega v$. Then $v$ evolves like so:

$$v(t) = e^{i\omega t}v(0)$$

Remember that $v$ and $cv$ specify the same state for any non-zero complex $c$. So the state doesn't change. (I am also assuming that the energy operator does not change with time. A changing $E$ calls for a different equation for $v(t)$.)

Bohr's states are ``almost'' stationary. Perturbation theory deals with this ambiguity. Express the energy operator as a sum of two parts:

$$E = H+V$$

where $V$, the perturbation, is ``small'' relative to $H$. Eigenstates of $H$ will not usually be eigenstates of $E$, but they will change ``slowly'' because $V$ is ``small''¹³. Start the system in an eigenstate of $H$, $v(0)$ say. Wait awhile. The system is now in a superposition of eigenstates of $H$ - say, for simplicity, $v(1)=c_1v(0)+c_2u$. Here $u$ is another eigenvector of $H$, and $c_1$ and $c_2$ are complex numbers. If we measure $H$, we have a certain probability of finding the system in the state $u$.

In other words, the system has made a transition from one state to another, under the influence of a perturbation.

Bohr's transition picture is a special case of this. The perturbation $V$ is the electromagnetic term, representing the interaction between the electromagnetic field and the atom. In other words, $V$ is due to the ability of the atom to make transitions by absorbing or emitting a photon. Everything else is stuffed into $H$: the attraction of the nucleus, spin-orbit coupling, the constant magnetic field of the Zeeman effect. (Classical electromagnetism gives an unambiguous way to separate this constant magnetic field from the travelling field of light.) Bohr's stationary states are the eigenstates of $H$.

What has become of Bohr's notion of quantum number? For Bohr and Sommerfeld, stationary states had quantum numbers. Translated into Hilbert space language, this asserts that eigenvectors of $H$ are eigenvectors of all other operators of physical importance. Alas, life is not so easy. If operators $Q$ and $H$ commute, one can generally pick a basis of eigenvectors of both $Q$ and $H$. (Trivial exercise: prove the converse.) Some phenomena do submit to such an approach. The Paschen-Back effect does not-- as we will soon see.

© 2001 Michael Weiss

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