Hermitian operator

An operator with domain D_{\hat{O}} is Hermitian if

\langle\boldsymbol{\mathit{m}}\vert\hat{O}\vert\boldsymbol{\mathit{n}}\rangle=\langle\boldsymbol{\mathit{n}}\vert\hat{O}\vert\boldsymbol{\mathit{m}}\rangle^{*}\; \; \; \; \; \; \; \; 35

for all \boldsymbol{\mathit{m}},\boldsymbol{\mathit{n}}\in D_{\hat{O}}.

Eq35 can also be expressed as \langle f_m\vert\hat{O}\vert f_n\rangle=\langle\hat{O}f_m\vert f_n\rangle because

\langle f_m\vert\hat{O}\vert f_n\rangle=\int f_{m}^{*}\hat{O}f_n\: d\tau =\int f_n(\hat{O}f_{m})^{*}\: d\tau=\langle\hat{O}f_m\vert f_n\rangle\; \; \; \; \; \; \; \; 36

where we have used the Hermitian property of the operator in the 2nd equality.

Another way of defining a Hermitian operator is by noting that the average value of a physical property \langle O\rangle must be a real number, i.e. \langle O\rangle=\langle O\rangle^{*} or

\int \psi^{*}\hat{O}\psi\, d\tau =\left ( \int \psi^{*}\hat{O}\psi\: d\tau\right )^{*}=\int \psi(\hat{O}\psi)^{*}d\tau=\int(\hat{O}\psi)^{*}\psi\, d\tau= \langle\hat{O}\psi\vert\psi\rangle\; \; \; \; \; \; \; \; 37



Show that eq37 is equivalent to eq36.


Substitute \psi=f_m+af_n where a is a constant in \int \psi^{*}\hat{O}\psi\, d\tau=\int \psi(\hat{O}\psi)^{*}\, d\tau of eq37 and simplify to give:

a^{*}\int f_{n}^{*}\hat{O}f_m\: d\tau+a\int f_{m}^{*}\hat{O}f_n\: d\tau=a\int f_n(\hat{O}f_{m})^{*}\: d\tau+a^{*}\int f_m(\hat{O}f_{n})^{*}\: d\tau\; \; \; \; \; \; \; \; 38

Since \psi can be expressed as any linear combination of the basis functions f_m and f_n, a can be any number. If we carry out the following:

  1. Substitute a=1 in eq38
  2. Subsitute a=i in eq38 and divide the resultant equation by i
  3. Sum the two resultant equations

we have eq36.


Hermitian operators have the following properties:

  1. Eigenfunctions of a Hermitian operator form a complete set
  2. Eigenvalues of a Hermitian operator are real
  3. Eigenfunctions of a Hermitian operator are orthogonal (if they have distinct eigenvalues) or can be chosen to be orthogonal (if they describe a degenerate state)

The first property is a postulate of quantum mechanics. For the 2nd property, we assume two eigenvalue equations \hat{O}f_m=a_mf_m and \hat{O}f_n=a_nf_n. Multiplying the first equation on the left by f_{n}^{*}, multiplying the complex conjugate of the 2nd equation on the left by f_m, and then subtracting the two resultant equations and rearranging, we have:


\int f_{n}^{*}\hat{O}f_m\, d\tau-\int f_m\hat{O}^{*}f_{n}^{*}\, d\tau=(a_m-a_{n}^{*})\int f_{n}^{*}f_m \, d\tau

Using the 2nd equality of eq36,

(a_m-a_{n}^{*})\int f_{n}^{*}f_m \, d\tau=0\; \; \; \; \; \; \; \; 39

To show that eigenvalues of a Hermitian operator are real, let m=n

(a_n-a_{n}^{*})\int \vert f_{n}\vert^{2}\, d\tau=0

The integral is zero if and only if f_n were zero over its entire domain. However, an eigenfunction is defined as a non-zero function. Therefore, a_n=a_{n}^{*}.

For the 3rd property, we rewrite eq39 as

(a_m-a_{n})\int f_{n}^{*}f_m \, d\tau=0

If m\neq n, we have a_m-a_n\neq 0 and so, \int f_{n}^{*}f_m\, d\tau=0.

Examples of Hermitian operators are \hat{x} and \hat{p}.



Show that \hat{x} and \hat{p}_x are Hermitian.


We need to show that \int \psi^{*}\hat{O}\psi\, d\tau=\left ( \int \psi^{*}\hat{O}\psi\, d\tau\right )^{*}.

For \hat{x}, we have

\int \psi^{*}\hat{x}\psi\, d\tau=\int \psi^{*}x\psi\, d\tau=\int \psi x\psi^{*} \, d\tau=\left ( \int \psi^{*}x\psi\, d\tau\right )^{*}=\left ( \int \psi^{*}\hat{x}\psi\, d\tau\right )^{*}

where, for the 3rd equality, we used the fact that x is the position of a particle which must be real (x^{*}=x).

For \hat{p}_x, we integrate \int_{-\infty}^{\infty}\psi^{*}(x)\frac{\hbar}{i}\frac{d}{dx}\psi(x)dx by parts to give:

\int_{-\infty}^{\infty}\psi^{*}(x)\frac{\hbar}{i}\frac{d}{dx}\psi(x)dx=\left [\frac{\hbar}{i}\psi^{*}(x)\psi(x)\right ]_{-\infty}^{\infty}+\int_{-\infty}^{\infty}\psi(x)\left (- \frac{\hbar}{i}\right )\frac{d}{dx}\psi^{*}(x)dx

Since a well-behaved wavefunction that is square-integrable is defined as \int_{-\infty}^{\infty}\left | \psi(x)\right |^{2}dx< \infty, i.e. it vanishes at x=\pm \infty, the 1st term on the RHS of the above equation vanishes and we have:

\int_{-\infty}^{\infty}\psi^{*}(x)\frac{\hbar}{i}\frac{d}{dx}\psi(x)dx=\int_{-\infty}^{\infty}\psi(x)\left (- \frac{\hbar}{i}\right )\frac{d}{dx}\psi^{*}(x)dx=\left [ \int_{-\infty}^{\infty}\psi^{*}(x)\frac{\hbar}{i}\frac{d}{dx}\psi(x)dx\right ]^{*}


Finally, two operators \hat{A} and \hat{B} are defined as a Hermitian conjugate pair if




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