Quantum Eigenstate of an Observable Definition

Scientific claim

Quantum Eigenstate of an Observable Definition

Claim

Let $A$ be an observable physical quantity represented by a self-adjoint operator $\hat A$ on a complex Hilbert space $\mathcal H$ .

A pure quantum state represented by the ray $[a]$ is an eigenstate of the observable $A$ with eigenvalue $a\in\mathbb R$ if any nonzero representative $|a\rangle$ of the ray satisfies

\hat A|a\rangle=a|a\rangle.

The vector $|a\rangle$ is an eigenvector of $\hat A$ , while the ray

[a]=\{c|a\rangle:c\in\mathbb C,\ c\neq0\}

is the corresponding quantum eigenstate.

Theoretical: Not applicableExperimental: Not applicable

Assumptions

The quantum system is associated with a complex Hilbert space $\mathcal H$ .
Pure quantum states are represented by rays in $\mathcal H$ .
The observable $A$ is represented by a self-adjoint operator $\hat A$ .
The vector $|a\rangle$ is nonzero and belongs to the domain of $\hat A$ .
The value $a$ is an eigenvalue in the discrete spectrum of $\hat A$ .

Proof

Let $A$ be an observable physical quantity represented by a self-adjoint operator

\hat A:D(\hat A)\rightarrow\mathcal H,

where $D(\hat A)\subseteq\mathcal H$ is the domain of $\hat A`.

From the mathematical definition of an eigenvalue–eigenvector pair, a nonzero vector $|a\rangle\in D(\hat A)$ is an eigenvector of $\hat A$ with eigenvalue $a$ if

\hat A|a\rangle=a|a\rangle.

Applying $\hat A$ to $|a\rangle$ therefore does not move the vector into a new independent direction in the Hilbert space. It only multiplies the vector by the scalar $a`.

Because $\hat A$ is self-adjoint, its eigenvalues are real. To see this, suppose $|a\rangle$ is normalized:

\langle a|a\rangle=1.

Using the eigenvalue equation,

\langle a|\hat A|a\rangle = a\langle a|a\rangle = a.

Since $\hat A$ is self-adjoint, its expectation value is real:

\langle a|\hat A|a\rangle\in\mathbb R.

Therefore,

a\in\mathbb R.

A physical pure state is represented by a ray rather than by one particular vector. If $c\in\mathbb C$ and $c\neq0$ , then linearity gives

\hat A(c|a\rangle) = c\hat A|a\rangle.

Using the eigenvalue equation,

\hat A(c|a\rangle) = ca|a\rangle = a(c|a\rangle).

Therefore, every nonzero scalar multiple of $|a\rangle$ is an eigenvector with the same eigenvalue $a`.

Consequently, the eigenstate is properly identified with the entire ray

[a]=\{c|a\rangle:c\in\mathbb C,\ c\neq0\},

rather than with a particular representative vector.

This establishes that a quantum eigenstate of the observable $A$ is a pure-state ray whose representative vectors are eigenvectors of the operator $\hat A`.

Examples

For a spin- $\frac12$ particle, the observable $S_z$ represents spin along the $z$ -axis. Its operator is
$\hat S_z = \frac{\hbar}{2} \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix}.$
Consider the normalized state vector
$|{\uparrow_z}\rangle = \begin{pmatrix} 1\\ 0 \end{pmatrix}.$
Applying the operator gives
$\hat S_z|{\uparrow_z}\rangle = \frac{\hbar}{2} \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} \begin{pmatrix} 1\\ 0 \end{pmatrix}.$
Therefore,
$\hat S_z|{\uparrow_z}\rangle = \frac{\hbar}{2} \begin{pmatrix} 1\\ 0 \end{pmatrix} = \frac{\hbar}{2}|{\uparrow_z}\rangle.$
Thus, the ray $[\uparrow_z]$ is an eigenstate of the observable $S_z$ with eigenvalue
$a=\frac{\hbar}{2}.$
Consider the normalized state vector
$|{\downarrow_z}\rangle = \begin{pmatrix} 0\\ 1 \end{pmatrix}.$
Applying the spin operator gives
$\hat S_z|{\downarrow_z}\rangle = \frac{\hbar}{2} \begin{pmatrix} 1&0\\ 0&-1 \end{pmatrix} \begin{pmatrix} 0\\ 1 \end{pmatrix}.$
Therefore,
$\hat S_z|{\downarrow_z}\rangle = -\frac{\hbar}{2} \begin{pmatrix} 0\\ 1 \end{pmatrix} = -\frac{\hbar}{2}|{\downarrow_z}\rangle.$
Thus, the ray $[\downarrow_z]$ is an eigenstate of $S_z$ with eigenvalue
$a=-\frac{\hbar}{2}.$
Let the observable $E$ represent the total energy of a quantum system, and let $\hat H$ be its Hamiltonian operator.
A state vector $|E_n\rangle$ is an energy eigenvector if
$\hat H|E_n\rangle=E_n|E_n\rangle.$
The corresponding ray $[E_n]$ is an energy eigenstate with energy eigenvalue $E_n`.
In the position representation, the same equation becomes
$\hat H\psi_n(x)=E_n\psi_n(x),$
where
$\psi_n(x)=\langle x|E_n\rangle.$
This is the time-independent Schrödinger equation.

Evidence

01

Eigenstates and EigenvaluesRichard Fitzpatrick · 2015-05-19 · supporting

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