I am reading quantum mechanics and worrying about the question that,

Are the eigenvectors of a Hermitian matrix(operator) always orthogonal?

If they are not always orthogonal, please explain when they are and when not.Thank you

  • $\begingroup$ For the finite case always. It's the Spectral theorem. See Wiki. $\endgroup$ – Dog_69 May 27 '18 at 5:50
  • $\begingroup$ Consider the identity operator. $\endgroup$ – WimC May 27 '18 at 5:51

Eigenvectors corresponding to distinct eigenvalues of a Hermitian matrix are always orthogonal, to wit:


$H^\dagger = H, \tag 0$

and that

$H \vec v_1 = \mu_1 \vec v_1, \tag 1$

$H \vec v_2 = \mu_2 \vec v_2; \tag 2$


$H \vec v = \mu \vec v, \; v \ne 0, \tag 3$

we have

$\mu \langle v, v \rangle = \langle v, \mu v \rangle = \langle v, H v \rangle = \langle H^\dagger v, v \rangle$ $= \langle Hv, v \rangle = \langle \mu v, v \rangle = \overline{\langle v, \mu v \rangle} = \bar \mu \overline{\langle v, v \rangle} = \bar \mu \langle v, v \rangle; \tag 4$

since $v \ne 0$, $\langle v, v \rangle \ne 0$ so we may divide it out of (4) and see that

$\mu = \bar \mu \tag 5$

for any operator satisfying (0); this means that the eigenvalues of any operator satisfying (0) are real; therefore we may write

$\mu_1 \langle v_1, v_2 \rangle = \langle \mu_1 v_1, v_2 \rangle = \langle H v_1, v_2 \rangle$ $= \langle v_1, H^\dagger v_2 \rangle = \langle v_1, H v_2 \rangle = \langle v_1, \mu_2 v_2 \rangle = \mu_2 \langle v_1, v_2 \rangle, \tag 6$


$(\mu_1 - \mu_2) \langle v_1, v_2 \rangle = 0, \tag 7$

whence, assuming $\mu_1 \ne \mu_2$,

$\langle v_1, v_2 \rangle = 0, \tag 8$

and thus the vectors $v_1$, $v_2$ are orthogonal.

If $\mu_1 = \mu_2$ but $v_1$ and $v_2$ are linearly independent, then any vector in $\text{span}\{v_1, v_2 \}$ is an eigenvector for $\mu$:

$H(av_ 1 + bv_2) = aHv_1 + bHv_2 = a\mu v_1 + b \mu v_2) = \mu(a v_1 + b v_2); \tag 9$

in this case, we won't in general have $\langle v_1, v_2 \rangle = 0$, but we are free to choose $v_1$, $v_2$ suchly if we so desire; this is often the wise choice, since a set of orthogonal eigenvectors is often convenient for applications. But here the orthogonality is a choice, not a necessity.

The preceding remarks bind over any complex inner product space whenever (0), (1) and (2) hold; $H$ needn't even be bounded provided its action is restricted to a subspace for which (0), (1) and (2) are meaningful.

Note Addded in Edit, Tuesday 5 May 2020 3:06 PM PST: Данило Клименко asked, in a comment to this answer, if the analogous result holds for skew-Hermitian operators, that is for operators $\Sigma$ such that

$\Sigma^\dagger = -\Sigma; \tag{10}$

this query my be answered in the affirmative via an argument parallel to that presented above; we first show that with

$\Sigma v = \sigma v, \tag{11}$

$\sigma$ is purely imaginary. For in light of (11) we have

$\sigma \langle v, v \rangle = \langle v, \sigma v \rangle = \langle v, \Sigma v \rangle = \langle \Sigma^\dagger v, v \rangle = \langle -\Sigma v, v \rangle = -\langle \Sigma v, v \rangle$ $= -\langle \sigma v, v \rangle = -\overline{\langle v, \sigma v \rangle} = -\bar \sigma \overline{\langle v, v \rangle} = -\bar \sigma \langle v, v \rangle; \tag{12}$

with $\langle v, v \rangle \ne 0$ we find

$\sigma = -\bar \sigma, \tag{13}$

i.e., $\sigma$ is a purely imaginary number. Now if

$\Sigma v_1 = \sigma_1 v_1 \tag{14}$


$\Sigma v_2 = \sigma_2 v_2, \tag{15}$

in parallel with (6)-(8):

$\sigma_2 \langle v_1, v_2 \rangle = \langle v_1, \sigma_2 v_2 \rangle$ $= \langle v_1, \Sigma v_2 \rangle = \langle \Sigma^\dagger v_1, v_2 \rangle = \langle -\Sigma v_1, v_2 \rangle $ $= \langle -\sigma_1 v_1, v_2 \rangle = -\langle \sigma_1 v_1, v_2 \rangle = -\bar \sigma_1 \langle v_1, v_2 \rangle = \sigma _1 \langle v_1, v_2 \rangle\tag{16}$


$(\sigma_1 - \sigma_2)\langle v_1, v_2 \rangle = 0, \tag{17}$

and now if we assume

$\sigma_1 \ne \sigma_2 \tag{18}$

we conclude that

$\langle v_1, v_2 \rangle = 0, \tag{19}$

the desired result; that is, eigenvectors corresponding to distinct eigenvalues of skew-Hermitian operators are in fact orthogonal.

This may in fact be see directly from the above ((0)-(9)) discussion concerning Hermitian operators if we observe that (10) yields

$(i\Sigma)^\dagger = \bar i \Sigma^\dagger = -i(-\Sigma) = i\Sigma, \tag{20}$

that is, $i\Sigma$ is Hermitian; then (14) and (15) imply

$i\Sigma v_1 = i\sigma_1 v_1, \tag{21}$


$i\Sigma v_2 = i\sigma_2 v_2, \tag{22}$

and thus invoking our previous result we infer that

$\langle v_1, v_2 \rangle = 0 \tag{23}$

provided $\sigma_1 \ne \sigma_2$.

Lastly, the analog of (9) binds; indeed, (9) and its surrounding text apply for an operator $H$; the assertion is not restricted to Hermitian, skew-Hermitian, or any other class of matrices or operators. End of Note.

  • 1
    $\begingroup$ does it work for skew-Hermitian matrix? $\endgroup$ – Dan Klymenko May 5 '20 at 12:55
  • $\begingroup$ @ДанилоКлименко: I modified my answer to address your question. Cheers! $\endgroup$ – Robert Lewis May 5 '20 at 23:55

If $Ax = \lambda x$ and $Ay = \mu y$ then $$ \lambda\langle x,y\rangle = \langle Ax,y\rangle = \langle x,Ay\rangle = \mu\langle x,y\rangle $$ which leads for $\lambda \neq \mu$ to $\langle x,y\rangle = 0$. (Remember that eigenvalues of the Hermitian operator are always real).

  • $\begingroup$ very concise, but it requires that no two eigenvalues are equal to each other. $\endgroup$ – Quantum Guy 123 May 12 at 16:40

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