Hermitian Operator

A Hermitian Operator $\tilde L$ is one which satisfies

$$\int_a^b v^*\tilde Lu\,dx = \int_a^b u\tilde Lv^*\, dx.$$ (1)

As shown in Sturm-Liouville Theory, if $\tilde L$ is Self-Adjoint and satisfies the boundary conditions

$$[v^*pu']_{x=a} = [v^*pu']_{x=b},$$ (2)

then it is automatically Hermitian. Hermitian operators have Real Eigenvalues, Orthogonal Eigenfunctions, and the corresponding Eigenfunctions form a Complete set when $\tilde L$ is second-order and linear. In order to prove that Eigenvalues must be Real and Eigenfunctions Orthogonal, consider

$$\tilde Lu_i+\lambda_iwu_i=0.$$ (3)

Assume there is a second Eigenvalue $\lambda_j$ such that

$$\tilde Lu_j+\lambda_jwu_j=0$$ (4)

$$\tilde L{u_j}^*+{\lambda_j}^*w{u_j}^*=0.$$ (5)

Now multiply (3) by ${u_j}^*$ and (5) by $u_i$

$${u_j}^*\tilde Lu_i+{u_j}^*\lambda_i wu_i=0$$ (6)

$${u_i}\tilde L{u_j}^*+u_i{\lambda_j}^*w{u_j}^*=0$$ (7)

$${u_j}^*\tilde Lu_i-u_i\tilde L{u_j}^*=({\lambda_j}^*-\lambda_i)wu_i{u_j}^*.$$ (8)

Now integrate

$$\int_a^b {u_j}^*\tilde Lu_i-\int_a^b u_i\tilde L{u_j}^*=({\lambda_j}^*-\lambda_i)\int_a^b wu_i{u_j}^*.$$ (9)

But because $\tilde L$ is Hermitian, the left side vanishes.

$$({\lambda_j}^*-\lambda_i)\int_a^b wu_i{u_j}^*=0.$$ (10)

If Eigenvalues $\lambda_i$ and $\lambda_j$ are not degenerate, then $\int_a^b wu_i{u_j}^*=0$, so the Eigenfunctions are Orthogonal. If the Eigenvalues are degenerate, the Eigenfunctions are not necessarily orthogonal. Now take $i=j$.

$$({\lambda_i}^*-\lambda_i)\int_a^b wu_i{u_i}^*=0.$$ (11)

The integral cannot vanish unless $u_i=0$, so we have ${\lambda_i}^*=\lambda_i$ and the Eigenvalues are real.

For a Hermitian operator $\tilde O$,

$$\left\langle{\phi\vert\tilde O\psi}\right\rangle{} = \left\l... ...ngle{}^* = \left\langle{\tilde O\phi\vert\psi}\right\rangle{}.$$ (12)

In integral notation,

$$\int(\tilde A\phi)^*\psi\,dx =\int\phi^*\tilde A\psi\,dx.$$ (13)

Given Hermitian operators $\tilde A$ and $\tilde B$,

$$\left\langle{\phi\vert\tilde A\tilde B\psi}\right\rangle{} =... ... \left\langle{\phi\vert\tilde B\tilde A\psi}\right\rangle{}^*.$$ (14)

Because, for a Hermitian operator $\tilde A$ with Eigenvalue $a$,

$$\left\langle{\psi\vert\tilde A\psi}\right\rangle{} = \left\langle{\tilde A\psi\vert\psi}\right\rangle{}$$ (15)

$$a\left\langle{\psi\vert\psi}\right\rangle{} = a^*\left\langle{\psi\vert\psi}\right\rangle{}.$$ (16)

Therefore, either $\left\langle{\psi\vert\psi}\right\rangle{} = 0$ or $a = a^*$. But $\left\langle{\psi\vert\psi}\right\rangle{} = 0$ Iff $\psi = 0$, so

$$\left\langle{\psi\vert\psi}\right\rangle{}\not= 0,$$ (17)

for a nontrivial Eigenfunction. This means that $a = a^*$, namely that Hermitian operators produce Real expectation values. Every observable must therefore have a corresponding Hermitian operator. Furthermore,

$$\left\langle{\psi_n\vert\tilde A \psi_m}\right\rangle{} = \left\langle{\tilde A\psi_n\vert\psi_m}\right\rangle{}$$ (18)

$$a_m\left\langle{\psi_n\vert\psi_m}\right\rangle{} = {a_n}^*\... ...\rangle{} = a_n\left\langle{\psi_n\vert\psi_m}\right\rangle{},$$ (19)

since $a_n = {a_n}^*$. Then

$$(a_m-a_n)\left\langle{\psi_n\vert\psi_m}\right\rangle{} = 0$$ (20)

For $a_m\not= a_n$ (i.e., $\psi_n\not= \psi_m$),

$$\left\langle{\psi_n\vert\psi_m}\right\rangle{} = 0.$$ (21)

For $a_m = a_n$ (i.e., $\psi_n = \psi_m$),

$$\left\langle{\psi_n\vert\psi_m}\right\rangle{} = \left\langle{\psi_n\vert\psi_n}\right\rangle{}\equiv 1.$$ (22)

Therefore,

$$\left\langle{\psi_n\vert\psi_m}\right\rangle{} =\delta_{nm},$$ (23)

so the basis of Eigenfunctions corresponding to a Hermitian operator are Orthonormal. Given two Hermitian operators $\tilde A$ and $\tilde B$,

$$(\tilde A \tilde B )^\dagger =\tilde B^\dagger\tilde A^\dagger = \tilde B\tilde A =\tilde A \tilde B +[\tilde B ,\tilde A ],$$ (24)

the operator $\tilde A\tilde B$ equals $(\tilde A\tilde B)^\dagger$, and is therefore Hermitian, only if

$$[\tilde B, \tilde A]= 0.$$ (25)

Given an arbitrary operator $\tilde A$,

$$\left\langle{\psi_1\vert(\tilde A+\tilde A^\dagger)\psi_2}\right\rangle{} = \left\langle{(\tilde A^\dagger+\tilde A)\psi_1\vert\psi_2}\right\rangle{}$$

$$= \left\langle{(\tilde A+\tilde A^\dagger)\psi_1\vert\psi_2}\right\rangle{},$$ (26)

so $\tilde A+\tilde A^\dagger$ is Hermitian.

$$\left\langle{\psi_1\vert i(\tilde A-\tilde A^\dagger)\psi_2}\right\rangle{} = \left\langle{-i(\tilde A^\dagger-\tilde A)\psi_1\vert\psi_2}\right\rangle{}$$

$$= \left\langle{i(\tilde A-\tilde A^\dagger)\psi_1\vert\psi_2}\right\rangle{},$$ (27)

so $i(\tilde A-\tilde A^\dagger)$ is Hermitian. Similarly,

$$\left\langle{\psi_1\vert(\tilde A\tilde A^\dagger)\psi_2}\ri... ...e{(\tilde A\tilde A^\dagger)\psi_1\vert\psi_2}\right\rangle{},$$ (28)

so $\tilde A \tilde A^\dagger$ is Hermitian.

Define the Hermitian conjugate operator $\tilde A^\dagger$ by

$$\left\langle{\tilde A \psi \vert\psi}\right\rangle{}\equiv \left\langle{\psi \vert\tilde A^\dagger\psi}\right\rangle{}.$$ (29)

For a Hermitian operator, $\tilde A=\tilde A^\dagger$. Furthermore, given two Hermitian operators $\tilde A$ and $\tilde B$,

$$\left\langle{\psi_2\vert(\tilde A\tilde B)^\dagger\psi_1}\right\rangle{} = \left\langle{(\tilde A\tilde B)\psi_2\vert\psi_1}\right\rangle{} = \left\langle{\tilde B\psi_2\vert\tilde A^\dagger\psi_1}\right\rangle{}$$

$$= \left\langle{\psi_2\vert\tilde B^\dagger\tilde A^\dagger\psi_1}\right\rangle{},$$ (30)

so

$$(\tilde A \tilde B)^\dagger =\tilde B^\dagger\tilde A^\dagger.$$ (31)

By further iterations, this can be generalized to

$$(\tilde A \tilde B \cdots\tilde Z)^\dagger =\tilde Z^\dagger\cdots\tilde B^\dagger\tilde A^\dagger.$$ (32)

See also Adjoint Operator, Hermitian Matrix, Self-Adjoint Operator, Sturm-Liouville Theory

References

Arfken, G. ``Hermitian (Self-Adjoint) Operators.'' §9.2 in Mathematical Methods for Physicists, 3rd ed. Orlando, FL: Academic Press, pp. 504-506 and 510-516, 1985.