Shuo Mi's Blog: Good Quantum Number

Now we are going to review a basic concept, i.e. 'Good quantum number'.

From "http://theworldofsmall.blogspot.nl/2009/01/what-are-good-coordinates.html", I quote:

In classical mechanics, we can assign any number of variables to specify the state of a system. For example, for a free particle, its position, momentum, etc,. can be known and can be assigned simultaneously to specify the state of the free particle. But the story of quantum mechanics is different. If the position of a particle at some instant is correctly known, its momentum will be highly/maximally uncertain. This means, we can't assign both the position and momentum to specify the quantum mechanical state of the particle simultaneously. In this sense, the quantities (observables) which can be assigned simultaneously to a quantum system at any time to specify its state at that time are called the good quantum mumbers of the system. A set of good quantum numbers is also called a complete set of commuting observables (C.S.C.O.). These coordiantes commute each other, which means that their measurements can be made simultaneously.

From wikipedia:

In quantum mechanics, given a particular Hamiltonian $H$ and an operator $O$ with corresponding eigenvalues and eigenvectors given by $O|q_j\rangle=q_j|q_j\rangle$ then the physical quantities $q_j$ are said to be "good quantum numbers" if every eigenvector $|q_j\rangle$ remains an eigenvector of $O$ with the same eigenvalue as time evolves.
Hence, if: $O|q_j\rangle=O\sum_k c_k(0) |e_k\rangle = q_j |q_j\rangle$
then we require

$O\sum_k c_k(0) \exp(-i e_k t/\hbar)\,|e_k\rangle=q_j\sum_k c_k(0) \exp(-i e_k t/\hbar)\,|e_k\rangle$

for all eigenvectors $|q_j\rangle$ in order to call $q$ a good quantum number.
A necessary and sufficient condition for q to be good is that $O$ commute with the Hamiltonian $H$ . Proof:
Assume $[O,\,H]=0$ . If $|\psi_0\rangle$ is an eigenvector of $O$ , then we have (by definition) that $O |\psi_0\rangle=q_j | \psi_0\rangle$ , and so :

$O|\psi_t\rangle=O\,T(t)\,|\psi_0\rangle$

$=O e^{-itH/\hbar}|\psi_0\rangle$

$= O \sum_{n=0}^{\infty} \frac{1}{n!} (-i H t/\hbar)^{n} |\psi_0\rangle$

$= \sum_{n=0}^{\infty} \frac{1}{n!} (-i H t/\hbar)^{n} O |\psi_0\rangle$

$=q_j |\psi_t\rangle$

It can be said that a good classical analogue is that good quantum numbers are the equivalent of conserved quantities in classical mechanics, as their values do not change over time.In non-relativistic treatment,l and s are good quantum numbers but in relativistic quantum mechanics they are no longer good quantum numbers as L and S do not commute with H(in Dirac theory). J=L+S is a good quantum number in relativistic quantum mechanics as J commutes with H.

Shuo Mi's Blog

Monday, July 9, 2012

Good Quantum Number

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