# 4.1: Expanding to Three Dimensions

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## Extending the Hilbert Space

When we extend our work on quantum mechanics in one dimension to three dimensions, we have to take into account what happens to the Hilbert space we have used to describe quantum states. The main difference that arises is that we still need to get to a single number (probability) from three times as many dimensions as before. Instead of describing Hilbert space "unit vectors" for position in terms of a single number \(x\), we require three numbers:

\[ \begin{array}{c} & \text{position eigenstate} & & \text{position-space wave function} \\ \\ \text{1-dimension:} & \left|\;x\;\right> & & \psi\left(x,t\right) = \left<\;x\;|\;\Psi\left(t\right)\;\right> \\ \\ \text{3-dimensions:} & \left|\;x,y,z\;\right>=\left|\;\overrightarrow r\;\right> & & \psi\left(x,y,z,t\right)=\psi\left(\overrightarrow r,t\right) = \left<\;\overrightarrow r\;|\;\Psi\left(t\right)\;\right> \end{array} \]

As before, the magnitude-squared of the state vector is a probability density, but in three dimensions, this becomes a volume density instead of a line density. Computations of probabilities in three dimensions requires integration over a three dimensional space:

The tiny volume element \(dV\) takes on different forms depending upon the coordinate system used. We will only be using cartesian and spherical coordinates:

\[ \begin{array}{c} & & \text{cartesian} & & \text{spherical} \\ dV & = & dx\;dy\;dz & & r^2\;dr\;\sin\theta \;d\theta \;d\phi \end{array} \]

Naturally the normalization condition is (in position space):

\[ \left<\;\Psi\left(t\right)\;|\;\Psi\left(t\right)\;\right> = \int \limits_{all\;space} \left|\psi\left(\overrightarrow r,t\right)\right|^2dV = 1\]

Written out explicitly for cartesian and spherical coordinates, this is:

As before, we can use eigenstates of momentum as our "unit vectors," and the relationship between the \(y\) and \(z\) components of momentum with the \(y\) and \(z\) coordinates is exactly the same as it was for the case of \(x\) in one dimension, which means that position space and momentum space wave functions are related through a fourier transform, which now must be performed in all three directions.

One interesting effect arises from the extension to three dimensions. With three independent variables, identifying just one of these three quantities does not define the state of the particle as it did in one dimension, when knowing \(x\) uniquely determined \(\psi\left(x\right)\). While it is clear that knowing the three quantities \(x\), \(y\), and \(z\) will define the state, this carries through into other descriptions of the state. For example, for bound states, it was enough to know the integer energy level \(n\) to have all the information about the state of the particle, but for bound states in three dimensions, three such integers are required (yes, the bound state is quantized in all three of these measurements). These integer labels are called *quantum numbers*, and are frequently used to label eigenstates in Hilbert space in the same way that \(n\) labeled the energy eigenstates in one dimension: \(\left|\;E_n\;\right>\).

## Quantum Numbers and "Unit Vectors"

In one dimension, for bound states described by a single quantum number, we said that the "unit vector" states are (of course) normalized, and also are orthogonal to each other. For example, for a one-dimensional energy eigenstate we wrote:

\[ \left<\;E_n\;|\;E_n'\;\right> = \delta_{nn'} \]

When we move to three dimensions and bound states involve three quantum numbers, these states are still all mutually orthogonal, so the inner product comes out to zero if *any* pair of comparable quantum numbers don't match. Abstractly, we can write this as:

## Schrödinger's Equation

Schrödinger's equation is easy to expand to three dimensions. All that is required is to use the kinetic and potential energy operators in three dimensions in the hamiltonian. In position space (and cartesian coordinates), we have:

\[ \widehat H = \frac{1}{2m}\left(\widehat p_x^2+\widehat p_y^2+\widehat p_z^2\right) + \widehat V = -\dfrac{\hbar^2}{2m}\left(\dfrac{\partial^2}{\partial x^2} + \dfrac{\partial^2}{\partial y^2} + \dfrac{\partial^2}{\partial z^2}\right) + V\left(x,y,z\right)\]

We can write this without reference to a choice of coordinate system by replacing the set of second derivatives with the laplacian operator (which we can look up for whatever coordinate system we like):

\[ \widehat H = -\dfrac{\hbar^2}{2m}\nabla^2 + V\left(\overrightarrow r\right) \]

Once again, we are faced with a partial differential equation, this time in *four* variables. Just as before, we can use the separation of variables method to select-out the stationary-state subset of solutions:

\[ \psi_E\left(\overrightarrow r,t\right) = \psi_E\left(\overrightarrow r\right)e^{-i\omega t},\;\;\;\;\; where:\;\;\; -\dfrac{\hbar^2}{2m}\nabla^2\psi_E\left(\overrightarrow r\right) + V\left(\overrightarrow r\right)\psi_E\left(\overrightarrow r\right)=E\;\psi_E\left(\overrightarrow r\right),\;\;\; \omega=\dfrac{E}{\hbar} \]

Okay, now we have a differential equation with three variables to solve. If the separation of variables trick is so effective, why not just use it again? Not so fast! Unlike separating the time variable from the spatial variables, separating the spatial variables from each other is trickier business, because we can choose any set of coordinates we like. That is, there is nothing about the Schrödinger equation written above in cartesian coordinates that is any more "correct" than if it is written in spherical coordinates, so how do we know whether we can (or should) do a separation of variables in cartesian coordinates?

The answer is utility – we trust that like the case of separation of the time variable, we will be able to construct more general solutions from linear combinations of whatever solutions we arrive at after separating variables, so we are free to try the separation in any coordinate system. So the system we choose should be the one that is easiest to work with for the physical situation given. If the potential is easy to work with in one coordinate system, then that is the one to use. Ultimately whatever separation we choose will lead to its own set of three quantum numbers, as we will see with our two cases of cartesian and spherical coordinates.