Operators (quantum mechanics)

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

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

Linear operators have the following properties:

1. $\hat{O}(\boldsymbol{u}+\boldsymbol{v})=\hat{O}\boldsymbol{u}+\hat{O}\boldsymbol{v}$
2. $\hat{O}(c\boldsymbol{u})=c\hat{O}\boldsymbol{u}$
3. $(\hat{O}_1\hat{O}_2)\boldsymbol{u}=\hat{O}_1(\hat{O}_2\boldsymbol{u})$

For property 1, the input state $\boldsymbol{u}+\boldsymbol{v}$ is a linear combination of two basis states. Suppose $\hat{O}=\begin{pmatrix}0&1\\0&0\end{pmatrix}$ , $\boldsymbol{u}=\begin{pmatrix}1\\0\end{pmatrix}$ and $\boldsymbol{v}=\begin{pmatrix}0\\1\end{pmatrix}$ , then the output state is $\boldsymbol{u}$ . This has implications for quantum mechanics (see diagram below for an illustration).

Conversely, the operator $\hat{O}=\begin{pmatrix}1&1\\1&-1\end{pmatrix}$ can transform $\boldsymbol{u}$ into a linear combination of $\boldsymbol{u}$ and $\boldsymbol{v}$ :

$\hat{O}\boldsymbol{u}=\begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}+\begin{pmatrix}0\\1\end{pmatrix}=\boldsymbol{u}+\boldsymbol{v}$

where the resulting state can be normalised to $\frac{\boldsymbol{u}+\boldsymbol{v}}{\sqrt{2}}$ .

$\boldsymbol{u}$ can also be transformed (or flipped) into $\boldsymbol{v}$ using $\hat{O}=\begin{pmatrix}0&0\\0&1\end{pmatrix}$ , while a phase can be added to $\boldsymbol{u}$ using $\hat{O}=\begin{pmatrix}e^{i\theta}&0\\0&0\end{pmatrix}$ , where $\theta$ is the phase angle.

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.

Operators (quantum mechanics)

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