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beginner · Physics · Observables, Operators & Measurement

Eigenstates and Definite Values

When you measure an observable, quantum mechanics says the result must be one of the operator's eigenvalues. But which one you get can be probabilistic — unless the state happens to be an eigenstate of that observable. In an eigenstate the outcome is completely determined: the same eigenvalue every single time, with zero spread.

The eigenvalue equation

An operator A^\hat{A} acting on a state ∣ψ⟩|\psi\rangle generally produces a different state. The special case where the output is simply a scalar multiple of the input is the eigenvalue equation:

A^∣ψ⟩=a∣ψ⟩.\hat{A}|\psi\rangle = a|\psi\rangle.

Here aa is a real number called the eigenvalue and ∣ψ⟩|\psi\rangle is the corresponding eigenstate (or eigenvector). Because observables are Hermitian, their eigenvalues are always real — which is exactly what we need since measurement results must be real numbers.

Definite values and zero variance

The connection to measurement is direct. If the system is in eigenstate ∣ψ⟩|\psi\rangle of A^\hat{A}, then every measurement of A^\hat{A} returns aa without exception. Formally, the expectation value is

⟨A^⟩=⟨ψ∣A^∣ψ⟩=⟨ψ∣a∣ψ⟩=a⟨ψ∣ψ⟩=a,\langle\hat{A}\rangle = \langle\psi|\hat{A}|\psi\rangle = \langle\psi|a|\psi\rangle = a\langle\psi|\psi\rangle = a,

and the variance ⟨A^2⟩−⟨A^⟩2\langle\hat{A}^2\rangle - \langle\hat{A}\rangle^2 is zero:

⟨A^2⟩=⟨ψ∣A^2∣ψ⟩=a2⟨ψ∣ψ⟩=a2,\langle\hat{A}^2\rangle = \langle\psi|\hat{A}^2|\psi\rangle = a^2\langle\psi|\psi\rangle = a^2,

so ΔA=0\Delta A = 0. A state with zero variance in A^\hat{A} is said to have a definite value for that observable.

A concrete example

Consider the diagonal matrix

A^=(300−1)\hat{A} = \begin{pmatrix} 3 & 0 \\ 0 & -1 \end{pmatrix}

acting on the two-dimensional state space. Apply it to the vector ∣v⟩=(10)|v\rangle = \begin{pmatrix}1\\0\end{pmatrix}:

A^∣v⟩=(300−1)(10)=(30)=3(10).\hat{A}|v\rangle = \begin{pmatrix} 3 & 0 \\ 0 & -1 \end{pmatrix}\begin{pmatrix}1\\0\end{pmatrix} = \begin{pmatrix}3\\0\end{pmatrix} = 3\begin{pmatrix}1\\0\end{pmatrix}.

The output is exactly 33 times the input, confirming that ∣v⟩|v\rangle is an eigenstate with eigenvalue λ=3\lambda = 3. A system prepared in this state and measured in the A^\hat{A} basis will always return the value 33.

The second basis vector ∣w⟩=(01)|w\rangle = \begin{pmatrix}0\\1\end{pmatrix} is a separate eigenstate with eigenvalue −1-1, since A^∣w⟩=(0−1)=−1(01)\hat{A}|w\rangle = \begin{pmatrix}0\\-1\end{pmatrix} = -1\begin{pmatrix}0\\1\end{pmatrix}. Any superposition α∣v⟩+β∣w⟩\alpha|v\rangle + \beta|w\rangle (with both α≠0\alpha\neq 0 and β≠0\beta\neq 0) is not an eigenstate: measuring A^\hat{A} yields 33 with probability ∣α∣2|\alpha|^2 and −1-1 with probability ∣β∣2|\beta|^2.

Try it

This is a numerical exercise — return a number, not a circuit. Given the observable A^=(300−1)\hat{A} = \begin{pmatrix}3 & 0\\0 & -1\end{pmatrix} and the candidate eigenstate ∣v⟩=(10)|v\rangle = \begin{pmatrix}1\\0\end{pmatrix}, compute A^∣v⟩\hat{A}|v\rangle, check that it equals λ∣v⟩\lambda|v\rangle for some scalar, and return λ\lambda.

Run your code to see the quantum state.

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