# Postulates of quantum mechanics: Schrödinger equation and propagators

References: Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 4.3.

The first three postulates of quantum mechanics concern the properties of a quantum state. The fourth postulate concerns how states evolve with time. The postulate simply states that in non-relativistic quantum mechanics, a state satisfies the Schrödinger equation:

$\displaystyle i\hbar\frac{\partial}{\partial t}\left|\psi\right\rangle =H\left|\psi\right\rangle \ \ \ \ \ (1)$

where ${H}$ is the Hamiltonian, which is obtained from the classical Hamiltonian by means of the other postulates of quantum mechanics, namely that we replace all references to the position ${x}$ by the quantum position operator ${X}$ with matrix elements (in the ${x}$ basis) of

$\displaystyle \left\langle x^{\prime}\left|X\right|x\right\rangle =\delta\left(x-x^{\prime}\right) \ \ \ \ \ (2)$

and all references to classical momentum ${p}$ by the momentum operator ${P}$ with matrix elements

$\displaystyle \left\langle x^{\prime}\left|P\right|x\right\rangle =-i\hbar\delta^{\prime}\left(x-x^{\prime}\right) \ \ \ \ \ (3)$

Although we’ve posted many articles based on Griffiths’s book in which we solved the Schrödinger equation, the approach taken by Shankar is a bit different and, in some ways, a lot more elegant. We begin with a Hamiltonian that does not depend explicitly on time, and then by observing that, since the Schrödinger equation contains only the first derivative with respect to time, The time evolution of a state can be uniquely determined if we specify only the initial state ${\left|\psi\left(0\right)\right\rangle }$. [A differential equation that is second order in time, such as the wave equation, requires both the initial position and initial velocity to be specified.]

The solution of the Schrödinger equation is then found in analogy to the approach we used in solving the coupled masses problem earlier. We find the eigenvalues and eigenvectors of the Hamiltonian in some basis and use these to construct the propagator ${U\left(t\right)}$. We can then write the solution as

$\displaystyle \left|\psi\left(t\right)\right\rangle =U\left(t\right)\left|\psi\left(0\right)\right\rangle \ \ \ \ \ (4)$

For the case of a time-independent Hamiltonian, we can actually construct ${U\left(t\right)}$ in terms of the eigenvectors of ${H}$. The eigenvalue equation is

$\displaystyle H\left|E\right\rangle =E\left|E\right\rangle \ \ \ \ \ (5)$

where ${E}$ is an eigenvalue of ${H}$ and ${\left|E\right\rangle }$ is its corresponding eigenvector. Since the eigenvectors form a vector space, we can expand the wave function in terms of them in the usual way

 $\displaystyle \left|\psi\left(t\right)\right\rangle$ $\displaystyle =$ $\displaystyle \sum\left|E\right\rangle \left\langle E\left|\psi\left(t\right)\right.\right\rangle \ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle \equiv$ $\displaystyle \sum a_{E}\left(t\right)\left|E\right\rangle \ \ \ \ \ (7)$

The coefficient ${a_{E}\left(t\right)}$ is the component of ${\left|\psi\left(t\right)\right\rangle }$ along the ${\left|E\right\rangle }$ vector as a function of time. We can now apply the Schrödinger equation 1 to get (a dot over a symbol indicates a time derivative):

 $\displaystyle i\hbar\frac{\partial}{\partial t}\left|\psi\left(t\right)\right\rangle$ $\displaystyle =$ $\displaystyle i\hbar\sum\dot{a}_{E}\left(t\right)\left|E\right\rangle \ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle H\left|\psi\left(t\right)\right\rangle \ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum a_{E}\left(t\right)H\left|E\right\rangle \ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum a_{E}\left(t\right)E\left|E\right\rangle \ \ \ \ \ (11)$

Since the eigenvectors ${\left|E\right\rangle }$ are linearly independent (as they form a basis for the vector space), each term in the sum in the first line must be equal to the corresponding term in the sum in the last line, so we have

$\displaystyle i\hbar\dot{a}_{E}\left(t\right)=a_{E}\left(t\right)E \ \ \ \ \ (12)$

The solution is

 $\displaystyle a_{E}\left(t\right)$ $\displaystyle =$ $\displaystyle a_{E}\left(0\right)e^{-iEt/\hbar}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{-iEt/\hbar}\left\langle E\left|\psi\left(0\right)\right.\right\rangle \ \ \ \ \ (14)$

The general solution 7 is therefore

$\displaystyle \left|\psi\left(t\right)\right\rangle =\sum e^{-iEt/\hbar}\left|E\right\rangle \left\langle E\left|\psi\left(0\right)\right.\right\rangle \ \ \ \ \ (15)$

from which we can read off the propagator:

$\displaystyle U\left(t\right)=\sum e^{-iEt/\hbar}\left|E\right\rangle \left\langle E\right| \ \ \ \ \ (16)$

Thus if we can determine the eigenvalues and eigenvectors of ${H}$, we can write the propagator in terms of them and get the general solution. We can see from this that ${U\left(t\right)}$ is unitary:

 $\displaystyle U^{\dagger}U$ $\displaystyle =$ $\displaystyle \sum_{E^{\prime}}\sum_{E}e^{-i\left(E-E^{\prime}\right)t/\hbar}\left|E\right\rangle \left\langle E\left|E^{\prime}\right.\right\rangle \left\langle E^{\prime}\right|\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{E^{\prime}}\sum_{E}e^{-i\left(E-E^{\prime}\right)t/\hbar}\left|E\right\rangle \delta_{EE^{\prime}}\left\langle E^{\prime}\right|\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{E}\left|E\right\rangle \left\langle E\right|\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (20)$

This derivation uses the fact that the eigenvectors are orthonormal and form a complete set, so that ${\left\langle E\left|E^{\prime}\right.\right\rangle =\delta_{EE^{\prime}}}$ and ${\sum_{E}\left|E\right\rangle \left\langle E\right|=1}$. Since a unitary operator doesn’t change the norm of a vector, we see from 4 that if ${\left|\psi\left(0\right)\right\rangle }$ is normalized, then so is ${\left|\psi\left(t\right)\right\rangle }$ for all times ${t}$. Further, the probability that the state will be measured to be in eigenstate ${\left|E\right\rangle }$ is constant over time, since this probability is given by

$\displaystyle \left|a_{E}\left(t\right)\right|^{2}=\left|e^{-iEt/\hbar}\left\langle E\left|\psi\left(0\right)\right.\right\rangle \right|^{2}=\left|\left\langle E\left|\psi\left(0\right)\right.\right\rangle \right|^{2} \ \ \ \ \ (21)$

This derivation assumed that the spectrum of ${H}$ was discrete and non-degenerate. If the possible eigenvalues ${E}$ are continuous, then the sum is replaced by an integral

$\displaystyle U\left(t\right)=\int e^{-iEt/\hbar}\left|E\right\rangle \left\langle E\right|dE \ \ \ \ \ (22)$

If the spectrum is discrete and degenerate, then we need to find an orthonormal set of eigenvectors that spans each degenerate subspace, and sum over these sets. For example, if ${E_{1}}$ is degenerate, then we find a set of eigenvectors ${\left|E_{1},\alpha\right\rangle }$ that spans the subspace for which ${E_{1}}$ is the eigenvalue. The index ${\alpha}$ runs from 1 up to the degree of degeneracy of ${E_{1}}$, and the propagator is then

$\displaystyle U\left(t\right)=\sum_{\alpha}\sum_{E_{i}}e^{-iE_{i}t/\hbar}\left|E_{i},\alpha\right\rangle \left\langle E_{i},\alpha\right| \ \ \ \ \ (23)$

The sum over ${E_{i}}$ runs over all the distinct eigenvalues, and the sum over ${\alpha}$ runs over the eigenvectors for each different ${E_{i}}$.

Another form of the propagator can be written directly in terms of the time-independent Hamiltonian as

$\displaystyle U\left(t\right)=e^{-iHt/\hbar} \ \ \ \ \ (24)$

This relies on the concept of the function of an operator, so that ${e^{-iHt/\hbar}}$ is a matrix whose elements are power series of the exponent ${-\frac{iHt}{\hbar}}$. The power series must, of course, converge for this solution to be valid. Since ${H}$ is Hermitian, ${U\left(t\right)}$ is unitary. We can verify that the solution using this form of ${U\left(t\right)}$ satisfies the Schrödinger equation:

 $\displaystyle \left|\psi\left(t\right)\right\rangle$ $\displaystyle =$ $\displaystyle U\left(t\right)\left|\psi\left(0\right)\right\rangle \ \ \ \ \ (25)$ $\displaystyle$ $\displaystyle =$ $\displaystyle e^{-iHt/\hbar}\left|\psi\left(0\right)\right\rangle \ \ \ \ \ (26)$ $\displaystyle i\hbar\left|\dot{\psi}\left(t\right)\right\rangle$ $\displaystyle =$ $\displaystyle i\hbar\frac{d}{dt}\left(e^{-iHt/\hbar}\right)\left|\psi\left(0\right)\right\rangle \ \ \ \ \ (27)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\left(-\frac{i}{\hbar}\right)He^{-iHt/\hbar}\left|\psi\left(0\right)\right\rangle \ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle He^{-iHt/\hbar}\left|\psi\left(0\right)\right\rangle \ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle H\left|\psi\left(t\right)\right\rangle \ \ \ \ \ (30)$

The derivative of ${U\left(t\right)}$ can be calculated from the derivatives of its matrix elements, which are all power series.

# Lagrangian for the Schrödinger equation

References: W. Greiner & J. Reinhardt, Field Quantization, Springer-Verlag (1996), Chapter 3, Section 3.1.

As a prelude to ‘proper’ quantum field theory, we’ll look first at turning the non-relativistic quantum theory based on the Schrödinger equation into a field theory. Before we develop a quantum field theory of the Schrödinger equation, we’ll first look at this equation treating the wave function ${\psi\left(\mathbf{x},t\right)}$ as a classical (that is, non-quantum) field. The Schrödinger equation is

$\displaystyle i\hbar\frac{\partial\psi}{\partial t}=-\frac{\hbar^{2}}{2m}\nabla^{2}\psi+V\left(\mathbf{x},t\right)\psi \ \ \ \ \ (1)$

where ${V\left(\mathbf{x},t\right)}$ is, as usual, the potential function.

In order to apply the techniques of classical field theory, we need a Lagrangian density ${\mathcal{L}}$. There doesn’t seem to be any way of actually deriving Lagrangian densities; presumably they are found through trial and error, with perhaps a bit of physical intuition. In any case, the Lagrangian density for the Schrödinger equation turns out to be

$\displaystyle \mathcal{L}\left(\psi,\nabla\psi,\dot{\psi}\right)=i\hbar\psi^*\dot{\psi}-\frac{\hbar^{2}}{2m}\nabla\psi^*\cdot\nabla\psi-V\left(\mathbf{x},t\right)\psi^*\psi \ \ \ \ \ (2)$

As ${\psi}$ is a complex function, it has real and imaginary parts, so we can treat ${\psi}$ and ${\psi^*}$ as independent fields. As we saw earlier, we can derive the Euler-Lagrange equations for multiple fields from the principle of least action and end up with

$\displaystyle \frac{\partial\mathcal{L}}{\partial\phi^{r}}-\frac{\partial}{\partial q^{\mu}}\left(\frac{\partial\mathcal{L}}{\partial\phi_{,\mu}^{r}}\right)=0 \ \ \ \ \ (3)$

where the ${\phi^{r}}$ are the fields and ${q^{\mu}=\left(\mathbf{x},t\right)}$. In this case, the two fields are ${\psi}$ and ${\psi^*}$ and we get the two equations

 $\displaystyle \frac{\partial\mathcal{L}}{\partial\psi}-\frac{\partial}{\partial x^{i}}\frac{\partial\mathcal{L}}{\partial\psi_{,i}}-\frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial\dot{\psi}}$ $\displaystyle =$ $\displaystyle \frac{\partial\mathcal{L}}{\partial\psi}-\nabla\cdot\frac{\partial\mathcal{L}}{\partial\nabla\psi}-\frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial\dot{\psi}}=0\ \ \ \ \ (4)$ $\displaystyle \frac{\partial\mathcal{L}}{\partial\psi^*}-\frac{\partial}{\partial x^{i}}\frac{\partial\mathcal{L}}{\partial\psi_{,i}^*}-\frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial\dot{\psi}^*}$ $\displaystyle =$ $\displaystyle \frac{\partial\mathcal{L}}{\partial\psi^*}-\nabla\cdot\frac{\partial\mathcal{L}}{\partial\nabla\psi^*}-\frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial\dot{\psi}^*}=0 \ \ \ \ \ (5)$

The second term in each row just introduces the gradient sign ${\nabla}$ as a shorthand for the ${\frac{\partial}{\partial x^{i}}\frac{\partial\mathcal{L}}{\partial\psi_{,i}}}$ and ${\frac{\partial}{\partial x^{i}}\frac{\partial\mathcal{L}}{\partial\psi_{,i}^*}}$ terms.

We can plug 2 into these two equations to verify that we recover the original Schrödinger equation 1 and its complex conjugate. From 4 we have

 $\displaystyle \frac{\partial\mathcal{L}}{\partial\psi}$ $\displaystyle =$ $\displaystyle -V\left(\mathbf{x},t\right)\psi^*\ \ \ \ \ (6)$ $\displaystyle \nabla\cdot\frac{\partial\mathcal{L}}{\partial\nabla\psi}$ $\displaystyle =$ $\displaystyle -\frac{\hbar^{2}}{2m}\nabla^{2}\psi^*\ \ \ \ \ (7)$ $\displaystyle \frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial\dot{\psi}}$ $\displaystyle =$ $\displaystyle i\hbar\dot{\psi}^*\ \ \ \ \ (8)$ $\displaystyle \frac{\partial\mathcal{L}}{\partial\psi}-\nabla\cdot\frac{\partial\mathcal{L}}{\partial\nabla\psi}-\frac{\partial}{\partial t}\frac{\partial\mathcal{L}}{\partial\dot{\psi}}$ $\displaystyle =$ $\displaystyle -i\hbar\dot{\psi}^*+\frac{\hbar^{2}}{2m}\nabla^{2}\psi^*-V\left(\mathbf{x},t\right)\psi^*=0\ \ \ \ \ (9)$ $\displaystyle -i\hbar\dot{\psi}^*$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{2m}\nabla^{2}\psi^*-V\left(\mathbf{x},t\right)\psi^* \ \ \ \ \ (10)$

which is the complex conjugate of 1. Plugging 2 into 5 just reproduces 1.

The conjugate momentum density ${\pi}$ can be calculated for the two fields ${\psi}$ and ${\psi^*}$. We get

 $\displaystyle \pi_{1}\left(\mathbf{x},t\right)$ $\displaystyle =$ $\displaystyle \frac{\partial\mathcal{L}}{\partial\dot{\psi}}=i\hbar\psi^*\left(\mathbf{x},t\right)\ \ \ \ \ (11)$ $\displaystyle \pi_{2}\left(\mathbf{x},t\right)$ $\displaystyle =$ $\displaystyle \frac{\partial\mathcal{L}}{\partial\dot{\psi}^*}=0 \ \ \ \ \ (12)$

The Hamiltonian density is defined as

 $\displaystyle \mathcal{H}$ $\displaystyle =$ $\displaystyle \sum_{r}\pi_{r}\dot{\phi}^{r}-\mathcal{L}\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle i\hbar\psi^*\dot{\psi}-\left[i\hbar\psi^*\dot{\psi}-\frac{\hbar^{2}}{2m}\nabla\psi^*\cdot\nabla\psi-V\left(\mathbf{x},t\right)\psi^*\psi\right]\ \ \ \ \ (14)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{2m}\nabla\psi^*\cdot\nabla\psi+V\left(\mathbf{x},t\right)\psi^*\psi \ \ \ \ \ (15)$

The total Hamiltonian is the integral of this over 3-d space:

 $\displaystyle H$ $\displaystyle =$ $\displaystyle \int d^{3}x\;\mathcal{H}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}x\left[\frac{\hbar^{2}}{2m}\nabla\psi^*\cdot\nabla\psi+V\left(\mathbf{x},t\right)\psi^*\psi\right] \ \ \ \ \ (17)$

We can integrate the first term by parts, by integrating the ${\nabla\psi^*}$ term and invoking the usual assumption that ${\psi^*\rightarrow0}$ fast enough at infinity that the integrated term is zero. We then get

 $\displaystyle H$ $\displaystyle =$ $\displaystyle \int d^{3}x\left[-\frac{\hbar^{2}}{2m}\psi^*\nabla^{2}\psi+V\left(\mathbf{x},t\right)\psi^*\psi\right]\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int d^{3}x\;\psi^*\left[-\frac{\hbar^{2}}{2m}\nabla^{2}\psi+V\left(\mathbf{x},t\right)\psi\right] \ \ \ \ \ (19)$

Referring back to quantum mechanics for a moment, we see that this last integral is just ${\left\langle \psi\left|\hat{H}\right|\psi\right\rangle }$, that is, the expectation value of the Hamiltonian operator, which is the total energy of the system.

Finally, we can write down the Poisson brackets, since these are general results for any field ${\psi}$ and its conjugate momentum ${\pi}$:

 $\displaystyle \left\{ \psi\left(\mathbf{x},t\right),\pi\left(\mathbf{x}^{\prime},t\right)\right\} _{PB}$ $\displaystyle =$ $\displaystyle \delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right)\ \ \ \ \ (20)$ $\displaystyle \left\{ \phi\left(\mathbf{x},t\right),\phi\left(\mathbf{x}^{\prime},t\right)\right\} _{PB}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (21)$ $\displaystyle \left\{ \pi\left(\mathbf{x},t\right),\pi\left(\mathbf{x}^{\prime},t\right)\right\} _{PB}$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (22)$

These brackets will be used later when we quantize the theory.

# Klein-Gordon equation: plane wave solutions

Reference: References: Robert D. Klauber, Student Friendly Quantum Field Theory, (Sandtrove Press, 2013) – Chapter 3, Problem 3.1.

The Klein-Gordon equation was one of the first attempts at producing a relativistic quantum theory. In natural units, the equation is

$\displaystyle \left(\partial_{\mu}\partial^{\mu}+m^{2}\right)\phi=0 \ \ \ \ \ (1)$

This equation also results from the Euler-Lagrange equation for a scalar field ${\phi}$with Lagrangian

$\displaystyle \mathcal{L}=\frac{1}{2}\left(\partial_{\mu}\phi\right)\left(\partial^{\mu}\phi\right)-\frac{1}{2}m^{2}\phi^{2} \ \ \ \ \ (2)$

This is the Lagrangian for zero potential ${V\left(\phi\right)=0}$.

To write solutions to equation 1, we can introduce some new notation. In natural units, the four-momentum is

 $\displaystyle p_{\mu}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} E & p_{i}\end{array}\right]\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} E & -p^{i}\end{array}\right] \ \ \ \ \ (4)$

The scalar product of four-momentum with a spacetime vector ${x^{\mu}}$ is therefore

$\displaystyle px\equiv p_{\mu}x^{\mu}=Et-\mathbf{p}\cdot\mathbf{x} \ \ \ \ \ (5)$

For a plane wave with angular frequency ${\omega}$, Planck’s relation is ${E=\hbar\omega=\omega}$, and the wave vector ${\mathbf{k}}$ has components in the three spatial directions of ${2\pi/\lambda_{i}}$, where ${\lambda_{i}}$ is the component of the wavelength in direction ${x_{i}}$. For example, a wave moving in the ${x_{1}}$ direction has ${\lambda_{2}=\lambda_{3}=\infty}$, so ${\mathbf{k}=\left[k_{1},0,0\right]}$. The four-vector ${k^{\mu}}$ is

$\displaystyle k^{\mu}=\left[\omega,\mathbf{k}\right] \ \ \ \ \ (6)$

and since ${\mathbf{p}=\mathbf{k}}$ in natural units, we have

$\displaystyle kx=k_{\mu}x^{\mu}=p_{\mu}x^{\mu}=px \ \ \ \ \ (7)$

A plane wave solution to 1 turns out to be

$\displaystyle \phi=\sum_{\mathbf{k}}\frac{1}{\sqrt{2V\omega_{\mathbf{k}}}}\left(A_{\mathbf{k}}e^{-ikx}+B_{\mathbf{k}}^{\dagger}e^{ikx}\right) \ \ \ \ \ (8)$

We can see this by direct substitution. Consider one term ${\phi_{\mathbf{k}}}$ from the sum. Then

 $\displaystyle \phi_{\mathbf{k}}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2V\omega_{\mathbf{k}}}}\left(A_{\mathbf{k}}e^{-ikx}+B_{\mathbf{k}}^{\dagger}e^{ikx}\right)\ \ \ \ \ (9)$ $\displaystyle \partial^{\mu}\phi_{\mathbf{k}}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2V\omega_{\mathbf{k}}}}\left(-ik_{\mu}A_{\mathbf{k}}e^{-ikx}+ik_{\mu}B_{\mathbf{k}}^{\dagger}e^{ikx}\right)\ \ \ \ \ (10)$ $\displaystyle \partial_{\mu}\partial^{\mu}\phi_{\mathbf{k}}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2V\omega_{\mathbf{k}}}}\left(-k_{\mu}k^{\mu}A_{\mathbf{k}}e^{-ikx}-k_{\mu}k^{\mu}B_{\mathbf{k}}^{\dagger}e^{ikx}\right) \ \ \ \ \ (11)$

However, using the invariant scalar ${p_{\mu}p^{\mu}}$ from relativity:

$\displaystyle k_{\mu}k^{\mu}=p_{\mu}p^{\mu}=E^{2}-p^{2}=m^{2} \ \ \ \ \ (12)$

Thus

$\displaystyle \partial_{\mu}\partial^{\mu}\phi_{\mathbf{k}}=-m^{2}\phi_{\mathbf{k}} \ \ \ \ \ (13)$

so 1 is true for a single component ${\phi_{\mathbf{k}}}$. Since the solution 8 is a linear combination of such solutions, and the original differential equation is linear, then 8 is also a solution. [The normalization factor ${1/\sqrt{2V\omega_{\mathbf{k}}}}$ is irrelevant in proving that 8 is a solution; it’s just there to make future calculations easier.]

Note that the first term (involving ${A_{\mathbf{k}}}$) is also a solution of the free-particle Schrödinger equation, which is

$\displaystyle i\frac{\partial\phi_{S}}{\partial t}=-\frac{1}{2m}\nabla^{2}\phi_{S} \ \ \ \ \ (14)$

If we take

$\displaystyle \phi_{S}=\sum_{\mathbf{k}}A_{\mathbf{k}}e^{-ikx} \ \ \ \ \ (15)$

then

 $\displaystyle i\frac{\partial\phi_{S}}{\partial t}$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}E_{\mathbf{k}}A_{\mathbf{k}}e^{-ikx}\ \ \ \ \ (16)$ $\displaystyle -\frac{1}{2m}\nabla^{2}\phi_{S}$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}\frac{k^{2}}{2m}A_{\mathbf{k}}e^{-ikx} \ \ \ \ \ (17)$

For a free particle, the energy ${E_{\mathbf{k}}=\frac{p^{2}}{2m}=\frac{k^{2}}{2m}}$, so the Schrödinger equation is satisfied. The second term (with ${B_{\mathbf{k}}^{\dagger}}$) does not satisfy the Schrödinger equation, since in that case we get

 $\displaystyle \phi_{S}$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}B_{\mathbf{k}}^{\dagger}e^{ikx}\ \ \ \ \ (18)$ $\displaystyle i\frac{\partial\phi_{S}}{\partial t}$ $\displaystyle =$ $\displaystyle -\sum_{\mathbf{k}}E_{\mathbf{k}}B_{\mathbf{k}}^{\dagger}e^{ikx}\ \ \ \ \ (19)$ $\displaystyle -\frac{1}{2m}\nabla^{2}\phi_{S}$ $\displaystyle =$ $\displaystyle \sum_{\mathbf{k}}\frac{k^{2}}{2m}B_{\mathbf{k}}^{\dagger}e^{ikx} \ \ \ \ \ (20)$

The extra minus sign means the two sides don’t match.

# Quantum field theory representation of non-relativistic quantum mechanics

References: Mark Srednicki, Quantum Field Theory, (Cambridge University Press, 2007) – Chapter 1, Problem 1.2.

One of the main problems faced in developing a relativistic quantum theory is that in non-relativistic quantum mechanics, position and time are not on an equal footing. Position is treated as an operator, while time is just a parameter that labels a particular instance of a state or wave function. In special relativity, space and time are treated equivalently, in the sense that they form equal components of four-dimensional spacetime. [Time and space aren’t the same, of course, since although we can travel to any point in space whenever we want, we can move only forward in time, even in relativity. However, space and time components are treated equally in the sense that they transform into each other in the Lorentz transformations.]

Attempts to develop a relativistic quantum theory therefore can take one of two paths in an attempt to solve this disparity. One way is to promote time to an operator, but this leads to complex theories (although they do work). The other way is to demote position from an operator to just a label, so its status is the same as that of time. This idea leads to quantum field theory.

The idea is that the position ${\mathbf{x}}$ becomes, like time, a label on an operator. We can define a set of operators ${\phi\left(\mathbf{x}\right)}$ such that at each point ${\mathbf{x}}$ in space, there is a separate operator. The position ${\mathbf{x}}$ becomes a label telling us which operator we’re dealing with. The set of all such operators (that is, the set of operators defined over all space) is called a quantum field, and hence we get quantum field theory by studying such sets of operators. In the general case, each operator is also a function of time so that a quantum field is actually made up of a set of operators ${\phi\left(\mathbf{x},t\right)}$.

To get an idea of how this works, we can rewrite non-relativistic quantum mechanics using a quantum field. Non-relativistic quantum mechanics is governed by the Schrödinger equation, which in its most general form is

$\displaystyle i\hbar\frac{\partial}{\partial t}\left|\psi,t\right\rangle =H\left|\psi,t\right\rangle \ \ \ \ \ (1)$

where ${H}$ is the hamiltonian. Now suppose we define a quantum field ${a\left(\mathbf{x}\right)}$ and its hermitian conjugate ${a^{\dagger}\left(\mathbf{x}\right)}$ that satisfy the commutation relations

 $\displaystyle \left[a\left(\mathbf{x}\right),a\left(\mathbf{x}^{\prime}\right)\right]$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (2)$ $\displaystyle \left[a^{\dagger}\left(\mathbf{x}\right),a^{\dagger}\left(\mathbf{x}^{\prime}\right)\right]$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (3)$ $\displaystyle \left[a\left(\mathbf{x}\right),a^{\dagger}\left(\mathbf{x}^{\prime}\right)\right]$ $\displaystyle =$ $\displaystyle \delta^{3}\left(\mathbf{x}-\mathbf{x}^{\prime}\right) \ \ \ \ \ (4)$

where ${\delta^{3}\left(\mathbf{x}\right)}$ is the 3-d Dirac delta function. These operators are similar to the raising and lowering operators we used to solve the harmonic oscillator, although in this case the operators are labelled by positions in space rather than energy states in an oscillator.

We require one additional property for this field: if ${\left|0\right\rangle }$ represents the vacuum state, that is, a state with no particles in it, then

$\displaystyle a\left(\mathbf{x}\right)\left|0\right\rangle =0 \ \ \ \ \ (5)$

That is, ${a\left(\mathbf{x}\right)}$ eliminates the vacuum state for all values of ${\mathbf{x}}$.

Now suppose we define a hamiltonian using this field, as follows:

 $\displaystyle H$ $\displaystyle =$ $\displaystyle \int d^{3}x\;a^{\dagger}\left(\mathbf{x}\right)\left(-\frac{\hbar^{2}}{2m}\nabla^{2}+U\left(\mathbf{x}\right)\right)a\left(\mathbf{x}\right)+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \frac{1}{2}\int d^{3}x\;d^{3}y\;V\left(\mathbf{x}-\mathbf{y}\right)a^{\dagger}\left(\mathbf{x}\right)a^{\dagger}\left(\mathbf{y}\right)a\left(\mathbf{y}\right)a\left(\mathbf{x}\right) \ \ \ \ \ (6)$

Here, ${U}$ is an external potential energy and ${V}$ is an interaction energy between two particles at locations ${\mathbf{x}}$ and ${\mathbf{y}}$.

Also, suppose we have a time-dependent quantum state defined by

$\displaystyle \left|\psi,t\right\rangle =\int d^{3}x_{1}\ldots d^{3}x_{n}\psi\left(\mathbf{x}_{1},\ldots,\mathbf{x}_{n};t\right)a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle \ \ \ \ \ (7)$

If we require 1 to be true, what condition does this place on the function ${\psi\left(\mathbf{x}_{1},\ldots,\mathbf{x}_{n};t\right)}$ inside the integral? To determine this, we need to apply ${H}$ to ${\left|\psi,t\right\rangle }$ by using the commutation relations. Consider the first integral in 6. We can propagate the operator ${a\left(\mathbf{x}\right)}$ through the list of operators ${a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)}$ in 7 by applying the commutator 4. We get

 $\displaystyle a\left(\mathbf{x}\right)a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)$ $\displaystyle =$ $\displaystyle \left[\delta^{3}\left(\mathbf{x}-\mathbf{x}_{1}\right)+a^{\dagger}\left(\mathbf{x}_{1}\right)a\left(\mathbf{x}\right)\right]a^{\dagger}\left(\mathbf{x}_{2}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta^{3}\left(\mathbf{x}-\mathbf{x}_{1}\right)a^{\dagger}\left(\mathbf{x}_{2}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle a^{\dagger}\left(\mathbf{x}_{1}\right)\left[\delta^{3}\left(\mathbf{x}-\mathbf{x}_{2}\right)+a^{\dagger}\left(\mathbf{x}_{2}\right)a\left(\mathbf{x}\right)\right]a^{\dagger}\left(\mathbf{x}_{3}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \delta^{3}\left(\mathbf{x}-\mathbf{x}_{1}\right)a^{\dagger}\left(\mathbf{x}_{2}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)+\ldots+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n-1}\right)\left[\delta^{3}\left(\mathbf{x}-\mathbf{x}_{n}\right)+a^{\dagger}\left(\mathbf{x}_{n}\right)a\left(\mathbf{x}\right)\right] \ \ \ \ \ (10)$

When we apply this expansion to the vacuum state ${\left|0\right\rangle }$, the last term in the final bracket vanishes because of 5. Doing the integral over ${x}$ in 6 results in

 $\displaystyle \int d^{3}x\;a^{\dagger}\left(\mathbf{x}\right)\left(-\frac{\hbar^{2}}{2m}\nabla^{2}+U\left(\mathbf{x}\right)\right)a\left(\mathbf{x}\right)\left|\psi,t\right\rangle$ $\displaystyle =\nonumber$ $\displaystyle \sum_{i}\int d^{3}x_{1}\ldots d^{3}x_{n}a^{\dagger}\left(\mathbf{x}_{i}\right)\left(-\frac{\hbar^{2}}{2m}\nabla_{i}^{2}+U\left(\mathbf{x}_{i}\right)\right)\psi a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{i-1}\right)a^{\dagger}\left(\mathbf{x}_{i+1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle \ \ \ \ \ (11)$ $\displaystyle =$ $\displaystyle \sum_{i}\int d^{3}x_{1}\ldots d^{3}x_{n}\left(-\frac{\hbar^{2}}{2m}\nabla_{i}^{2}+U\left(\mathbf{x}_{i}\right)\right)\psi a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)$

where the last line follows from 3.

We can do a similar calculation for the second integral in 6, although it’s a bit more complicated because we have to integrate over both ${x}$ and ${y}$. Applying ${a\left(\mathbf{x}\right)}$ to ${a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)}$ gives the result 10. Applying ${a\left(\mathbf{y}\right)}$ to the first term of this result gives

 $\displaystyle a\left(\mathbf{y}\right)\delta^{3}\left(\mathbf{x}-\mathbf{x}_{1}\right)a^{\dagger}\left(\mathbf{x}_{2}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle$ $\displaystyle =$ $\displaystyle \delta^{3}\left(\mathbf{x}-\mathbf{x}_{1}\right)\times\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \left[\delta^{3}\left(\mathbf{y}-\mathbf{x}_{2}\right)a^{\dagger}\left(\mathbf{x}_{3}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)+\delta^{3}\left(\mathbf{y}-\mathbf{x}_{n}\right)a^{\dagger}\left(\mathbf{x}_{2}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n-1}\right)\right]\left|0\right\rangle \ \ \ \ \ (12)$

with similar terms arising from the other terms in 10. When we integrate over ${x}$ and ${y}$ this first term gives us

$\displaystyle \sum_{j=2}^{n}\int d^{3}x_{1}\ldots d^{3}x_{n}V\left(\mathbf{x}_{1}-\mathbf{x}_{j}\right)\psi\left(\mathbf{x}_{1},\ldots,\mathbf{x}_{n};t\right)a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle \ \ \ \ \ (13)$

The other terms in the expansion 10 each contribute a sum

$\displaystyle \sum_{\begin{array}{c} j=1\\ j\ne i \end{array}}^{n}\int d^{3}x_{1}\ldots d^{3}x_{n}V\left(\mathbf{x}_{i}-\mathbf{x}_{j}\right)\psi\left(\mathbf{x}_{1},\ldots,\mathbf{x}_{n};t\right)a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle \ \ \ \ \ (14)$

so the overall result for the second integral is

 $\displaystyle \frac{1}{2}\int d^{3}x\;d^{3}y\;V\left(\mathbf{x}-\mathbf{y}\right)a^{\dagger}\left(\mathbf{x}\right)a^{\dagger}\left(\mathbf{y}\right)a\left(\mathbf{y}\right)a\left(\mathbf{x}\right)\left|\psi,t\right\rangle$ $\displaystyle =\nonumber$ $\displaystyle \sum_{j=1}^{n}\sum_{i=1}^{j-1}\int d^{3}x_{1}\ldots d^{3}x_{n}V\left(\mathbf{x}_{i}-\mathbf{x}_{j}\right)\psi\left(\mathbf{x}_{1},\ldots,\mathbf{x}_{n};t\right)a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle \ \ \ \ \ (15)$

Requiring 11 plus 15 to satisfy 1 gives us (from equating the integrands on both sides):

$\displaystyle i\hbar\frac{\partial\psi}{\partial t}=\sum_{i=1}^{n}\left(-\frac{\hbar^{2}}{2m}\nabla_{i}^{2}+U\left(\mathbf{x}_{i}\right)\right)\psi+\sum_{j=1}^{n}\sum_{i=1}^{j-1}V\left(\mathbf{x}_{i}-\mathbf{x}_{j}\right)\psi \ \ \ \ \ (16)$

This is just the Schrödinger equation in its more traditional form, for a hamiltonian containing kinetic energy, overall potential energy ${U}$ and particle-particle interaction energy ${V}$ for a collection of ${n}$ particles. Thus the state ${a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle }$ corresponds to a state containing ${n}$ particles at locations ${\mathbf{x}_{i}}$, so the hermitian conjugate field operators ${a^{\dagger}\left(\mathbf{x}_{i}\right)}$ act as creation operators, with each operator creating its particle at the location ${\mathbf{x}_{i}}$ used to label the operator.

The operator

$\displaystyle N\equiv\int d^{3}x\;a^{\dagger}\left(\mathbf{x}\right)a\left(\mathbf{x}\right) \ \ \ \ \ (17)$

counts the particles present, in effect by annihilating a particle at location ${\mathbf{x}}$ (if one exists there), then creating it again in the same location. More formally, if we apply ${N}$ to the state ${a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle }$ then the annihilation operator ${a\left(\mathbf{x}\right)}$ acting on the state produces the result 10, so we get

 $\displaystyle Na^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle$ $\displaystyle =$ $\displaystyle \int d^{3}x\;a^{\dagger}\left(\mathbf{x}\right)\delta^{3}\left(\mathbf{x}-\mathbf{x}_{1}\right)a^{\dagger}\left(\mathbf{x}_{2}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle +\ldots+\nonumber$ $\displaystyle$ $\displaystyle$ $\displaystyle \int d^{3}x\;a^{\dagger}\left(\mathbf{x}\right)a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n-1}\right)\delta^{3}\left(\mathbf{x}-\mathbf{x}_{n}\right)\left|0\right\rangle \ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sum_{i=1}^{n}a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle \ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle n\;a^{\dagger}\left(\mathbf{x}_{1}\right)\ldots a^{\dagger}\left(\mathbf{x}_{n}\right)\left|0\right\rangle \ \ \ \ \ (20)$

# Adding a constant to the potential introduces a phase factor

Required math: algebra, calculus (partial derivatives and integration by parts), complex numbers

Required physics: Schrödinger equation, probability density

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 1.8.

The time-independent Schrödinger equation in one dimension can be separated into two equations as follows:

 $\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}\psi(x)}{dx^{2}}+V(x)\psi(x)$ $\displaystyle =$ $\displaystyle E\psi(x)\ \ \ \ \ (1)$ $\displaystyle i\hbar\frac{d\Xi(t)}{dt}$ $\displaystyle =$ $\displaystyle E\Xi(t) \ \ \ \ \ (2)$

and the general solution is

$\displaystyle \Psi\left(x,t\right)=\psi\left(x\right)\Xi\left(t\right) \ \ \ \ \ (3)$

The time component can be solved as

$\displaystyle \Xi\left(t\right)=Ce^{-iEt/\hbar} \ \ \ \ \ (4)$

where ${C}$ is the constant of integration.

If we add a constant (in both space and time) ${V_{0}}$ to the potential, then the original Schrödinger equation becomes

 $\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}\Psi}{dx^{2}}+V(x)\Psi+V_{0}\Psi$ $\displaystyle =$ $\displaystyle i\hbar\frac{\partial\Psi}{\partial t}\ \ \ \ \ (5)$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}\Psi}{dx^{2}}+V(x)\Psi$ $\displaystyle =$ $\displaystyle i\hbar\frac{\partial\Psi}{\partial t}-V_{0}\Psi \ \ \ \ \ (6)$

Applying separation of variables gives us

 $\displaystyle -\frac{\hbar^{2}}{2m}\frac{1}{\psi(x)}\frac{\partial^{2}\psi(x)}{\partial x^{2}}+V(x)$ $\displaystyle =$ $\displaystyle E\ \ \ \ \ (7)$ $\displaystyle i\hbar\frac{1}{\Xi(t)}\frac{\partial\Xi}{\partial t}-V_{0}$ $\displaystyle =$ $\displaystyle E \ \ \ \ \ (8)$

[Since ${V_{0}}$ is independent of both ${x}$ and ${t}$, we could put it in either the ${\psi\left(x\right)}$ or the ${\Xi\left(t\right)}$ equation, but putting it in the ${\Xi}$ equation eliminates it from the more complex ${\psi}$ equation, so we’ll do that.]

The solution to 8 is now

$\displaystyle \Xi\left(t\right)=Ce^{-i\left(E+V_{0}\right)t/\hbar} \ \ \ \ \ (9)$

so we’ve introduced a phase factor ${e^{-iV_{0}t/\hbar}}$ into the overall wave function ${\Psi}$. For the time-independent Schrödinger equation, all quantities of physical interest involve multiplying the complex conjugate ${\Psi^*}$ by some operator ${\hat{Q}\left(x\right)}$ that depends only on ${x}$, operating on ${\Psi}$. That is, we’re interested only in quantities of the form

 $\displaystyle \Psi^*\left[\hat{Q}\left(x\right)\Psi\right]$ $\displaystyle =$ $\displaystyle \left|C\right|^{2}e^{+i\left(E+V_{0}\right)t/\hbar}e^{-i\left(E+V_{0}\right)t/\hbar}\psi^*\left[\hat{Q}\left(x\right)\psi\right]\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left|C\right|^{2}\psi^*\left[\hat{Q}\left(x\right)\psi\right] \ \ \ \ \ (11)$

Thus the phase factor disappears when calculating any physical quantity.

# Integral form of the Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 11.8.

As a prelude to the Born approximation in quantum scattering, we need to look at the integral form of the time-independent Schrödinger equation. The equation in its original differential equation form is

$\displaystyle -\frac{\hbar^{2}}{2m}\nabla^{2}\psi+V\psi=E\psi \ \ \ \ \ (1)$

which can be written as

 $\displaystyle \left(\nabla^{2}+k^{2}\right)\psi$ $\displaystyle =$ $\displaystyle Q\ \ \ \ \ (2)$ $\displaystyle k$ $\displaystyle \equiv$ $\displaystyle \frac{\sqrt{2mE}}{\hbar}\ \ \ \ \ (3)$ $\displaystyle Q$ $\displaystyle \equiv$ $\displaystyle \frac{2m}{\hbar^{2}}V\psi \ \ \ \ \ (4)$

To convert this to an integral equation, we need to define a Green’s function ${G\left(\mathbf{r}\right)}$ which satisfies the differential equation

$\displaystyle \left(\nabla^{2}+k^{2}\right)G\left(\mathbf{r}\right)=\delta^{3}\left(\mathbf{r}\right) \ \ \ \ \ (5)$

Using this function we can write ${\psi}$ as an integral equation

$\displaystyle \psi\left(\mathbf{r}\right)=\int G\left(\mathbf{r}-\mathbf{r}_{0}\right)Q\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0} \ \ \ \ \ (6)$

We can show this works by plugging in ${G}$ from 5:

 $\displaystyle \left(\nabla^{2}+k^{2}\right)\psi\left(\mathbf{r}\right)$ $\displaystyle =$ $\displaystyle \int\left(\nabla^{2}+k^{2}\right)G\left(\mathbf{r}-\mathbf{r}_{0}\right)Q\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0}\ \ \ \ \ (7)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int\delta\left(\mathbf{r}-\mathbf{r}_{0}\right)Q\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0}\ \ \ \ \ (8)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Q\left(\mathbf{r}\right) \ \ \ \ \ (9)$

which gives us back 2.

This isn’t a solution of the Schrödinger equation, of course, because ${Q}$ contains ${\psi}$, so we’d need to actually know ${\psi}$ in advance in order to work out the integral with the Green’s function. Rather, it’s just a different way of writing the Schrödinger equation which proves useful in scattering theory.

Because 5 doesn’t depend on the potential ${V}$, we can work out the Green’s function which is valid for every potential. The process is rather involved, but Griffiths goes through the details in section 11.4.1, so I won’t reproduce them here, apart from noting that the solution uses what is, to me, one of the most beautiful theorems in mathematics: Cauchy’s theorem on contour integration. Maybe I’ll return to it later.

Anyway, the Green’s function turns out to be

$\displaystyle G\left(\mathbf{r}\right)=-\frac{e^{ikr}}{4\pi r} \ \ \ \ \ (10)$

We can verify this is in fact a solution by plugging it back into 5. We need the Laplacian of ${G}$ which we can get by calculating the divergence of the gradient. Taking the gradient first, we use the product rule for gradients:

$\displaystyle \nabla\left(fg\right)=f\nabla g+g\nabla f \ \ \ \ \ (11)$

We get

 $\displaystyle \nabla G$ $\displaystyle =$ $\displaystyle -\frac{1}{4\pi}\left(\frac{1}{r}\nabla e^{ikr}+e^{ikr}\nabla\frac{1}{r}\right) \ \ \ \ \ (12)$

To calculate the divergence of the gradient, we use the identity for the divergence of the product of a scalar and a vector:

$\displaystyle \nabla\cdot\left(f\mathbf{A}\right)=\mathbf{A}\cdot\nabla f+f\nabla\cdot\mathbf{A} \ \ \ \ \ (13)$

We therefore have

 $\displaystyle \nabla^{2}G$ $\displaystyle =$ $\displaystyle -\frac{1}{4\pi}\left[\left(\nabla\frac{1}{r}\right)\cdot\left(\nabla e^{ikr}\right)+\frac{1}{r}\nabla^{2}e^{ikr}+\left(\nabla e^{ikr}\right)\cdot\left(\nabla\frac{1}{r}\right)+e^{ikr}\nabla^{2}\frac{1}{r}\right] \ \ \ \ \ (14)$

The last term turns out to be a delta function:

$\displaystyle \nabla^{2}\frac{1}{r}=\nabla\cdot\left(\nabla\frac{1}{r}\right)=-\nabla\cdot\left(\frac{\hat{\mathbf{r}}}{r^{2}}\right)=-4\pi\delta^{3}\left(\mathbf{r}\right) \ \ \ \ \ (15)$

To work out the second term, we use the formula for the Laplacian in spherical coordinates, for a function that depends only on ${r}$:

$\displaystyle \nabla f\left(r\right)=\frac{1}{r^{2}}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial f}{\partial r}\right) \ \ \ \ \ (16)$

We get

 $\displaystyle \nabla^{2}e^{ikr}$ $\displaystyle =$ $\displaystyle \frac{1}{r^{2}}\frac{\partial}{\partial r}\left(ikr^{2}e^{ikr}\right)\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2ik}{r}e^{ikr}-k^{2}e^{ikr} \ \ \ \ \ (18)$

Putting this back into 14 we get

 $\displaystyle \nabla^{2}G$ $\displaystyle =$ $\displaystyle -\frac{1}{4\pi}\left[2\left(\nabla\frac{1}{r}\right)\cdot\left(\nabla e^{ikr}\right)+\frac{2ik}{r^{2}}e^{ikr}-\frac{k^{2}}{r}e^{ikr}-4\pi\delta^{3}\left(\mathbf{r}\right)e^{ikr}\right]\ \ \ \ \ (19)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{4\pi}\left[-\frac{2ik}{r^{2}}e^{ikr}+\frac{2ik}{r^{2}}e^{ikr}-\frac{k^{2}}{r}e^{ikr}-4\pi\delta^{3}\left(\mathbf{r}\right)e^{ikr}\right]\ \ \ \ \ (20)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{1}{4\pi}\left[-\frac{k^{2}}{r}e^{ikr}-4\pi\delta^{3}\left(\mathbf{r}\right)\right]\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{k^{2}}{4\pi r}e^{ikr}+\delta^{3}\left(\mathbf{r}\right)\ \ \ \ \ (22)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -k^{2}G\left(\mathbf{r}\right)+\delta^{3}\left(\mathbf{r}\right)\ \ \ \ \ (23)$ $\displaystyle \left(\nabla^{2}+k^{2}\right)G\left(\mathbf{r}\right)$ $\displaystyle =$ $\displaystyle \delta^{3}\left(\mathbf{r}\right) \ \ \ \ \ (24)$

where we dropped the ${e^{ikr}}$ from the last term in the third line since the delta function is zero except when ${\mathbf{r}=0}$.

Using this Green’s function, the integral form of the Schrödinger equation is

$\displaystyle \psi\left(\mathbf{r}\right)=\psi_{0}\left(\mathbf{r}\right)-\frac{m}{2\pi\hbar^{2}}\int\frac{e^{ik\left|\mathbf{r}-\mathbf{r}_{0}\right|}}{\left|\mathbf{r}-\mathbf{r}_{0}\right|}V\left(\mathbf{r}_{0}\right)\psi\left(\mathbf{r}_{0}\right)d^{3}\mathbf{r}_{0} \ \ \ \ \ (25)$

where ${\psi_{0}}$ is a solution of the free particle Schrödinger equation

$\displaystyle \left(\nabla^{2}+k^{2}\right)\psi_{0}\left(\mathbf{r}\right)=0 \ \ \ \ \ (26)$

# Infinite square well – minimum energy

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Problem 2.3.

We’ve seen that the energy of a system must always be greater than the minimum of the potential function. As a specific example of this we can look at the Schrödinger equation for the square well, between ${x=0}$ and ${x=a}$:

$\displaystyle \frac{d^{2}\psi}{dx^{2}}=-\frac{2m}{\hbar^{2}}E\psi \ \ \ \ \ (1)$

If ${E=0}$, ${\psi"=0}$. Integrating gives ${\psi=Ax+B.}$ Attempting to satisfy the boundary conditions, we get ${\psi(0)=0}$ giving ${B=0}$. Then the condition ${\psi(a)=0}$ gives ${A=0}$, thus ${\psi(x)=0}$ and cannot be normalized.

If ${E<0}$, we solve the equation

$\displaystyle \psi"=-\frac{2mE}{\hbar^{2}}\psi\equiv k^{2}\psi \ \ \ \ \ (2)$

with ${k=\sqrt{-2mE/\hbar^{2}}}$. Since ${E<0}$, ${k}$ is real. The general solution is ${\psi(x)=Ae^{kx}+Be^{-kx}}$. Applying boundary conditions, we get ${\psi(0)=0=A+B}$, so ${A=-B}$. At ${x=a}$, we have

 $\displaystyle \psi(a)$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle Ae^{ka}+Be^{-ka}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle AC+\frac{B}{C}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle A\left(C-\frac{1}{C}\right) \ \ \ \ \ (6)$

where ${C\equiv e^{ka}>0}$. Since ${C}$ is strictly positive (being an exponential) the only way we can get ${C-1/C=0}$ is for ${C=1}$, implying ${ka=0}$. However, neither ${k}$ nor ${a}$ is zero here, so ${C\neq1}$, so ${A=0=B}$ and ${\psi(x)=0}$ again. Thus if ${E\leq0}$, the wave function cannot be normalized and satisfy the boundary conditions.

# Hydrogen atom – series solution and Bohr energy levels

Required math: calculus

Required physics: Schrödinger equation in 3-d

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Sec 4.2.1.

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 13, Exercises 13.1.1 – 13.1.2.

[This page follows the derivation given in Griffiths. The discussion in Shankar’s chapter 13 is similar, but he uses Gaussian units, so the answer looks different. However, I can’t be bothered going through the whole derivation again with different units, since the steps are essentially the same.]

We saw in an earlier post that the radial part of the three-dimensional Schrödinger equation for the hydrogen atom can be reduced to the differential equation

$\displaystyle \rho\frac{d^{2}v}{d\rho^{2}}+2(l+1-\rho)\frac{dv}{d\rho}+(\rho_{0}-2l-2)v=0 \ \ \ \ \ (1)$

where

 $\displaystyle u(\rho)$ $\displaystyle =$ $\displaystyle \rho^{l+1}e^{-\rho}v(\rho)\ \ \ \ \ (2)$ $\displaystyle u(r)$ $\displaystyle \equiv$ $\displaystyle rR(r)\ \ \ \ \ (3)$ $\displaystyle \rho$ $\displaystyle =$ $\displaystyle \kappa r\ \ \ \ \ (4)$ $\displaystyle \rho_{0}$ $\displaystyle =$ $\displaystyle \frac{me^{2}}{2\pi\epsilon_{0}\hbar^{2}\kappa}\ \ \ \ \ (5)$ $\displaystyle \kappa$ $\displaystyle =$ $\displaystyle \frac{\sqrt{-2mE}}{\hbar} \ \ \ \ \ (6)$

and ${R(r)}$ is the radial part of the three-dimensional wave function.

Our task here is to solve 1 by using the same method as for the harmonic oscillator. We propose a solution of the form

$\displaystyle v(\rho)=\sum_{j=0}^{\infty}c_{j}\rho^{j} \ \ \ \ \ (7)$

and attempt to determine the coefficients ${c_{j}}$. The two derivatives needed in the equation are

 $\displaystyle \frac{dv}{d\rho}$ $\displaystyle =$ $\displaystyle \sum_{j=0}^{\infty}jc_{j}\rho^{j-1}\ \ \ \ \ (8)$ $\displaystyle \frac{d^{2}v}{d\rho^{2}}$ $\displaystyle =$ $\displaystyle \sum_{j=0}^{\infty}j(j-1)c_{j}\rho^{j-2} \ \ \ \ \ (9)$

We now plug these back into 1 and fiddle with the summation indexes so that every term in every sum is a multiple of ${\rho^{j}}$.

$\displaystyle \sum_{j=0}^{\infty}j(j-1)c_{j}\rho^{j-1}+2(l+1)\sum_{j=0}^{\infty}jc_{j}\rho^{j-1}-2\sum_{j=0}^{\infty}jc_{j}\rho^{j}+(\rho_{0}-2l-2)\sum_{j=0}^{\infty}c_{j}\rho^{j}=0 \ \ \ \ \ (10)$

The two terms containing ${\rho^{j-1}}$ can be converted to sums over ${\rho^{j}}$ by shifting the summation index from ${j}$ to ${j+1}$. This means that the sum becomes

$\displaystyle \sum_{j=-1}^{\infty}(j+1)jc_{j+1}\rho^{j}+2(l+1)\sum_{j=-1}^{\infty}(j+1)c_{j+1}\rho^{j}-2\sum_{j=0}^{\infty}jc_{j}\rho^{j}+(\rho_{0}-2l-2)\sum_{j=0}^{\infty}c_{j}\rho^{j}=0 \ \ \ \ \ (11)$

Note that the term with ${j=-1}$ in the first two sums is zero because of the ${(j+1)}$ factor, so we can start the sum at ${j=0}$. Since ${\rho^{j}}$ is now a common factor in all sums we can write the overall sum as

$\displaystyle \sum_{j=0}^{\infty}\left[(j+1)jc_{j+1}+2(l+1)(j+1)c_{j+1}-2jc_{j}+(\rho_{0}-2l-2)c_{j}\right]\rho^{j}=0 \ \ \ \ \ (12)$

Because each power series is unique (a mathematical theorem), the only way this sum can be valid for all values of ${\rho}$ is if all the coefficients are zero. That is

$\displaystyle (j+1)jc_{j+1}+2(l+1)(j+1)c_{j+1}-2jc_{j}+(\rho_{0}-2l-2)c_{j}=0 \ \ \ \ \ (13)$

This can be rewritten as a recursion relation:

$\displaystyle c_{j+1}=\frac{2(j+l+1)-\rho_{0}}{(j+1)(j+2(l+1))}c_{j} \ \ \ \ \ (14)$

[This equation is essentially the same as Shankar’s 13.1.11 if you replace ${j\rightarrow k}$ and use Gaussian units in ${\rho_{0}}$.]

The argument at this point is again similar to that for the harmonic oscillator: we examine the behaviour for large ${j}$. In that case, we can ignore the ${l+1}$ and ${\rho_{0}}$ terms and write

 $\displaystyle c_{j+1}$ $\displaystyle \sim$ $\displaystyle \frac{2j}{j(j+1)}c_{j}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{2}{j+1}c_{j} \ \ \ \ \ (16)$

(We could also ignore the 1 in the denominator, but keeping it makes the argument easier, as we will see.) If we took this as an exact recursion relation, then starting with some initial constant ${c_{0}}$, we get

 $\displaystyle c_{1}$ $\displaystyle =$ $\displaystyle \frac{2}{1}c_{0}\ \ \ \ \ (17)$ $\displaystyle c_{2}$ $\displaystyle =$ $\displaystyle \frac{2^{2}}{2\times1}c_{0}\ \ \ \ \ (18)$ $\displaystyle c_{3}$ $\displaystyle =$ $\displaystyle \frac{2^{3}}{3\times2\times1}c_{0}\ \ \ \ \ (19)$ $\displaystyle c_{j}$ $\displaystyle =$ $\displaystyle \frac{2^{j}}{j!}c_{0}\ \ \ \ \ (20)$ $\displaystyle v(\rho)$ $\displaystyle =$ $\displaystyle c_{0}\sum_{j=0}^{\infty}\frac{2^{j}}{j!}\rho^{j}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}e^{2\rho} \ \ \ \ \ (22)$

In the last line we used the series expansion for the exponential function.

Returning for a moment to the original definition of ${v(\rho)}$, we get

 $\displaystyle u(\rho)$ $\displaystyle =$ $\displaystyle \rho^{l+1}e^{-\rho}v(\rho)\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}\rho^{l+1}e^{\rho} \ \ \ \ \ (24)$

Thus the infinite series solution gives a value for ${u}$ that increases exponentially for large ${\rho}$, which isn’t normalizable, so isn’t a valid solution. The only way to resolve this problem is again the same as in the harmonic oscillator case, which is to require the series to terminate after a finite number of terms. That is, we must have, for some value of ${j}$,

$\displaystyle 2(j+l+1)=\rho_{0} \ \ \ \ \ (25)$

That is, ${\rho_{0}}$ must be an even integer, which we can define as ${2n}$. Recalling the definition of ${\rho_{0}}$ from above, we therefore have the condition which quantizes the energy levels in the hydrogen atom:

 $\displaystyle \rho_{0}$ $\displaystyle =$ $\displaystyle \frac{me^{2}}{2\pi\epsilon_{0}\hbar^{2}\kappa}\ \ \ \ \ (26)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2n \ \ \ \ \ (27)$

so

$\displaystyle \kappa=\frac{me^{2}}{4\pi\epsilon_{0}\hbar^{2}n} \ \ \ \ \ (28)$

But ${\kappa=\frac{\sqrt{-2mE}}{\hbar}}$, so for the energy levels, we get

$\displaystyle E=-\frac{1}{n^{2}}\frac{me^{4}}{2\hbar^{2}(4\pi\epsilon_{0})^{2}} \ \ \ \ \ (29)$

This is the Bohr formula (although Bohr got the formula without using the Schrödinger equation) for the energy levels of hydrogen. [Again, this is equivalent to Shankar’s 13.1.16 if you use Gaussian units, so that the ${(4\pi\epsilon_{0})^{2}}$ factor becomes 1.]

The degeneracy of each energy level is found by noting that for a given value of ${n}$, any value of ${l}$ is possible such that ${j+l+1=n}$. Since ${j}$ is just the index on the series coefficient ${c_{j}}$, this means that ${l}$ can be any value from 0 up to ${n-1}$. For each ${l}$, the ${z}$ component of angular momentum can have any value from ${m=-l}$ up to ${m=+l}$, which gives ${2l+1}$ possibilities for each ${l}$. Thus the degeneracy for energy state ${E_{n}}$ is

 $\displaystyle d\left(n\right)$ $\displaystyle =$ $\displaystyle \sum_{l=0}^{n-1}\left(2l+1\right)\ \ \ \ \ (30)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\frac{1}{2}\left(n-1\right)n+n\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle n^{2} \ \ \ \ \ (32)$

where we’ve used the formula

$\displaystyle \sum_{l=1}^{N}l=\frac{1}{2}N\left(N+1\right) \ \ \ \ \ (33)$

Before leaving the series solution, we need to point out that the polynomials produced by 14, with the constraint that ${\rho_{0}=2n}$, are known mathematically as the associated Laguerre polynomials. They can be written as derivatives. First we define the ordinary Laguerre polynomials ${L_{q}}$:

$\displaystyle L_{q}(x)=e^{x}\frac{d^{q}}{dx^{q}}(e^{-x}x^{q}) \ \ \ \ \ (34)$

Now the associated Laguerre polynomials ${L_{q-p}^{p}}$ which depend on two parameters can be defined in terms of the ordinary Laguerre polynomials:

$\displaystyle L_{q-p}^{p}(x)=(-1)^{p}\frac{d^{p}}{dx^{p}}(L_{q}(x)) \ \ \ \ \ (35)$

A more useful formula for the associated Laguerre polynomials is

$\displaystyle L_{n}^{k}(x)=\sum_{j=0}^{n}\frac{(-1)^{j}(n+k)!}{(n-j)!(k+j)!j!}x^{j} \ \ \ \ \ (36)$

In terms of associated Laguerre polynomials, the solution of 1 is (apart from normalization)

$\displaystyle v(\rho)=L_{n-l-1}^{2l+1}(2\rho) \ \ \ \ \ (37)$

We can verify that this is the solution of 1 by direct substitution. First, we plug in the correct indexes into 36:

$\displaystyle L_{n-l-1}^{2l+1}(2\rho)=\sum_{j=0}^{n-l-1}\frac{(-1)^{j}2^{j}(n+l)!}{(n-l-j-1)!(2l+j+1)!j!}\rho^{j} \ \ \ \ \ (38)$

Now we define the coefficients in the polynomial and show that the recurrence relation 14 is valid:

 $\displaystyle c_{j}$ $\displaystyle =$ $\displaystyle \frac{(-1)^{j}2^{j}(n+l)!}{(n-l-j-1)!(2l+j+1)!j!}\ \ \ \ \ (39)$ $\displaystyle \frac{c_{j+1}}{c_{j}}$ $\displaystyle =$ $\displaystyle \frac{-2(n-l-1-j)}{(j+1)(2l+j+2)} \ \ \ \ \ (40)$

This is the same recurrence relation provided ${\rho_{0}=2n}$. However, this isn’t enough to verify the solution since other definitions of ${c_{j}}$ would give the same relation (for example, we could leave out the ${(n+l)!}$ factor in the numerator and still get the same recurrence relation). To verify that the polynomials are in fact solutions, we can work out their derivatives and plug them into 1 directly.

We get

 $\displaystyle \sum_{j=0}^{n-l-1}\left[c_{j}(j-1)j\rho^{j-1}+2(l+1-\rho)c_{j}j\rho^{j-1}+2(n-l-1)c_{j}\rho^{j}\right]$ $\displaystyle =$ $\displaystyle \sum_{j=0}^{n-l-1}\left[c_{j}(j-1)j\rho^{j-1}+2(l+1)c_{j}j\rho^{j-1}+-2jc_{j}\rho^{j}+2(n-l-1)c_{j}\rho^{j}\right] \ \ \ \ \ (41)$

We can now shift the summation index for the first two terms so that we sum over ${j+1}$ instead of ${j.}$ This results in

$\displaystyle \sum_{j=-1}^{n-l-2}\left[c_{j+1}j(j+1)+2(l+1)(j+1)c_{j+1}\right]\rho^{j}+\sum_{j=0}^{n-l-1}\left[-2jc_{j}+2(n-l-1)c_{j}\right]\rho^{j} \ \ \ \ \ (42)$

In the first sum, the ${j=-1}$ term is zero due to the ${(j+1)}$ factor, so we can start both sums from ${j=0}$. Thus for all values of ${j}$ from 0 to ${n-l-2}$, we can examine the coefficient of ${\rho^{j}}$:

$\displaystyle c_{j+1}(j+1)(j+2l+2)+c_{j}(-2j+2n-2l-2) \ \ \ \ \ (43)$

Using the relation between ${c_{j}}$ and ${c_{j+1}}$ above, we get

 $\displaystyle \frac{c_{j+1}}{c_{j}}(j+1)(j+2l+2)+(-2j+2n-2l-2)$ $\displaystyle =$ $\displaystyle 2(j+l+1-n)+2(-j+n-l-1)\ \ \ \ \ (44)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (45)$

For the one remaining term in the second sum where ${j=n-l-1}$ we note that this term is zero on its own, since ${(-j+n-l-1)=0}$ in this case. Thus the overall sum satisfies the original differential equation 1.

# Hydrogen atom – radial equation

Required math: calculus

Required physics: Schrödinger equation in 3-d

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Sec 4.2.1.

We’ve seen that we can solve the three-dimensional Schrödinger equation by separation of variables, provided that the potential is a function of ${r}$ only. In that case, the angular parts of the equation can be solved in general in terms of spherical harmonics, so the wave function has the form ${\psi(r,\theta,\phi)=R(r)Y(\theta,\phi)}$, where the ${Y}$ functions are the spherical harmonics, and ${R(r)}$ is the, as yet unsolved, radial function, which satisfies the differential equation

$\displaystyle \frac{d}{dr}\left(r^{2}\frac{dR}{dr}\right)-\frac{2mr^{2}}{\hbar^{2}}(V(r)-E)R=l(l+1)R \ \ \ \ \ (1)$

By making the further substitution

$\displaystyle u(r)\equiv rR \ \ \ \ \ (2)$

we can convert the above equation into a differential equation for ${u(r)}$:

$\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}u}{dr^{2}}+\left(V+\frac{\hbar^{2}}{2m}\frac{l(l+1)}{r^{2}}\right)u=Eu \ \ \ \ \ (3)$

This equation has the same form as the original Schrödinger equation except that the potential has picked up an extra so-called centrifugal term. We must now solve this equation when ${V(r)}$ is the potential found in the hydrogen atom.

The hydrogen atom consists of a proton and an electron. The proton is, in the first approximation, taken to be fixed, since its mass is more than a thousand times that of the electron. The force between the two particles can be taken as solely electric, since the gravitational force is many orders of magnitude smaller and will have essentially no effect. In this case, the potential is

$\displaystyle V(r)=-\frac{e^{2}}{4\pi\epsilon_{0}}\frac{1}{r} \ \ \ \ \ (4)$

where ${e}$ is the elementary charge and ${1/4\pi\epsilon_{0}}$ is the Coulomb constant. The equation to be solved is thus:

$\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}u}{dr^{2}}+\left(-\frac{e^{2}}{4\pi\epsilon_{0}}\frac{1}{r}+\frac{\hbar^{2}}{2m}\frac{l(l+1)}{r^{2}}\right)u=Eu \ \ \ \ \ (5)$

The solution of this equation follows a similar method as was used in solving the harmonic oscillator. We first investigate the asymptotic behaviour of the equation for large and small ${r}$, factor out this behaviour and then use a series to try to find the solution of what’s left.

First, we can introduce a couple of symbol changes. If we define

$\displaystyle \kappa\equiv\frac{\sqrt{-2mE}}{\hbar} \ \ \ \ \ (6)$

(note that since ${E<0}$ for bound states, ${\kappa}$ is real), then we can rewrite 5 as

$\displaystyle \frac{1}{\kappa^{2}}\frac{d^{2}u}{dr^{2}}=\left[1-\frac{me^{2}}{2\pi\epsilon_{0}\hbar^{2}\kappa}\frac{1}{\kappa r}+\frac{l(l+1)}{(\kappa r)^{2}}\right]u \ \ \ \ \ (7)$

Since ${r}$ occurs always multiplied by ${\kappa}$, we can try using a new variable

$\displaystyle \rho\equiv\kappa r \ \ \ \ \ (8)$

and this results in

$\displaystyle \frac{d^{2}u}{d\rho^{2}}=\left[1-\frac{me^{2}}{2\pi\epsilon_{0}\hbar^{2}\kappa}\frac{1}{\rho}+\frac{l(l+1)}{\rho^{2}}\right]u \ \ \ \ \ (9)$

We can simplify the notation a bit more by defining a constant

$\displaystyle \rho_{0}\equiv\frac{me^{2}}{2\pi\epsilon_{0}\hbar^{2}\kappa} \ \ \ \ \ (10)$

giving us the equation

$\displaystyle \frac{d^{2}u}{d\rho^{2}}=\left[1-\frac{\rho_{0}}{\rho}+\frac{l(l+1)}{\rho^{2}}\right]u \ \ \ \ \ (11)$

Now we can investigate the asymptotic behaviour. First, for large ${\rho}$, the two terms in the brackets that depend inversely on ${\rho}$ become negligible, so we get in this limit:

$\displaystyle \frac{d^{2}u}{d\rho^{2}}=u \ \ \ \ \ (12)$

This has the general solution

$\displaystyle u=Ae^{-\rho}+Be^{\rho} \ \ \ \ \ (13)$

and only the first term is acceptable, since the term ${Be^{\rho}}$ becomes infinite for large ${\rho}$. So for large ${\rho}$, we must have

$\displaystyle u(\rho)\sim Ae^{-\rho} \ \ \ \ \ (14)$

At the other end, when ${\rho}$ is very small, the term in ${\rho^{-2}}$ becomes the largest, so the approximate equation to solve is

$\displaystyle \frac{d^{2}u}{d\rho^{2}}=\frac{l(l+1)}{\rho^{2}}u \ \ \ \ \ (15)$

[This argument fails if ${l=0}$, but all we’re after here is looking at asymptotic behaviour in an attempt to factor this behaviour out of the overall solution. As we’ll see when we finally get the solution, it is valid for ${l=0}$ as well.]

In this case, the general solution is

$\displaystyle u(\rho)=C\rho^{l+1}+D\rho^{-l} \ \ \ \ \ (16)$

This can be verified by direct substitution:

 $\displaystyle \frac{d^{2}u}{d\rho^{2}}$ $\displaystyle =$ $\displaystyle Cl(l+1)\rho^{l-1}+D(-l)(-l-1)\rho^{-l-2}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{l(l+1)}{\rho^{2}}u \ \ \ \ \ (18)$

In this case, the term ${D\rho^{-l}}$ becomes infinite as ${\rho\rightarrow0}$, so ${D=0}$ and

$\displaystyle u(\rho)\sim C\rho^{l+1} \ \ \ \ \ (19)$

So now we know the behaviours at the two extremes, and we can factor both of these out, hoping to solve for what is left over. That is, we can write

$\displaystyle u(\rho)=\rho^{l+1}e^{-\rho}v(\rho) \ \ \ \ \ (20)$

where ${v(\rho)}$ is what we must find. Note that we have absorbed the two constants ${A}$ and ${C}$ into ${v(\rho)}$.

The idea is to plug 20 back into 11 and see what sort of equation we get for ${v(\rho)}$ as a result. We need the second derivative of ${u}$ in terms of ${v}$. We need to use the product rule a few times to get it.

 $\displaystyle \frac{du}{d\rho}$ $\displaystyle =$ $\displaystyle (l+1)\rho^{l}e^{-\rho}v-\rho^{l+1}e^{-\rho}v+\rho^{l+1}e^{-\rho}\frac{dv}{d\rho}\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \rho^{l}e^{-\rho}\left((l+1-\rho)v+\rho\frac{dv}{d\rho}\right)\ \ \ \ \ (22)$ $\displaystyle \frac{d^{2}u}{d\rho^{2}}$ $\displaystyle =$ $\displaystyle l\rho^{l-1}e^{-\rho}\left((l+1-\rho)v+\rho\frac{dv}{d\rho}\right)-\rho^{l}e^{-\rho}\left((l+1-\rho)v+\rho\frac{dv}{d\rho}\right)+\ \ \ \ \ (23)$ $\displaystyle$ $\displaystyle$ $\displaystyle \qquad\rho^{l}e^{-\rho}\left(-v+(l+1-\rho)\frac{dv}{d\rho}+\frac{dv}{d\rho}+\rho\frac{d^{2}v}{d\rho^{2}}\right)\ \ \ \ \ (24)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \rho^{l}e^{-\rho}\left[\left(\frac{l(l+1)}{\rho}+\rho-2l-2\right)v+2(l+1-\rho)\frac{dv}{d\rho}+\rho\frac{d^{2}v}{d\rho^{2}}\right] \ \ \ \ \ (25)$

Plugging this back into 11 and collecting terms we get

 $\displaystyle \rho^{l}e^{-\rho}\left[\left(\frac{l(l+1)}{\rho}+\rho-2l-2\right)v+2(l+1-\rho)\frac{dv}{d\rho}+\rho\frac{d^{2}v}{d\rho^{2}}\right]$ $\displaystyle =$ $\displaystyle \left[1-\frac{\rho_{0}}{\rho}+\frac{l(l+1)}{\rho^{2}}\right]\rho^{l+1}e^{-\rho}v\ \ \ \ \ (26)$ $\displaystyle \rho^{l}e^{-\rho}\left[\left(\frac{l(l+1)}{\rho}+\rho-2l-2\right)v+2(l+1-\rho)\frac{dv}{d\rho}+\rho\frac{d^{2}v}{d\rho^{2}}\right]$ $\displaystyle =$ $\displaystyle \left[\rho-\rho_{0}+\frac{l(l+1)}{\rho}\right]\rho^{l}e^{-\rho}v\ \ \ \ \ (27)$ $\displaystyle \left[\left(\frac{l(l+1)}{\rho}+\rho-2l-2\right)v+2(l+1-\rho)\frac{dv}{d\rho}+\rho\frac{d^{2}v}{d\rho^{2}}\right]$ $\displaystyle =$ $\displaystyle \left[\rho-\rho_{0}+\frac{l(l+1)}{\rho}\right]v\ \ \ \ \ (28)$ $\displaystyle \rho\frac{d^{2}v}{d\rho^{2}}+2(l+1-\rho)\frac{dv}{d\rho}+(\rho_{0}-2l-2)v$ $\displaystyle =$ $\displaystyle 0 \ \ \ \ \ (29)$

This version of the differential equation may not look any friendlier than the original, but we can now try to solve it by expressing ${v(\rho)}$ as a series in ${\rho}$, which we will do in the next post.

# Schrödinger equation in three dimensions – the radial equation

Required math: calculus

Required physics: Schrödinger equation

Reference: Griffiths, David J. (2005), Introduction to Quantum Mechanics, 2nd Edition; Pearson Education – Sec 4.1.3.

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.6.2.

[If some equations are too small to read easily, use your browser’s magnifying option (Ctrl + on Chrome, probably something similar on other browsers).]

When we considered the solution of the Schrödinger equation in three dimensions, we found that the general solution separated neatly into a product of three functions, one for each variable in spherical coordinates.

The Schrödinger equation in three dimensions can be written as

$\displaystyle -\frac{\hbar^{2}}{2m}\nabla^{2}\Psi+V\Psi=i\hbar\frac{\partial\Psi}{\partial t} \ \ \ \ \ (1)$

If we assume that the potential ${V=V(x,y,z)}$ is independent of time, we can use the same separation of variables method that we used in one dimension to split off the time part of the solution to get

$\displaystyle \Psi(x,y,z,t)=\psi(x,y,z)e^{-iEt/\hbar} \ \ \ \ \ (2)$

where, as before, the energy ${E}$ takes on a set of discrete values for the bound states and a set of continuous values for the scattering, or unbound, states. The spatial wave function ${\psi}$ satisfies the time-independent Schrödinger equation:

$\displaystyle -\frac{\hbar^{2}}{2m}\nabla^{2}\psi+V\psi=E\psi \ \ \ \ \ (3)$

So far, the analysis is the same as that for one dimension.

Using separation of variables in the form ${\psi(r,\theta,\phi)=R(r)Y(\theta,\phi)}$ we got two separated equations:

 $\displaystyle \frac{1}{R}\frac{\partial}{\partial r}\left(r^{2}\frac{\partial R}{\partial r}\right)-\frac{2mr^{2}}{\hbar^{2}}(V-E)$ $\displaystyle =$ $\displaystyle l(l+1)\ \ \ \ \ (4)$ $\displaystyle \frac{1}{Y\sin\theta}\frac{\partial}{\partial\theta}\left(\sin\theta\frac{\partial Y}{\partial\theta}\right)+\frac{1}{Y\sin^{2}\theta}\left(\frac{\partial^{2}Y}{\partial\phi^{2}}\right)$ $\displaystyle =$ $\displaystyle -l(l+1) \ \ \ \ \ (5)$

where ${l(l+1)}$ is a constant term.

We found that the angular equation could be solved and that the solutions were the spherical harmonics:

$\displaystyle Y_{l}^{m}(\theta,\phi)=\left[\frac{2l+1}{4\pi}\frac{(p-m)!}{(p+m)!}\right]^{1/2}e^{im\phi}P_{l}^{m}(\cos\theta) \ \ \ \ \ (6)$

They obey the normalization condition

$\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}(Y_{l}^{m})^*Y_{l'}^{m'}\sin\theta d\theta d\phi=\delta_{ll'}\delta_{mm'} \ \ \ \ \ (7)$

Returning to the radial function we find that we can actually make one further transformation of the equation that makes it a bit easier to solve in some cases. We can rewrite the equation using total derivatives, since ${R(r)}$ depends only on ${r}$:

$\displaystyle \frac{d}{dr}\left(r^{2}\frac{dR}{dr}\right)-\frac{2mr^{2}}{\hbar^{2}}(V-E)R=l(l+1)R \ \ \ \ \ (8)$

We can now make the substitution

 $\displaystyle u(r)$ $\displaystyle \equiv$ $\displaystyle rR\ \ \ \ \ (9)$ $\displaystyle R$ $\displaystyle =$ $\displaystyle \frac{u}{r}\ \ \ \ \ (10)$ $\displaystyle \frac{dR}{dr}$ $\displaystyle =$ $\displaystyle -\frac{u}{r^{2}}+\frac{u'}{r}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{r^{2}}(ru'-u)\ \ \ \ \ (12)$ $\displaystyle \frac{d}{dr}\left(r^{2}\frac{dR}{dr}\right)$ $\displaystyle =$ $\displaystyle u'+ru^{\prime\prime}-u'\ \ \ \ \ (13)$ $\displaystyle$ $\displaystyle =$ $\displaystyle ru^{\prime\prime} \ \ \ \ \ (14)$

The radial equation then becomes

 $\displaystyle r\frac{d^{2}u}{dr^{2}}-\frac{2mr}{\hbar^{2}}(V-E)u$ $\displaystyle =$ $\displaystyle l(l+1)\frac{u}{r}\ \ \ \ \ (15)$ $\displaystyle -\frac{\hbar^{2}}{2m}\frac{d^{2}u}{dr^{2}}+\left(V+\frac{\hbar^{2}}{2m}\frac{l(l+1)}{r^{2}}\right)u$ $\displaystyle =$ $\displaystyle Eu \ \ \ \ \ (16)$

In this form, the equation looks like the original one-dimensional Schrödinger equation with the wave function given by ${u}$ and the potential given by

$\displaystyle V_{rad}=V+\frac{\hbar^{2}}{2m}\frac{l(l+1)}{r^{2}} \ \ \ \ \ (17)$

The extra term ${\frac{\hbar^{2}}{2m}\frac{l(l+1)}{r^{2}}}$ is called the centrifugal term. Classically, the force due to this term is:

 $\displaystyle F_{cent}$ $\displaystyle =$ $\displaystyle -\frac{d}{dr}\frac{\hbar^{2}}{2m}\frac{l(l+1)}{r^{2}}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\hbar^{2}}{m}\frac{l(l+1)}{r^{3}} \ \ \ \ \ (19)$

which is a force that tends to repel the particle from the origin (the force gets larger the closer to the origin we are). Thus it is analogous to the pseudo-force known as the centrifugal force in classical physics.