# Quantum Mechanics in higher-dimensional spaces

(Part of the Wolverhampton Lectures of Physics's Quantum Physics Course)

Last year, we studied a lot the 1D Schrödinger equation:

$$i\hbar\partial_t\psi=H\psi$$

which was 1D because the Hamiltonian (i.e., energy) $H$ was 1D, namely, we studied:

$$H=\frac12mv^2+V$$

(kinetic + potential energy), and using the correspondence principle, that takes us from classical to quantum mechanics:

$$p\rightarrow-i\hbar\partial_x$$

so our Schrödinger equation from last year was:

$$i\hbar\partial_t\psi(x,t)=\left(-{\hbar^2\over2m}\partial^2_x+V\right)\psi(x,t)\,.$$

This Year, our program will be essentially to do quantum mechanics in higher-dimensional spaces. The nature of these extra dimensions can be of a different character. The most obvious case is that it could be that our particle is now allowed to move in a higher-dimensional space, like free space (so that's 3D). We'll spend much time studying a fascinating problem, the hydrogen atom, which is 3D. We will remind ourselves of the basic principles of the theory first and see how it extends to the higher-D case.

For such higher-dimensional cases of a single particle, the momentum $\vec p=m \vec v$ becomes a vector, with three components. The corresponding principle reads:

$$\vec p\rightarrow-i\hbar\nabla\,.$$

The wavefunction itself on which this applies is a function of the corresponding-dimension variable:

$$\psi(\vec r,t)$$

with normalization condition

$$\int|\psi(\vec r,t)|^2 d\vec r=1$$

and observables obtained through the high-D sandwich process

$$\langle\Omega\rangle=\int\psi^*(\vec r)\Omega\psi(\vec r)d\vec r\,.$$

Remember that $\Omega$ is an operator, and does not commute in general with the wavefunctions!

A typical observable is, for instance, the position of the particle:

\begin{align} \langle\vec r\rangle&=\langle x\hat\imath+y\hat\jmath+z\hat k\rangle\\ &=\langle x\hat\imath\rangle+ \langle y\hat\jmath\rangle+\langle z\hat k\rangle\\ &=\langle x\rangle\hat\imath+ \langle y\rangle\hat\jmath+\langle z\rangle\hat k \end{align}

Let us now look at the kinetic energy, $K=\frac12mv^2$. It is a scalar (energy is a scalar) but $v^2$ comes from a 3D velocity, which is, clearly:

$$v^2=\vec v\cdot\vec v=v_x^2+v_y^2+v_z^2\,.$$

From the corresponding principle, we will, like in 1D, use momentum rather than velocity:

$$\displaystyle K={p^2\over 2m}={\vec p\cdot\vec p\over 2m}={p_x^2\over 2m}+{p_y^2\over 2m}+{p_z^2\over 2m}\,.$$

So our 3D Schrödinger equation finally reads:

$$i\hbar\partial_t\psi(\vec r,t)=\left(-{\hbar^2\over2m}\nabla^2+V\right)\psi(\vec{r},t)$$

That is a general result. Like in 1D, things will become particular when we specify the potential energy, and we'll also look at familiar cases. For the hydrogen atom, one of our main targets, this will be Coulomb's energy (that is, the potential which gradient gives Coulomb's Force), and our Hamiltonian will be

$$H=K+V$$

with $K$ as above and $V$ as:

$$\displaystyle V(r)=-{e^2\over4\pi\epsilon_0}{1\over r}\,.$$

The general time-dependent Schrödinger equation can still be solved by separation of (time and space) variables, which yields:

$$H\psi=E\psi\,.$$

But in the spirit of starting with easy things first, we will start with the simpler problem of a Cartesian problem $(x, y, z)$ that is separable in this coordinate system, starting with the so-called "particle in a box", which we know well in 1D, and that we will now study in 3D, although this will mainly be a review indeed as in this case, the 3D Schrödinger equation really is a 1D equation. Indeed, from:

$$\displaystyle-{\hbar^2\nabla^2\over2m}(\partial_x^2+\partial_y^2+\partial_z^2)\psi(x,y,z)=E\psi(x,y,z)$$

by introducing $\psi(x,y,z)=X(x)Y(y)Z(z)$, we get to:

$$\displaystyle-{\hbar^2\over2m}\left({\partial_x^2 X\over X}+{\partial_y^2 Y\over Y}+{\partial_z^2 Z\over Z}\right)=E$$

which, by the method of separation of variables, turn into three equations:

\begin{align} -{\hbar^2\over2m}\left({\partial_x^2X\over X}\right)&=E_x\\ -{\hbar^2\over2m}\left({\partial_y^2Y\over Y}\right)&=E_y\\ -{\hbar^2\over2m}\left({\partial_z^2Z\over Z}\right)&=E_z \end{align}

with

$$E_x+E_y+E_z=E$$

and, each of these equation is a 1D Schrödinger equation:

$$-{\hbar^2\over2m}\partial_x^2X=E_xX$$

which we know how to solve. In the box, the solutions are:

$$X_n=\sqrt{2\over L}\sin\left({n_x\pi x\over L}\right)$$

with

$$\displaystyle E_{n_x}={n_x^2\pi^2\hbar^2\over 2mL^2}$$

and similarly for $x$ and $y$, with obvious substitutions.

The ground state is the state with $n_x=n_y=n_z=1$ with energy $3E_0$ where $E_0\equiv{\pi^2\hbar^2\over 2mL^2}$.

The first excited state is not unique, since we find that (1,1,2), (1,2,1) & (2,1,1) all have the same energy: $6E_0$. These three states are said to be degenerate. The second excited state also has a degeneracy of three (meaning, 3 different states which result in the same energy, namely (1,2,2), (2,1,2) & (2,2,1)) with energy $9E_0$, and so is the 3rd state... Do we see a pattern there? The 4th excited state is not degenerate: (2,2,2) with energy $12E_0$. And the fifth excited state has degeneracy 6 ((3,2,1) (3,1,2) (2,3,1) (2,1,3) (1,3,2) (1,2,3)) with energy $14E_0$. Such counting is important in statistical mechanics, where one looks at the possible ways to distribute energy into the available excited states.