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# 4.2: Cartesian Symmetry

## Separation of Variables

Potentials with particular properties encourage us to separate variables in the cartesian coordinate system, so let's look at how this works. We seek solutions to the stationary-state Schrödinger equation that admit wave functions which can be written as a product of three functions of single variables:

$\psi_E\left(x,y,z\right) = X\left(x\right)Y\left(y\right)Z\left(z\right)$

Plugging this into the stationary-state Schrödinger equation, and dividing the whole equation by the wave function separates it into terms that are functions of only $$x$$, $$y$$, and $$z$$, along with a potential that so far we have not restricted:

$-\dfrac{\hbar^2}{2m}\left(\dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2} + \dfrac{\partial^2}{\partial z^2}\right)X\left(x\right)Y\left(y\right)Z\left(z\right) + V\left(x,y,z\right)X\left(x\right)Y\left(y\right)Z\left(z\right)= E\;X\left(x\right)Y\left(y\right)Z\left(z\right) \\ \Rightarrow \;\;\; \dfrac{1}{X}\dfrac{\partial^2X}{\partial x^2} + \dfrac{1}{Y}\dfrac{\partial^2Y}{\partial y^2} +\dfrac{1}{Z}\dfrac{\partial^2Z}{\partial z^2} -\dfrac{2m}{\hbar^2}V\left(x,y,z\right) = -\dfrac{2mE}{\hbar^2}$

For this method to be useful, we need to be able to separate the entire equation into a sum of terms that are exclusively functions of one variable at a time ($$x$$, $$y$$, or $$z$$). To see how this works, consider the following:

$f\left(x\right) + g\left(y\right) + h\left(z\right) = constant$

We can plug any $$x$$ value we like into $$f\left(x\right)$$, without changing any of the $$y$$ or $$z$$ values. For this equation to remain correct, it must mean that $$f\left(x\right)$$ is the same constant value for all choices of $$x$$. The same argument can be made for $$g\left(y\right)$$ and $$h\left(z\right)$$. This gives us three separate equations, each in a single variable.

The first three terms are already separated into $$x$$, $$y$$, and $$z$$, but the potential function poses a problem. This method is only really effective in two cases: When the potential is a constant (universally, or piecewise), or when it splits up into a sum of functions of single variables. We will look at examples of all these cases.

## Free Particle

If the potential is universally constant, then the particle is obviously free. As with the one-dimensional free particle, the stationary-state Schrödinger's equation gives us wave functions of energy eigenstates, but we can filter-out the momentum eigenstates (plane waves) if we wish. Plugging zero in for the potential allows us to separate the equation into three differential equations in single variables. The choices for the form of the constants below will become quickly apparent:

$\dfrac{1}{X}\dfrac{d^2X}{dx^2}=-k_x^2,\;\;\;\;\;\dfrac{1}{Y}\dfrac{d^2Y}{dy^2}=-k_y^2,\;\;\;\;\dfrac{1}{Z}\dfrac{d^2Z}{dz^2}=-k_z^2,\;\;\;\;\; k_x^2+k_y^2+k_z^2=\dfrac{2mE}{\hbar^2}$

The plane wave solutions of each of the separate differential equations are:

$X\left(x\right)=A_xe^{ik_x x},\;\;\;\;\; Y\left(y\right)=A_ye^{ik_y y},\;\;\;\;\; Z\left(z\right)=A_ze^{ik_z z}$

Reconstructing the full momentum eigenstate wave function, we get:

$\psi_k\left(x,y,z\right) = X\left(x\right)Y\left(y\right)Z\left(z\right) = Ae^{i\left(k_x x + k_y y + k_z z\right)}$

If we define the momentum vector $$\overrightarrow p$$ in terms of a wave vector $$\overrightarrow k = k_x \widehat i + k_y \widehat j + k_z \widehat k$$, we get the rather compact plane wave solution moving in a specific direction:

$\psi_k\left(\overrightarrow r\right) = Ae^{i\overrightarrow k \cdot \overrightarrow r},\;\;\;\;\;\;\;\; \overrightarrow k = \dfrac{\overrightarrow p}{\hbar},\;\;\;\;\;\;\;\; E=\dfrac{p^2}{2m}=\dfrac{\hbar^2}{2m}\left(\overrightarrow k \cdot \overrightarrow k\right)$

This plane wave is also an energy eigenstate, so the full time-dependent wave function can also be written:

$\psi_k\left(\overrightarrow r, t\right) = Ae^{i\left(\overrightarrow k \cdot \overrightarrow r - \omega t\right)},\;\;\;\;\;\;\;\; \omega = \dfrac{E}{\hbar}$

## Particle in a 3D Infinite Square Well

The infinite square well in three dimensions has the same property as the one-dimensional box – the potential is zero everywhere inside, and instantly becomes infinity at the boundaries. The one-dimensional case had a specified length, but we will not saddle this infinite well with the same width in all three directions, meaning we will confine the particle to a rectangular prism, not a cube. We will define our coordinate system so that the walls are parallel to the three planes, and (unlike what we chose for the one-dimensional case), we will place the origin at one of the box corners. The lengths of the walls along the $$x$$, $$y$$, and $$z$$ axes we will call $$L_x$$, $$L_y$$, $$L_z$$, respectively.

Figure 4.2.1 Three-Dimensional Infinite Square Well

Mathematically, the potential is written:

$V\left(x,y,z\right) = \left\{ \begin{array}{l} 0 & 0<x<L_x\;\;and\;\;0<y<L_y\;\;and\;\;0<z<L_z \\ \infty & elsewhere \end{array} \right.$

As we saw for the one-dimensional box, we can use a combination of two oppositely-moving plane waves (for each of the three axes) to construct a wave function of definite energy that vanishes at the walls (i.e. sinusoidal functions). The individual solutions to the differential equations in $$x$$, $$y$$, and $$z$$ are the same as before, with two exceptions: Each dimension involves a separate harmonic number $$n$$, and as we have chosen the origin to be at a wall (rather than centering it within the well), the wave functions are all sines:

$X\left(x\right) = A_x\sin\dfrac{n_x\pi \;x}{L_x} \;\;\;\;\; Y\left(y\right) = A_y\sin\dfrac{n_y\pi \;y}{L_y} \;\;\;\;\; Z\left(z\right)= A_z\sin\dfrac{n_z\pi \;z}{L_z},\;\;\;\;\; n_x,\;n_y,\;n_z=1,\;2,\;\dots$

Before we move on to the energy spectrum, let's construct the spatial full wave function by multiplying the partial wave functions. We also have to deal with normalizing the full wave function. Normalization does not tell us anything about the values of $$A_x$$, $$A_y$$, and $$A_z$$, but their product must equal the normalization constant for the full wave function. The normalization integral is over all three dimensions and integrals over $$x$$ will only affect $$X\left(x\right)$$, and similarly for $$y$$ and $$z$$, so the normalization constant for the full wave function turns out to be the same as the product of the normalization constants for the three separate one-dimensional wave functions.

$\psi_{n_xn_yn_z}\left(x,y,z\right) = \sqrt{\dfrac{8}{L_xL_yL_z}}\sin\dfrac{n_x\pi \;x}{L_x}\sin\dfrac{n_y\pi \;y}{L_y}\sin\dfrac{n_z\pi \;z}{L_z},\;\;\;\;\; n_x,\;n_y,\;n_z=1,\;2,\;\dots$

In keeping with our notation of labeling the wave function with the quantum numbers, we have labeled the energy eigenstate wave function accordingly.

Plugging the wave function back into the stationary-state Schrödinger equation, we get the following energy spectrum:

$E_{n_xn_yn_z} = \left(\dfrac{n_x^2}{L_x^2}+\dfrac{n_y^2}{L_y^2}+\dfrac{n_z^2}{L_z^2}\right)\dfrac{\pi^2\hbar^2}{2m},\;\;\;\;\; n_x,\;n_y,\;n_z=1,\;2,\;\dots$

One might be tempted to think that the ground state energy of this particle occurs when the $$n$$ along the longest dimension is 1, while the others are zero, but It should be emphasized that the minimum value of all three $$n$$ values is 1. None of the three modes can provide a zero contribution to the energy.

Another way that the three-dimensional case differs from the one-dimensional case is apparent if we consider the hierarchy of the energy spectrum. Suppose for example, we wanted to draw an energy-level diagram for this spectrum. We know that $$\psi_{111}$$ is the ground state, but which quantum state would be the first excited state? The answer depends upon the dimensions of the well. If the $$L_x$$ is greater than the other two box dimensions, then the smallest increase in the total energy will come from incrementing $$n_x$$ from 1 to 2, and the first excited state would be $$\psi_{211}$$. What about the second excited state? Well, now we need even more information. If $$L_x$$ is only slightly greater than $$L_y$$ dimension (and both are longer than $$L_z$$), then the second excited state would be $$\psi_{221}$$. But if $$L_x$$ is significantly longer than the other dimensions, then the second excited state would be $$\psi_{311}$$. In other words, with three quantum numbers, we have lost the ability (at least for this case) to express the energy levels with a single integer.

## The 3D Harmonic Oscillator

As our final example of a potential that allows for separation of variables in cartesian coordinate, we consider the three dimensional harmonic oscillator, which has a potential that is a sum of functions purely of $$x$$, $$y$$, and $$z$$. In general, the spring constants are different for each direction, so:

$V\left(x,y,z\right) = \frac{1}{2}\kappa_x x^2 + \frac{1}{2}\kappa_y y^2 + \frac{1}{2}\kappa_z z^2$

Plugging this into Equation 4.2.2 results in an equation with three separated functions again:

$\left[\dfrac{1}{X}\dfrac{\partial^2X}{\partial x^2} + \dfrac{\kappa_x m}{\hbar^2} x^2 \right] + \left[\dfrac{1}{Y}\dfrac{\partial^2Y}{\partial y^2} + \dfrac{\kappa_y m}{\hbar^2} y^2 \right] + \left[\dfrac{1}{Z}\dfrac{\partial^2Z}{\partial z^2} + \dfrac{\kappa_z m}{\hbar^2} z^2 \right] = -\dfrac{2mE}{\hbar^2}$

Following the same procedure as before gives us three separate differential equations. This decoupling maneuver once again leaves us with three wave function pieces, which are multiplied together to get the full wave function. As with the case of the square well, the energy contributions of the partial wave functions are added together to give the total energy of the state:

$E_{n_xn_yn_z} = \left(n_x + \frac{1}{2}\right)\hbar\sqrt{\dfrac{\kappa_x}{m}} + \left(n_y + \frac{1}{2}\right)\hbar\sqrt{\dfrac{\kappa_y}{m}} + \left(n_z + \frac{1}{2}\right)\hbar\sqrt{\dfrac{\kappa_z}{m}}$

While we frequently encounter interactions in the real world that approximate the harmonic oscillator potential, it is rare that the interactions between particles are different along different axes, so of particular interest is the isotropic harmonic oscillator, which involves equal spring constants ($$\kappa$$) in all three directions. In this case, the energy spectrum reduces to:

$E_{n_xn_yn_z} = \left(n_x + n_y + n_z + \frac{3}{2}\right)\hbar\omega_c,\;\;\;\;\;\;\;\; \omega_c = \sqrt{\dfrac{\kappa}{m}}$

Notice that even with the simplification of isotropy, three quantum numbers are required to define the state.

## Degeneracy

Notice that unlike one-dimensional potentials, in these cases a single quantum number does not define the energy. But there is even more to it than that. Looking at the case of the three-dimensional box again, suppose it has three equal sides: $$L_x=L_y=L_z=L$$. In this case, there exist three distinct quantum states that possess the same total energy, namely $$\psi_{211}$$, $$\psi_{121}$$, and $$\psi_{112}$$. These states clearly possess equal energies, and they are distinct because, for example, the states $$\psi_{211}$$ and $$\psi_{121}$$ yield different uncertainties in the $$x$$-component of the particle's position ($$\psi_{211}$$ has two antinodes along the $$x$$ axis, while $$\psi_{121}$$ has only one). A similar thing occurs (only more dramatically) for the isotropic harmonic oscillator, as any combination of $$n_x$$, $$n_y$$, and $$n_z$$ that gives the same sum will result in the same energy.

When multiple quantum states yield the same energy, they are said to be degenerate, and if there are a total of $$j$$ distinct states for the same energy, that energy level is said to be $$j$$-fold degenerate. Typically degeneracy comes about due to obvious symmetries, such as in the cases mentioned above. All we need to do is rename our axes, and the states morph into each other, so naturally the energies are the same. But occasionally degeneracies arise unexpectedly, through what can only really be described as a coincidence. These are called accidental degeneracies, the most famous of which arises for the hydrogen atom, as we will see later. An example of one of these for the three-dimensional square well arises for the states $$\psi_{511}$$, $$\psi_{151}$$, $$\psi_{115}$$, and $$\psi_{333}$$ – this energy level is 4-fold degenerate, rather than the "expected" 3-fold degeneracy. Naturally the first three states are not unexpectedly degenerate, but the fourth seems to come from left field.

Symmetric quantum systems are common in physics, and degeneracy follows them everywhere. This can cause difficulty in developing theory, as some internal structure can be obscured when different configurations result in the same energy spectrum. The trick then is to introduce an external perturbation that breaks the symmetry, thereby separating otherwise degenerate states. The analogous case for the cubical box would be squeezing or stretching one of the dimensions slightly. This also can work in the other direction – we might see unexpected additional spectral lines that indicate that there is additional structure present that breaks the symmetry we thought existed. So the analogous case for this is an infinite well that we think should be cubical, but provides a spectrum with energy levels landing between those that we compute.