Operators

An operator $\hat{O}$ in a vector space $V$ maps a set of vectors $\boldsymbol{\mathit{v}}_i$ to another set of vectors $\boldsymbol{\mathit{v}}_j$ (or transforms one function into another function), where $\boldsymbol{\mathit{v}}_i,\boldsymbol{\mathit{v}}_j\in V$ . For example, the operator $\frac{d^{2}}{dx^{2}}$ transform the function $f(x)$ into $f^{''}(x)$ :

$\frac{d^{2}}{dx^{2}}f(x)=f^{''}(x)$

Linear operators have the following properties:

$\hat{O}(\boldsymbol{\mathit{u}}+\boldsymbol{\mathit{v}})=\hat{O}\boldsymbol{\mathit{u}}+\hat{O}\boldsymbol{\mathit{v}}$
$\hat{O}(c\boldsymbol{\mathit{u}})=c\hat{O}\boldsymbol{\mathit{u}}$
$(\hat{O}_1\hat{O}_2)\boldsymbol{\mathit{u}}=\hat{O}_1(\hat{O}_2\boldsymbol{\mathit{u}})$

Two operators commonly encountered in quantum mechanics are the position and linear momentum operators. To construct these operators, we refer to probability theory, where the expectation value of the position $x$ of a particle in a 1-D box of length $L$ is

$\langle x\rangle=\int_{0}^{L}xP(x)dx=\int_{0}^{L}x\left |\psi(x) \right |^{2}dx=\int_{0}^{L}\psi(x)\, x\, \psi(x)dx$

where $P(x)$ is the probability of observing the particle at a particular position between $0$ and $L$ , and $\psi(x)$ is the particle’s wavefunction, which is assumed to be real.

Comparing the above equation with the expression of the expectation value of a quantum-mechanical operator, $\hat{x}=x$ .

One may infer that the linear momentum operator is $\hat{p}_x=p_x$ . However, we must find a form of $p_x$ that is a function of $x$ so that we can compute $\int_{0}^{L}\psi(x)p_x\psi(x)dx$ . If we compare the time-independent Schrodinger equation $\left [ -\frac{\hbar^{2}}{2m}\frac{\partial^{2}}{\partial x^{2}}+V(x) \right ]\psi(x)=E\psi(x)$ with the total energy equation $\frac{p_x^{\: 2}}{2m}+V(x)=E$ , we have $\hat{p}_x=\frac{\hbar}{i}\frac{d}{dx}$ .

To test the validity of $\hat{x}=x$ and $\hat{p}_x=\frac{\hbar}{i}\frac{d}{dx}$ , we compute $\langle x\rangle$ and $\langle p_x\rangle$ using the 1-D box wavefunction of $\sqrt\frac{2}{L}sin\frac{n\pi x}{L}$ and check if the results are reasonable with respect to classical mechanics.

Integrating $\langle x\rangle=\frac{2}{L}\int_{0}^{L}xsin^{2}\left ( \frac{n\pi x}{L} \right )dx$ by parts, we have $\langle x\rangle=\frac{L}{2}$ . In classical mechanics, the particle can be anywhere in the 1-D box with equal probability. Therefore, the average position of $\langle x\rangle=\frac{L}{2}$ is reasonable.

For the linear momentum operator, we have $\langle p_x\rangle=\frac{2\hbar}{iL}\int_{0}^{L}sin\left ( \frac{n\pi x}{L} \right )\frac{d}{dx}sin\left ( \frac{n\pi x}{L} \right )dx=0$ . Since $\langle x\rangle=\frac{L}{2}$ , we expect $\langle p_x\rangle=m\frac{d\langle x\rangle}{dt}=0$ . Therefore, $x$ and $\frac{\hbar}{i}\frac{d}{dx}$ are reasonable assignments of the position and linear momentum operators respectively. In 3-D, the position and linear momentum operators are:

$\hat{x}=x\; \; \; \; \;\hat{y}=y\; \; \; \; \;\hat{z}=z$

$\hat{p}_x=\frac{\hbar}{i}\frac{\partial}{\partial x}\; \; \; \; \;\hat{p}_y=\frac{\hbar}{i}\frac{\partial}{\partial y}\; \; \; \; \;\hat{p}_z=\frac{\hbar}{i}\frac{\partial}{\partial z}\; \; \; \; \;\; \; \; 4$

To see the proof that the position and linear momentum operators are Hermitian, read this article.

Next article: commuting operators

Previous article: kronecker product

Content page of quantum mechanics

Content page of advanced chemistry

Main content page

Leave a Reply Cancel reply