📂Quantum Mechanics

Two eigenvectors with different eigenvalues are orthogonal.

Summary

Any two eigenfunctions corresponding to distinct eigenvalues of an Hermitian operator $A$ are orthogonal to each other.

$$\begin{cases} A\psi_{n}=a_{n}\psi_{n} \\ A\psi_{m}=a_{m}\psi_{m} \end{cases}$$

If $a_{n} \ne a_{m}$,

$$\braket{\psi_{n} | \psi_{m}} = 0$$

Proof

Since eigenvalues of Hermitian operators are always real, the following holds:

$$\braket{A\psi_{n}|\psi_{m} } ={a_{n}}^{\ast}\braket{\psi_{n}|\psi_{m} } =a_{n}\braket{\psi_{n}|\psi_{m}}$$

Moreover, according to the definition of Hermitian operator,

$$\braket{A\psi_{n}|\psi_{m}}=\braket{\psi_{n}|A^{\dagger}\psi_{m}}=\braket{\psi_{n}|A\psi_{m}} = a_{m} \braket{\psi_{n} | \psi_{m}}$$

Therefore, subtracting the above two equations, we get:

$$0 = \braket{A\psi_{n}|\psi_{m}}- \braket{A\psi_{n}|\psi_{m}} = a_{m}\braket{\psi_{n}|\psi_{m}} - a_{n}\braket{\psi_{n}|\psi_{m}}$$ $$\implies (a_{m}-a_{n})\braket{\psi_{n}|\psi_{m}}=0$$

Given that $a_{m} \ne a_{n}$, we have $\braket{\psi_{n}|\psi_{m}}=0$.

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See Also

• Proof in Linear Algebra