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):


Linear operators have the following properties:

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



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