The Schrödinger equation in three dimensions · Quantum Mechanics

Intuition

In space the wavefunction depends on three coordinates, the momentum becomes a vector of three operators, and the second derivative of the line becomes the Laplacian. Nothing new in principle: the same equation, the same normalisation, the same current. What is new is room: a free particle now has a whole sphere of momenta with the same energy, and potentials can have shapes — spheres, cylinders, boxes — whose symmetries the next lessons exploit.

A guitar string vibrates along a line; a drum skin in a plane; the air in a room in three dimensions. The equation is the same kind in each, but in more dimensions there are many more ways to vibrate at the same pitch.

Wave vectors of one length drawn in the $k_{x}$ – $k_{y}$ plane. Every plane wave $e^{i\mathbf{k}\cdot\mathbf{r}}$ with $|\mathbf{k}|=k$ has the energy $\hbar^{2}k^{2}/2m$ , whatever its direction: in three dimensions a whole sphere of momenta shares each free energy.

The Schrödinger equation in space

The momentum is $\hat{\mathbf{p}}=-i\hbar\nabla$ , the kinetic energy $-\frac{\hbar^{2}}{2m}\nabla^{2}$ , and the wavefunction is normalised over all space.

\begin{aligned}-\frac{\hbar^{2}}{2m}\nabla^{2}\psi+V(\mathbf{r})\psi&=E\psi\\ \int|\psi(\mathbf{r})|^{2}\,d^{3}r&=1\end{aligned}

Properties

$[\hat{x}_{i},\hat{p}_{j}]=i\hbar\,\delta_{ij}$ : each coordinate fails to commute only with its own momentum.

Plane waves solve the free equation

\begin{aligned}&\psi=e^{i\mathbf{k}\cdot\mathbf{r}}\ \Longrightarrow\ -i\hbar\nabla\psi=\hbar\mathbf{k}\,\psi\\ &\phantom{\psi=e^{i\mathbf{k}\cdot\mathbf{r}}}\ \ \ -\frac{\hbar^{2}}{2m}\nabla^{2}\psi=\frac{\hbar^{2}k^{2}}{2m}\psi\end{aligned}

Each derivative of the exponential brings down $i$ times the matching component of $\mathbf{k}$ . One derivative gives the momentum; two give $-k^{2}$ , and the kinetic energy follows.

Proof steps

\nabla e^{i\mathbf{k}\cdot\mathbf{r}}=i\mathbf{k}\,e^{i\mathbf{k}\cdot\mathbf{r}}

$\partial_{x}e^{i(k_{x}x+k_{y}y+k_{z}z)}=ik_{x}e^{i\mathbf{k}\cdot\mathbf{r}}$ , and so for and .

-i\hbar\nabla e^{i\mathbf{k}\cdot\mathbf{r}}=\hbar\mathbf{k}\,e^{i\mathbf{k}\cdot\mathbf{r}}

A momentum eigenstate with eigenvalue $\hbar\mathbf{k}$ .

\nabla^{2}e^{i\mathbf{k}\cdot\mathbf{r}}=(i\mathbf{k})\cdot(i\mathbf{k})\,e^{i\mathbf{k}\cdot\mathbf{r}}=-k^{2}e^{i\mathbf{k}\cdot\mathbf{r}}

The divergence of the gradient.

-\frac{\hbar^{2}}{2m}\nabla^{2}e^{i\mathbf{k}\cdot\mathbf{r}}=\frac{\hbar^{2}k^{2}}{2m}\,e^{i\mathbf{k}\cdot\mathbf{r}}

The energy depends only on $|\mathbf{k}|$ .

Applications

Practice

Momentum as a Gradient

In three dimensions the momentum operator is $-i\hbar\nabla$ : one derivative for each component.

\hat{\mathbf{p}}=-i\hbar\nabla

Try it

What is $\hat{p}_{y}$ acting on a wavefunction $\psi(x,y,z)$ ?