# 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).

In other words, the eigenvalues $q_{j}$ are good quantum numbers if the corresponding operator $O$ is a constant of motion (commutes with the time evolution). Good quantum numbers are often used to label initial and final states in experiments. For example, in particle colliders:

1. Particles are initially prepared in approximate momentum eigenstates; the particle momentum being a good quantum number for non-interacting particles.

2. The particles are made to collide. At this point, the momentum of each particle is undergoing change and thus the particles’ momenta are not a good quantum number for the interacting particles during the collision.

3. A significant time after the collision, particles are measured in momentum eigenstates. Momentum of each particle has stabilized and is again a good quantum number a long time after the collision.

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 $[O,\,H]=0$ .

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 :
$O|\psi _{t}\rangle =O\,T(t)\,|\psi _{0}\rangle$ $=Oe^{-itH/\hbar }|\psi _{0}\rangle$ $=O\sum _{n=0}^{\infty }{\frac {1}{n!}}(-iHt/\hbar )^{n}|\psi _{0}\rangle$ $=\sum _{n=0}^{\infty }{\frac {1}{n!}}(-iHt/\hbar )^{n}O|\psi _{0}\rangle$ $=q_{j}|\psi _{t}\rangle .\quad \square$ ## Ehrenfest Theorem and Good Quantum Numbers

The Ehrenfest Theorem 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 the 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, and will be elaborated in more detail below.

## 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 the Hamiltonian $H$ . This implies that we can have common eigenstates of $A$ and $H$ . Assume that our system is in one of these common eigenstates. If we measure of $A$ , it will definitely yield an eigenvalue of $A$ (the good quantum number). Also, it is a well-known result that an eigenstate of the Hamiltonian is a stationary state, which means that even if the system is left to evolve for some time before the measurement is made, it will still yield the same eigenvalue. Therefore, If our system is in a common eigenstate, its eigenvalues of A (good quantum numbers) won't change with time.

Conclusion: If ${\textstyle [A,H]=0}$ and the system is in a common eigenstate of $A$ and $H$ , the eigenvalues of $A$ (good quantum numbers) don'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, ${\textstyle [A,H]=0}$ . But now the system is not in any of the common eigenstates of $H$ and $A$ . So the system must be in some linear combination of the basis formed by the 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, they are bound to yield the same result. In this case, a (weaker) statement of conservation holds: Using the Ehrenfest Theorem, ${\textstyle [A,H]=0}$ doesn't explicitly depend on time:

${\frac {d}{dt}}\langle A\rangle =\left\langle {\frac {\partial A}{\partial t}}\right\rangle +{\frac {1}{i\hbar }}\langle [A,H]\rangle =0$ This says that the expectation value of $A$ remains constant in time. When the measurement is made on identical systems again and again, it will generally yield different values, but the expectation value remains constant. This is a weaker conservation condition than the case when our system was a common eigenstate of $A$ and $H$ : The eigenvalues of $A$ are not ensured to remain constant, only its expectation value.

Conclusion: If ${\textstyle [A,H]=0}$ , $A$ doesn't explicitly depend on time and the system isn't in a common eigenstate of $A$ and $H$ , the expectation value of $A$ is conserved, but the conservation of the eigenvalues of $A$ is not ensured.

## Analogy with Classical Mechanics

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

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

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 $H$ .

#### 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 $l,j,m_{\text{l}},m_{s},m_{j}$ . 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 hydrogen 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.

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

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

$\beta =\beta (n,l)=Z^{4}{\mu _{0} \over 4{\pi }^{4}}g_{\text{s}}\mu _{\text{B}}^{2}{1 \over n^{3}a_{0}^{3}l(l+1/2)(l+1)}$ As we can see, the above expressions contain the good quantum numbers, namely $l,s,j$ 