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 2^nd 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

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

Question

Show that eq37 is equivalent to eq36.

Answer

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:

Substitute $a=1$ in eq38
Subsitute $a=i$ in eq38 and divide the resultant equation by $i$
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 2^nd property, consider 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 2^nd equation on the left by $f_m$ , and then subtracting the two resultant equations and rearranging, yields:

$f_{n}^{*}\hat{O}f_m-f_m\hat{O}^{*}f_{n}^{*}=(a_m-a_{n}^{*})f_{n}^{*}f_m$

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

Using the 2^nd 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 3^rd 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 $\hat{p}$ .

Question

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

Answer

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 3^rd 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 1^st 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

$\langle\boldsymbol{\mathit{m}}\vert\hat{A}\vert\boldsymbol{\mathit{n}}\rangle=\langle\boldsymbol{\mathit{n}}\vert\hat{B}\vert\boldsymbol{\mathit{m}}\rangle^{*}$

Hermitian operator

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