Single-particle states vs. basis states vs. eigenkets vs. quantum numbers

Question

I am confused between these terms that we usually find in quantum mechanics.

In my understanding, we can write any operator in a given Hilbert space in terms of the eigen-kets of any operator. For example, if we have an operator $\hat{A}$ which has eigen-kets $\{|a_i\rangle \}$, then any other operator $\hat{B}$ can be written in terms of $\{|a_i\rangle \}$ as $\hat{B}=\sum_{ij}|a_i\rangle\langle a_i|\hat{B}|a_j\rangle\langle a_j|$. Similarly, we could have eigen-kets of the Hamiltonian operator $\{|E_i\rangle\}$.

My confusions are:

1. Are eigen-kets $\{|a_i\rangle \}$ of any operator called the single-particle states? Or the eigen-kets of only Hamiltonian, $\{|E_i\rangle\}$, are called single-particle states?

2. Are eigen-kets $\{|a_i\rangle \}$ and "basis states" same thing?

3. What are quantum numbers? Are eigen-kets $\{|a_i\rangle \}$ the quantum numbers?

Answer 1

(Orthonormal) basis

A set of ket $\{|\psi_i\rangle:i\in I\}$ is a (orthonormal basis) if the kets are orthonormal $\langle \psi_i|\psi_j\rangle = \delta_{ij}$ and satisfy the completeness relation : $$1 = \sum_i |\psi_i\rangle\langle \psi_i|$$

Eigen-kets

Given a self-adjoint operator $\hat A$, a ket $|\psi\rangle$ is an eigen-ket of $\hat A$ if there is a complex number $a$ such that : $$\hat A|\psi\rangle =a|\psi\rangle $$

In this case, $a$ is called the eigenvalue.

The spectral theorem ensures that there exists an orthonormal basis $\{|\psi_i\rangle, i\in I\}$ whose kets are all eigenkets of $\hat A$ : $$\hat A|\psi_i\rangle =a_i|\psi_i\rangle$$

If $|\psi\rangle,|\phi\rangle$ are eigenkets of $\hat A$ with the same eigenvalue, then any linear combination of them is also an eigenkets. If $|\psi\rangle,|\phi\rangle$ are eigenkets of $\hat A$ with different eigenvalues, then they are orthogonal.

Therefore, if $\hat A$ has no degeneracy, we can construct an eigen-basis of $\hat A$ by choosing one eigenket $|a\rangle$ for any eigenvalue $a$.

Quantum numbers

Often, operators have degeneracy. In this case, we want to find a maximal set of commuting observables. This is a set $\hat A_1,\ldots, \hat A_n$ of self-adjoint operators which commute with each other (and therefore, by the spectral theorem, we can find a basis whose kets are eigen-kets of all the $A_k$) and such that for any set $(a_1,\ldots,a_n)$ of eigenvalue, there is at most one ket (up to normalization and phase) such that : $$\hat A_k|\psi\rangle = a_k|\psi\rangle$$

In this context, the eigenvalues $(a_1,\ldots,a_n)$ are called the quantum numbers of the state $|\psi\rangle$. Since they determine it uniquely, we often write $|\psi\rangle = |a_1,\ldots,a_n\rangle$

Single particle states

This only make sense for a Hilbert space which also contain multi-particle space (eg Fock spaces). In this case, there should be a particle number operator $\hat N$ and single-particle states are eigenstates with the eigenvalue $1$ : $$\hat N|\psi\rangle = |\psi\rangle$$

Answer 2

We can write any operator $\hat{B}$ in terms of $\{|\alpha_i\rangle\}$ because of the completeness relation the set of eigenkets of $\hat{A}$, namely $\{|\alpha_i\rangle\}$, satisfies, $$\sum_{i=1}^n|\alpha_i\rangle\langle\alpha_i|=\textbf{1}_{n\times n}$$ where $n$ is the dimensionality of the basis.

Having said that,

1. Neither $|\alpha_i\rangle$, nor $|E_i\rangle$ need be one-particle states. They can be many particle states. They can also be one-particle states. In particular, $|E_i\rangle$ can be a one-particle state if the Hamiltonian is a Hamiltonian that time-evolves a one-particle system.
2. If the eigenkets are orthonormal and they form a complete set (i.e. span the vector space), then yes, they are the same.
3. The quantum numbers are the eigenvalues of some operators. Usually, with the same eigenvalues we also label the corresponding arrays (or states), i.e. take as an example the total spin operator and its eigenstates of the spin operator $$S^2|s\rangle=\hbar s(s+1)|s\rangle$$ The label $s$ denotes the quantum number of spin.

I hope this helps. If there are any questions, please comment.

Answer 3

Firstly, not all operators have a well defined set of eigenstates. Even if they have right-eigenstates, they may not have left-eigenstates. Or the eigenstates may not be complete. The eigenstates of an operator are not necessarily single-particle state. Consider for example the annihilation operators. It has right-eigenstates, called the coherent states $$ \hat{a} |\alpha\rangle = |\alpha\rangle\alpha . $$ These coherent states are not in general single particle-states.

Hermitian operators do in general (as far as I know) always have complete sets of orthogonal eigenstates. (Even if it has degenerate eigenvalues, one can define an orthogonal basis). So if $\hat{A}$ is Hermitian, then we can represent it as $$ \hat{A} = \sum_m |m\rangle \lambda_m \langle m| , $$ where $|m\rangle$ is the eigenstates and $\lambda_m$ the eigenvalues. A complete set of orthogonal states can serve as a basis. Therefore, such a set of eigenstates can serve as a basis.

The quantum numbers are quantities that represent the degrees of freedom of a system. It could for instance be represented by the index $m$, depending on the context.