Good quantum number

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 numbers (or the eigenvalues) $q_{j}$ 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(-ie_{k}t/\hbar )\,|e_{k}\rangle =q_{j}\sum _{k}c_{k}(0)\exp(-ie_{k}t/\hbar )\,|e_{k}\rangle

for all eigenvectors $|q_{j}\rangle$ in order to call $q$ a good quantum number (where $e_{k}$ s represent the eigenvectors of the Hamiltonian)

Theorem: A necessary and sufficient condition for q (which is an eigenvalue of an operator O) to be good is that $O$ commutes with the Hamiltonian $H$

Proof: Assume $[O,\,H]=0$ .

|\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

=Oe^{{-itH/\hbar }}|\psi _{0}\rangle

=O\sum _{{n=0}}^{{\infty }}{\frac {1}{n!}}(-iHt/\hbar )^{{n}}|\psi _{0}\rangle

=\sum _{{n=0}}^{{\infty }}{\frac {1}{n!}}(-iHt/\hbar )^{{n}}O|\psi _{0}\rangle

=q_{j}|\psi _{t}\rangle

Ehrenfest Theorem and Good Quantum Numbers

Ehrenfest Theorem ^[1] gives the rate of change of the expectation value of operators. It reads as follows:

{\frac {d}{dt}}\langle A(t)\rangle =\left\langle {\frac {\partial A(t)}{\partial t}}\right\rangle +{\frac {1}{i\hbar }}\langle [A(t),H]\rangle

Commonly occurring operators don't depend explicitly on time. If such operators commute with the Hamiltonian, then their expectation value remains constant with time. Now, if the system is in one of the common eigenstates of the operator $A$ (and $H$ too), then system remains in this eigenstate as time progresses. Any measurement of the quantity $A$ will give us the eigenvalue (or the good quantum number) associated with the eigenstates in which the particle is. This is actually a statement of conservation in Quantum Mechanics.

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 $H$ .

Conservation in Quantum Mechanics

Case I: Stronger statement of conservation : When the system is in one of the common eigenstates of H and A

Let A be an operator which commutes with Hamiltonian. This implies that we can have an orthonormal basis of common eigenvectors of A and H in the vector space consisting of the states of our system.^[2] A measurement of A upon the system is bound to yield one of the eigenvalues of A.^[3] Assume that our system is in one of the common basis eigenstates of A and H. If we make a measurement of A upon the system, it will definitely yield the eigenvalue of A (or the good quantum number) corresponding to the eigenstate in which the system is. This is because the probability of getting an eigenvalue of an operator is the square of the coefficient attached with the corresponding eigenstate (which in this case is 1) that occurs in the expansion of the state of the system as a linear combination of the basis states. Even if the system is left to evolve for a certain time before the measurement is made, it will still yield the same eigenvalue.^[4] This is because the state of the system doesn't change physically (although it does change mathematically; a phase term gets attached). Such states are called stationary states. Now,after one measurement, if we go on measuring A again and again, on the same system, we keep on getting the same value; i.e. the same eigenvalue or the good quantum number. This indicates a kind of conservation in Quantum Mechanics. Here the standard deviation of A is zero. A short proof is as follows:^[5]

$\sigma ^{2}=\langle (A-\langle A\rangle )^{2}\rangle$
$=\langle \psi |(A-a)^{2}\psi \rangle$
(a is the eigenvalue of A for the state in which the system is. Now, $\langle A\rangle =a$ because every measurement of A always yields a )
$=\langle (A-a)\psi |(A-a)\psi \rangle =0$

(Use has been made of the fact that both A and an are hermitian; thus (A-a) is also hermitian; so we can transfer one (A-a) term to the 'bra' side in the inner product.)

Conclusions: The state doesn't change. So, almost every physical quantity is conserved (doesn't change with time)

Case II: Weaker statement of conservation : When the system is not in any of the common eigenstates of H and A

As assumed in case I, [A,H]=0. Also assume that the system is not in any of the common eigenstates of H and A. In this case, the system must be some linear combination of the basis (common) eigenstates of H and A. When a measurement of A is made, it can yield any of the eigenvalues of A. And then, if any number of subsequent measurements of A are made, it is bound to yield the value which was obtained on the first measurement of A. In this case, there holds a (weaker) statement of conservation. Using the Ehrenfest Theorem : ${\frac {d}{dt}}\langle A(t)\rangle =\left\langle {\frac {\partial A(t)}{\partial t}}\right\rangle +{\frac {1}{i\hbar }}\langle [A(t),H)]\rangle =0$
since the operator A commutes with H and we have assumed that A doesn't depend on time explicitly . It says that the expectation value of the operator A remains constant in time.^[6] When the measurement is made on identical systems again and again, in general, it will yield different eigenvalues, but for any such identical system the expectation value of A remains constant. It is this kind of conservation that is valid in this case. This is a weaker conservation condition than the case when our system was a common eigenstate of A and H (discussed above).

Conclusions: The system evolves in time. Here fewer quantities are conserved as compared to case I. The quantities that are conserved (expectation value of the operator A) don't have the value equal to the good quantum numbers.

Analogy with Classical Mechanics

In classical mechanics, the total time derivative of a physical quantity $A$ is given as:^[7]

\frac{dA}{dt} = \frac{\partial A}{\partial t} + \{A, H\}

where the curly braces refer to Poisson bracket of $A$ and $H$ . This bears a striking resemblance to the Ehrenfest Theorem. It implies that a physical quantity $A$ is conserved if its Poisson Bracket with the Hamiltonian vanishes and the quantity does not depend on time explicitly. This condition in classical mechanics is analogous to the condition in quantum mechanics for the conservation of an observable (as implied by Ehrenfest Theorem: Poisson bracket is replaced by commutator)

Systems which can be labelled by good quantum numbers

Systems which can be labelled by good quantum numbers are actually eigenstates of the Hamiltonian. They are also called stationary states.^[8] They are so called because the system remains in the same state as time elapses, in every observable way. The states changes mathematically, since the complex phase factor attached to it changes continuously with time, but it can't be observed.

Such a state satisfies:

{\hat H}|\Psi \rangle =E_{{\Psi }}|\Psi \rangle

where

$|\Psi \rangle$ is a quantum state, which is a stationary state;
$\hat H$ is the Hamiltonian operator;
$E_{{\Psi }}$ is the energy eigenvalue of the state $|\Psi \rangle$ .

The evolution of the state ket is governed by the Schrödinger Equation:

i\hbar {\frac {\partial }{\partial t}}|\Psi \rangle =E_{{\Psi }}|\Psi \rangle

It gives the time evolution of the state of the system as:

|\Psi (t)\rangle =e^{{-iE_{{\Psi }}t/\hbar }}|\Psi (0)\rangle

Examples

The hydrogen atom: no spin-orbit coupling

In the case of the hydrogen atom (with the assumption that there is no spin-orbit coupling), the observables that commute with Hamiltonian are the orbital angular momentum, spin angular momentum, the sum of the spin angular momentum and orbital angular momentum, and the $z$ components of the above angular momenta. Thus, the good quantum numbers in this case, (which are the eigenvalues of these observables) are $l,j,m_{{\text{l}}},m_{s},m_{j}$ .^[9] We have omitted $s$ , since it always is constant for an electron and carries no significance as far the labeling of states is concerned.

Good quantum numbers and CSCO

However, all the good quantum numbers in the above case of the hydrogen atom (with negligible spin-orbit coupling), namely $l,j,m_{{\text{l}}},m_{s},m_{j}$ can't be used simultaneously to specify a state. Here is when CSCO (Complete set of commuting observables) comes into play. Here are some general results which are of general validity :

1. A certain number of good quantum numbers can be used to specify uniquely a certain quantum state only when the observables corresponding to the good quantum numbers form a CSCO.

2. If the observables commute, but don't form a CSCO, then their good quantum numbers refer to a set of states. In this case they don't refer to a state uniquely.

3. If the observables don't commute they can't even be used to refer to any set of states, let alone refer to any unique state.

In the case of hygrogen atom, the $L^{2},J^{2},L_{z},J_{z}$ don't form a commuting set. But $n,l,m_{{\text{l}}},m_{s}$ are the quantum numbers of a CSCO. So, are in this case, they form a set of good quantum numbers. Similarly, $n,l,j,m_{{\text{j}}}$ too form a set of good quantum numbers.

The hydrogen atom: spin-orbit interaction included

If the spin orbit interaction is taken into account, we have to add an extra term in Hamiltonian which represents the magnetic dipole interaction energy.^[10]

\Delta H_{{\text{SO}}}=-{\boldsymbol {\mu }}\cdot {\boldsymbol {B}}.

Now, the new Hamiltonian with this new $\Delta H_{{\text{SO}}}$ term doesn't commute with ${\boldsymbol {L}}$ and ${\boldsymbol {S}}$ ; but it does commute with L², S² and ${\boldsymbol {J}}$ , which is the total angular momentum. In other words, $l,j,m_{{\text{l}}},m_{s}$ are no longer good quantum numbers, but $l,j,m_{{\text{j}}}$ are.

And since, good quantum numbers are used to label the eigenstates, the relevant formulae of interest are expressed in terms of them. For example, the spin-orbit interaction energy is given by^[11]

\Delta H_{{\text{SO}}}={\beta \over 2}(j(j+1)-l(l+1)-s(s+1))