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202203171422 Half Picture of Quantum Mechanics 01B

(This half picture is based on the second half of manuscript of Chapter One, 2015-2016 PHYS3351 Quantum Mechanics Lecture Notes.)

Section 01

Momentum

As the wavefunction describes the state of a quantum system, how do we find the information of velocity or momentum?

Expectation value \langle\mathbf{r}\rangle of position \mathbf{r}

\langle \mathbf{r}\rangle=\displaystyle{\iiint\mathrm{d}V\,\mathbf{r}|\Psi (\mathbf{r},t)|^2}

is a function of time t.

\begin{aligned} \frac{\mathrm{d}\langle\mathbf{r}\rangle}{\mathrm{d}t} & = \iiint\mathrm{d}V\,\mathbf{r}\frac{\partial}{\partial t}|\Psi |^2 \\ & = \frac{i\hbar}{2m}\iiint\mathrm{d}V\,\mathbf{r}\nabla\cdot [\Psi^*\nabla\Psi - (\nabla\Psi^* )\Psi ] \\ \dots &\textrm{ integration by parts }\dots \\ \frac{\mathrm{d}\langle\mathbf{r}\rangle}{\mathrm{d}t} & = -\frac{i\hbar}{2m}\iiint [\Psi^*\nabla\Psi - (\nabla\Psi^* )\Psi ]\,\mathrm{d}V \\ \dots &\textrm{ integration by parts }\dots \\ \underbrace{\frac{\mathrm{d}\langle\mathbf{r}\rangle}{\mathrm{d}t}}_{\textrm{velocity}} & =\iiint\mathrm{d}V\,\Psi^*\bigg( -\frac{i\hbar}{m}\nabla\bigg)\Psi \\ \end{aligned}

\displaystyle{\frac{\mathrm{d}\langle \mathbf{r}\rangle}{\mathrm{d}t}} is the “velocity” of the central position of the particle “cloud”; whereas

\displaystyle{m\frac{\mathrm{d}\langle\mathbf{r}\rangle}{\mathrm{d}t}=\iiint\mathrm{d}V\,\Psi^*(-i\hbar\nabla )\Psi}

is a very suggestive form for us to introduce the momentum operator:

\displaystyle{\hat{p}=\frac{\hbar}{i}\nabla}.

Expectation value:

\begin{aligned} \langle\hat{p}\rangle & = -i\hbar\iiint\mathrm{d}V\,\Psi^*\nabla\Psi \\ & = m\frac{\mathrm{d}\langle \mathbf{r}\rangle}{\mathrm{d}t} \\ \end{aligned}

Kinetic energy:

\hat{T}=\displaystyle{\frac{\hat{p}}{2m}=-\frac{\hbar^2}{2m}\nabla^2}

Angular momentum:

\hat{L}=\mathbf{r}\times \hat{p} = \displaystyle{\mathbf{r}\times \bigg(\frac{\hbar}{i}\nabla\bigg)}.

\begin{aligned} \langle\hat{T}\rangle & = -\frac{\hbar^2}{2m}\iiint\mathrm{d}V\,\Psi^*\nabla^2\Psi \\ \langle \hat{L}\rangle & = \iiint\mathrm{d}V\,\Psi^*\bigg(\mathbf{r}\times\frac{\hbar}{i}\nabla \bigg)\Psi \\ \end{aligned}

Summary.

In quantum mechanics, all observable quantities, such as momentum, position, or energy, are described by operators. When we describe a particle by the wavefunction \Psi (\mathbf{r},t), we are looking at the probability distribution at the position representation. In this representation the position operator takes the special form of a number: \mathbf{r}; whereas other observables such as momentum and energy involve the differential operator \nabla. For a particle described by a wavefunction, the expectation value of the observed quantity is given by

\langle\hat{S}\rangle = \displaystyle{\iiint\mathrm{d}V\,\Psi^*\hat{S}\Psi}

where \hat{S} denotes the operator form of the observable.

Section 02

Uncertainty principle

Like the standard deviation of position distribution, we can also evaluate the standard deviation of momentum:

\sigma_{p_x}^2=\langle\hat{p}_x^2\rangle - \langle\hat{p}_x\rangle^2.

For any possible wavefunction, one finds that

\sigma_x\sigma_{p_x}\geqslant\displaystyle{\frac{\hbar}{2}}.

Example: Gaussian wavepacket

\begin{aligned} \Psi (x,0) & = A\exp \bigg(-\frac{x^2}{4\sigma^2}+iKx\bigg) \\ A & = (2\pi\sigma^2)^{-1/4} \\ \hat{p}_x & = \frac{\hbar}{i}\frac{\partial}{\partial x} \\ \Psi (x,0) & = \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi (k)e^{ikx}\,\mathrm{d}k \\ \phi (k) & = (2\sigma^2/\pi )^{1/4}\exp [-\sigma^2(k-K)^2] \\ \end{aligned}

\begin{aligned} \langle \hat{p}_x^n\rangle & = \int_{-\infty}^{\infty}\bigg[ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi^*(k)e^{-ikx}\,\mathrm{d}k\bigg](\hat{p}_x)^n\bigg[ \frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi (k')e^{ik'x}\,\mathrm{d}k'\bigg] \\ & = \int_{-\infty}^{\infty}\mathrm{d}x\,\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi^* (k)e^{-ikx}\,\mathrm{d}k\frac{1}{\sqrt{2\pi}}\int_{-\infty}^{\infty}\phi (k')(\hbar k')^ne^{ik'x}\,\mathrm{d}k' \\ & = \int_{-\infty}^{\infty}\mathrm{d}k\,\phi^* (k)\int_{-\infty}^{\infty}\phi (k')\delta (k-k')(\hbar k')^n\,\mathrm{d}k' \\ & = \int_{-\infty}^{\infty}(\hbar k)^n|\phi (k)|^2\,\mathrm{d}k \\ \end{aligned}

so

\begin{aligned} \langle\hat{p}_x\rangle & = \int\mathrm{d}x\,\Psi^*\bigg( \frac{\hbar}{i}\frac{\partial}{\partial x}\bigg)\Psi = \hbar K \\ \Delta p_x & = \langle \hat{p}_x^2\rangle - \langle \hat{p}_x\rangle^2 = \frac{\hbar}{2\sigma} \\ \dots \textrm{ because }&\Delta x=\sigma \\ \Delta x\cdot\Delta p_x & = \frac{\hbar}{2} \\ \end{aligned}

\therefore Gaussian wavepackets satisfy the uncertainty relation. This relation tells us that the position and the momentum of a particle can \textrm{\scriptsize{NEVER}} be simultaneously determined with precision. This also shows the significance of the constant \hbar. Only when we talk about quantities with this scale, quantum effect is significant, that is, things cannot be described by classical mechanics. But if we cannot measure things with such precision, you won’t “notice” the quantum effect: that is why our daily experience is classical.

Von Neumann projection postulate:

Measurement collapses the wavefunction.

E.g., A position measurement with the outcome \mathbf{r}_0 collapses the wavefunction from \Psi (\mathbf{r},t) to \delta (\mathbf{r}-\mathbf{r}_0). So a second measurement immediately following will also find the particle at \mathbf{r}_0.

Note the subtle difference of expectation value from mean value in the usual sense.

(to be continued)