# Pauli matrices: trace

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 14, Exercise 14.3.3.

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The three components of the spin operator ${\mathbf{S}}$ for spin ${\frac{1}{2}}$ can be expressed in terms of the Pauli matrices

$\displaystyle \sigma_{x}=\left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right];\quad\sigma_{y}=\left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right];\quad\sigma_{z}=\left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right] \ \ \ \ \ (1)$

as

$\displaystyle S_{i}=\frac{\hbar}{2}\sigma_{i} \ \ \ \ \ (2)$

As the trace of a matrix is the sum of its diagonal elements, it’s obvious from their definitions that the ${\sigma_{i}}$ are traceless, but for some reason Shankar wants us to show this by a roundabout method.

We can show by direct calculation that the Pauli matrices anticommute with each other. For example

 $\displaystyle \sigma_{x}\sigma_{y}$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\ \ \ \ \ (3)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} i & 0\\ 0 & -i \end{array}\right]\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left[\begin{array}{cc} -i & 0\\ 0 & i \end{array}\right]\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\sigma_{y}\sigma_{x} \ \ \ \ \ (7)$

In general, we have, for ${i\ne j}$:

 $\displaystyle \left[\sigma_{i},\sigma_{j}\right]_{+}$ $\displaystyle =$ $\displaystyle 0\ \ \ \ \ (8)$ $\displaystyle \sigma_{i}\sigma_{j}$ $\displaystyle =$ $\displaystyle -\sigma_{j}\sigma_{i} \ \ \ \ \ (9)$

Also, by direct calculation (or by using the commutation relations for ${S_{i}}$) we can show that

 $\displaystyle \left[\sigma_{x},\sigma_{y}\right]$ $\displaystyle =$ $\displaystyle \sigma_{x}\sigma_{y}-\sigma_{y}\sigma_{x}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\sigma_{x}\sigma_{y}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2\left[\begin{array}{cc} i & 0\\ 0 & -i \end{array}\right]\ \ \ \ \ (12)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 2i\sigma_{z} \ \ \ \ \ (13)$

This gives the relation

$\displaystyle \sigma_{x}\sigma_{y}=i\sigma_{z} \ \ \ \ \ (14)$

and also for cyclic permutations of ${x}$, ${y}$ and ${z}$. Also by direct calculation we can see that

$\displaystyle \sigma_{i}^{2}=I \ \ \ \ \ (15)$

We can write this more generally as

$\displaystyle \sigma_{i}\sigma_{j}=\delta_{ij}I+i\sum_{k}\varepsilon_{ijk}\sigma_{k} \ \ \ \ \ (16)$

where ${\varepsilon_{ijk}}$ is the Levi-Civita antisymmetric tensor.

Returning to the trace, we can use the theorem for the trace of a product:

$\displaystyle \mbox{Tr}\left(AB\right)=\mbox{Tr}\left(BA\right) \ \ \ \ \ (17)$

Applying this to 9 we have

$\displaystyle \mbox{Tr}\left(\sigma_{x}\sigma_{y}\right)=\mbox{Tr}\left(\sigma_{y}\sigma_{x}\right)=-\mbox{Tr}\left(\sigma_{y}\sigma_{x}\right) \ \ \ \ \ (18)$

Any quantity equal to its negative must be zero, so

$\displaystyle \mbox{Tr}\left(\sigma_{x}\sigma_{y}\right)=0 \ \ \ \ \ (19)$

Thus from 14 we get

$\displaystyle \mbox{Tr}\sigma_{z}=0 \ \ \ \ \ (20)$

We can use the same argument for ${\sigma_{x}}$ and ${\sigma_{y}}$ by cyclic permutation.

# Any spinor is an eigenket of the spin operator for some direction of spin

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 14, Exercise 14.3.1.

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The eigenvectors of the spin ${\frac{1}{2}}$ matrix in an arbitrary direction are given by

 $\displaystyle \left|\hat{n}+\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} \cos\frac{\theta}{2}e^{-i\phi/2}\\ \sin\frac{\theta}{2}e^{i\phi/2} \end{array}\right]\ \ \ \ \ (1)$ $\displaystyle \left|\hat{n}-\right\rangle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} -\sin\frac{\theta}{2}e^{-i\phi/2}\\ \cos\frac{\theta}{2}e^{i\phi/2} \end{array}\right] \ \ \ \ \ (2)$

where the direction vector is given by

$\displaystyle \hat{\mathbf{n}}=\sin\theta\cos\phi\hat{\mathbf{x}}+\sin\theta\sin\phi\hat{\mathbf{y}}+\cos\theta\hat{\mathbf{z}} \ \ \ \ \ (3)$

The corresponding spin operator is given by the matrix

$\displaystyle \hat{\mathbf{n}}\cdot\mathbf{S}=\frac{\hbar}{2}\left[\begin{array}{cc} \cos\theta & \sin\theta e^{-i\phi}\\ \sin\theta e^{i\phi} & -\cos\theta \end{array}\right] \ \ \ \ \ (4)$

Any 2-component normalized spinor is an eigenvector of such a matrix. To see this, suppose we have an arbitrary spinor written as

 $\displaystyle \left|\chi\right\rangle$ $\displaystyle =$ $\displaystyle \rho_{1}e^{i\phi_{1}}\left[\begin{array}{c} 1\\ 0 \end{array}\right]+\rho_{2}e^{i\phi_{2}}\left[\begin{array}{c} 0\\ 1 \end{array}\right]\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{c} \rho_{1}e^{i\phi_{1}}\\ \rho_{2}e^{i\phi_{2}} \end{array}\right] \ \ \ \ \ (6)$

where ${\rho_{1,2}}$ and ${\phi_{1,2}}$ are arbitrary real numbers (so that the coefficients on the RHS are arbitrary complex numbers). From normalization we have

$\displaystyle \left\langle \chi\left|\chi\right.\right\rangle =1=\left[\begin{array}{cc} \rho_{1}e^{-i\phi_{1}} & \rho_{2}e^{-i\phi_{2}}\end{array}\right]\left[\begin{array}{c} \rho_{1}e^{i\phi_{1}}\\ \rho_{2}e^{i\phi_{2}} \end{array}\right]=\rho_{1}^{2}+\rho_{2}^{2} \ \ \ \ \ (7)$

Thus we can write ${\rho_{1}}$ and ${\rho_{2}}$ as the sine and cosine of some angle, which we’ll call ${\frac{\theta}{2}}$, giving

$\displaystyle \left|\chi\right\rangle =\left[\begin{array}{c} \cos\frac{\theta}{2}e^{i\phi_{1}}\\ \sin\frac{\theta}{2}e^{i\phi_{2}} \end{array}\right] \ \ \ \ \ (8)$

We can put this in the form 1 as follows. Since an overall phase doesn’t affect the physics of the spinor, we can write

$\displaystyle \left|\chi\right\rangle =e^{i\alpha}\left[\begin{array}{c} \cos\frac{\theta}{2}e^{-i\phi/2}\\ \sin\frac{\theta}{2}e^{i\phi/2} \end{array}\right]=\left[\begin{array}{c} \cos\frac{\theta}{2}e^{i\phi_{1}}\\ \sin\frac{\theta}{2}e^{i\phi_{2}} \end{array}\right] \ \ \ \ \ (9)$

We have the conditions

 $\displaystyle \phi_{1}$ $\displaystyle =$ $\displaystyle \alpha-\frac{\phi}{2}\ \ \ \ \ (10)$ $\displaystyle \phi_{2}$ $\displaystyle =$ $\displaystyle \alpha+\frac{\phi}{2} \ \ \ \ \ (11)$

Solving, we get

 $\displaystyle \alpha$ $\displaystyle =$ $\displaystyle \frac{\phi_{1}+\phi_{2}}{2}\ \ \ \ \ (12)$ $\displaystyle \phi$ $\displaystyle =$ $\displaystyle \phi_{2}-\phi_{1} \ \ \ \ \ (13)$

giving

$\displaystyle \left|\chi\right\rangle =e^{i\left(\phi_{1}+\phi_{2}\right)/2}\left[\begin{array}{c} \cos\frac{\theta}{2}e^{-i\left(\phi_{2}-\phi_{1}\right)/2}\\ \sin\frac{\theta}{2}e^{i\left(\phi_{2}-\phi_{1}\right)/2} \end{array}\right] \ \ \ \ \ (14)$

Thus ${\left|\chi\right\rangle }$ as given by 6 is an eigenvector of the operator 4, where

$\displaystyle \hat{\mathbf{n}}=\sin\theta\cos\left(\phi_{2}-\phi_{1}\right)\hat{\mathbf{x}}+\sin\theta\sin\left(\phi_{2}-\phi_{1}\right)\hat{\mathbf{y}}+\cos\theta\hat{\mathbf{z}} \ \ \ \ \ (15)$

# Kinematics of spin: Hilbert space for an electron

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Section 14.3.

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We’ve looked at quantum mechanical spin before but Shankar’s treatment is quite different to that of Griffiths, so it’s worth another look. Shankar begins with a thought experiment in which an electron is prepared (don’t ask how!) in a state with zero momentum. Since its momentum is known precisely, its position is completely uncertain, so we can take the wave function in position space to be a constant, independent of position. Since the angular momentum operator ${\mathbf{L}}$ that we’ve met before is defined by replacing classical quantities by quantum operators in the classical relation ${\mathbf{L}=\mathbf{r}\times\mathbf{p}}$, an electron in this state must have ${\mathbf{L}=0}$. (We can see this also from the relation ${L_{z}=-i\hbar\frac{\partial}{\partial\phi}}$ which gives zero since the wave function is independent of position.) However, if we measure the angular momentum along some direction such as ${z}$ of an electron in such a state, we find that it always has the values ${\pm\frac{\hbar}{2}}$.

If we want to construct a wave function that describes the electron, we therefore need to consider a function that has a component that is independent of position, but which has eigenvalues ${\pm\frac{\hbar}{2}}$ when operated on by some operator with the properties of an angular momentum operator. The key to finding such a wave function is given by our example of a vector wave function. In that example, which considered the behaviour of a 2-d vector valued function under an infinitesimal rotation, we found that there were two effects of such a rotation. (It might help the reader to refer back to our earlier discussion at this point.) First, the rotation carries a vector ${\mathbf{V}}$ from its initial location ${A}$ to some other point ${B}$. Second, the original vector ${\mathbf{V}}$ also gets rotated through the infinitesimal angle so that it now points in a slightly different direction, giving the rotated vector ${\mathbf{V}^{\prime}}$. The components of ${\mathbf{V}^{\prime}}$ are linear combinations of the components of the unrotated vector ${\mathbf{V}}$.

The first effect (that of rotating the function at ${A}$ into the new position ${B}$) is generated by the original angular momentum operator ${\mathbf{L}}$. This rotation depends on the position coordinates, since we must know the coordinates of the two points ${A}$ and ${B}$ to calculate the effect of the rotation. The second effect (that of rotating the vector so it points in a different direction) does not depend on the positions; rather, it depends only on the angle of rotation. As we showed in the previous example (working in 2-d), the second effect is generated by a ${2\times2}$ matrix ${S_{z}}$. We need a matrix rather than just a single number since we need to form a linear combination of the two components of the original vector to get the rotated vector.

In the special case of the 2-d rotation, the combined effect of these two types of rotation are given by

 $\displaystyle J_{z}$ $\displaystyle =$ $\displaystyle L_{z}+S_{z}\ \ \ \ \ (1)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[\begin{array}{cc} L_{z} & 0\\ 0 & L_{z} \end{array}\right]+\left[\begin{array}{cc} 0 & -i\hbar\\ i\hbar & 0 \end{array}\right] \ \ \ \ \ (2)$

The transformation then becomes

$\displaystyle \left[\begin{array}{c} V_{1}^{\prime}\\ V_{2}^{\prime} \end{array}\right]=\left[I-\frac{i\varepsilon_{z}}{\hbar}J_{z}\right]\left[\begin{array}{c} V_{1}\\ V_{2} \end{array}\right] \ \ \ \ \ (3)$

This equation is valid for the rotation of a vector wave function with two components through an angle ${\varepsilon_{z}}$ about the ${z}$ axis, in two dimensions.

At this point, we can generalize this result to 3-d, and also to a wave function with some arbitrary number ${n}$ components. The dimension of the matrix is determined by ${n}$, so if we again consider a rotation about the ${z}$ axis by some angle ${\varepsilon}$, we get

 $\displaystyle \left[\begin{array}{c} \psi_{1}^{\prime}\\ \vdots\\ \psi_{n}^{\prime} \end{array}\right]$ $\displaystyle =$ $\displaystyle \left[I-\frac{i\varepsilon_{z}}{\hbar}J_{z}\right]\left[\begin{array}{c} \psi_{1}\\ \vdots\\ \psi_{n} \end{array}\right]\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\left[\begin{array}{ccc} 1 & 0 & 0\\ 0 & \ddots & 0\\ 0 & 0 & 1 \end{array}\right]-\frac{i\varepsilon_{z}}{\hbar}\left[\begin{array}{ccc} -i\hbar\frac{\partial}{\partial\phi} & 0 & 0\\ 0 & \ddots & 0\\ 0 & 0 & -i\hbar\frac{\partial}{\partial\phi} \end{array}\right]-\frac{i\varepsilon_{z}}{\hbar}S_{z}\right)\left[\begin{array}{c} \psi_{1}\\ \vdots\\ \psi_{n} \end{array}\right] \ \ \ \ \ (5)$

Here, the actual form of ${S_{z}}$ is yet to be determined.

Although we worked out this for the special case of a rotation about the ${z}$ axis, we can generalize it to a rotation about an arbitrary axis, and thus get a vector operator:

$\displaystyle \mathbf{J}=\mathbf{L}+\mathbf{S} \ \ \ \ \ (6)$

The orbital angular momentum operator ${\mathbf{L}}$ operates on spatial coordinates (and is the same operator on each of the ${n}$ components), while the spin operator ${\mathbf{S}}$ mixes the ${n}$ components of the wave function (and not on the spatial coordinates). As these two operators operate on different quantities, they commute, which leads to the commutation relation

$\displaystyle \left[J_{i},J_{j}\right]=i\hbar\sum_{k}\varepsilon_{ijk}J_{k} \ \ \ \ \ (7)$

which separates into the same commutation relations for each component of ${\mathbf{J}}$ so we have

 $\displaystyle \left[L_{i},L_{j}\right]$ $\displaystyle =$ $\displaystyle i\hbar\sum_{k}\varepsilon_{ijk}L_{k}\ \ \ \ \ (8)$ $\displaystyle \left[S_{i},S_{j}\right]$ $\displaystyle =$ $\displaystyle i\hbar\sum_{k}\varepsilon_{ijk}S_{k} \ \ \ \ \ (9)$

Shankar worked out the matrices that satisfy 7 in his equations 12.5.22 to 12.5.24 so we won’t go through that again here. What’s important to remember is that these matrices are block diagonal matrices consisting of a series of blocks of dimension ${\left(2j+1\right)\times\left(2j+1\right)}$ for ${j=0,\frac{1}{2},1,\frac{3}{2},\ldots}$. Each of these blocks satisfies 7 on its own, so we can pick the block with the right dimension to satisfy the experimental result that the electron has two spin states: ${\pm\frac{\hbar}{2}}$. That is, for the electron, we have the number of components in the wave ${n=2}$, so we choose ${j=\frac{1}{2}}$ for the spin operators, which turn out to be the familiar ones we’ve met before:

 $\displaystyle S_{x}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2}\left[\begin{array}{cc} 0 & 1\\ 1 & 0 \end{array}\right]\ \ \ \ \ (10)$ $\displaystyle S_{y}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2}\left[\begin{array}{cc} 0 & -i\\ i & 0 \end{array}\right]\ \ \ \ \ (11)$ $\displaystyle S_{z}$ $\displaystyle =$ $\displaystyle \frac{\hbar}{2}\left[\begin{array}{cc} 1 & 0\\ 0 & -1 \end{array}\right] \ \ \ \ \ (12)$

The complete wave function of an electron is therefore the product of a function of position with a two-component vector, called a spinor, which represents the spin state. That is, we can write, in the position-spin basis

$\displaystyle \left|\psi\right\rangle =\psi_{+}\left(\mathbf{r}\right)\left[\begin{array}{c} 1\\ 0 \end{array}\right]+\psi_{-}\left(\mathbf{r}\right)\left[\begin{array}{c} 0\\ 1 \end{array}\right] \ \ \ \ \ (13)$

In terms of Hilbert space, the spatial components are essentially the infinite-dimensional vectors ${\psi_{+}}$ and ${\psi_{-}}$ which have values defined at each point in 3-d space, while the spinor components are 2-d vectors. Thus the complete Hilbert space ${\mathbb{V}_{e}}$ of an electron is defined as

$\displaystyle \mathbb{V}_{e}=\mathbb{V}_{0}\otimes\mathbb{V}_{s} \ \ \ \ \ (14)$

where ${\mathbb{V}_{0}}$ is the infinite-dimensional spatial component and ${\mathbb{V}_{s}}$ is the 2-d spinor component. In terms of this space, we can create a unit operator as a sum over a complete set of states:

$\displaystyle \mathbf{1}=\sum_{s_{z}}\int\left|xyzs_{z}\right\rangle \left\langle xyzs_{z}\right|dx\;dy\;dz \ \ \ \ \ (15)$

The normalization condition for a wave function thus becomes

 $\displaystyle 1=\left\langle \psi\left|\psi\right.\right\rangle$ $\displaystyle =$ $\displaystyle \sum_{s_{z}}\int\left\langle \psi\left|xyzs_{z}\right.\right\rangle \left\langle xyzs_{z}\left|\psi\right.\right\rangle dx\;dy\;dz\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \int\left(\left|\psi_{+}\right|^{2}+\left|\psi_{-}\right|^{2}\right)dx\;dy\;dz \ \ \ \ \ (17)$

where we get the last line by substituting 13. The term

$\displaystyle \int\left|\psi_{+}\right|^{2}dx\;dy\;dz \ \ \ \ \ (18)$

represents the probability that the electron will be found in the spin state ${+\frac{\hbar}{2}}$ anywhere in space.

The important point to remember from this derivation is that spin is an essentially new phenomenon with no classical analogue. Thus the wave function for a particle that has spin is necessarily an expanded Hilbert space where the extra subspace ${\mathbb{V}_{s}}$ is introduced to allow the extra spin states. The two spaces ${\mathbb{V}_{0}}$ and ${\mathbb{V}_{s}}$ are completely separate from each other, with each possessing its own operators, eigenstates and eigenvalues. Of course, it’s possible to construct operators that are composed of other operators from both spaces, but we’ll leave that until later.

One final point is worth making. The above derivation of the Hilbert space for an electron relied on the experimental result that the electron has exactly two spin states, thus leading to the ${2\times2}$ spin matrices. This is the best we can do in non-relativistic quantum theory. Shankar promises us that when we study the Dirac equation, which arose out of the need to introduce special relativity, the two spin states of the electron can actually be derived.

# Sizes of elementary particles

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 13, Exercises 13.3.1 – 13.3.2.

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Due to the position-momentum uncertainty principle, if we wish to determine the location of a particle to within a distance ${\Delta X}$, the momentum of the photon used to detect the particle must satisfy

$\displaystyle \Delta P\Delta X\ge\frac{\hbar}{2} \ \ \ \ \ (1)$

This relation is valid in non-relativistic quantum mechanics, where we are using position eigenkets ${\left|X\right\rangle }$ which define a particle’s position exactly. To do this, however, would require a photon of infinite energy. In relativistic quantum theory, if the energy of the photon is large enough, it is possible to convert the energy into mass by creating a particle-antiparticle pair. If we’re trying to determine the location of an electron, then if the energy of the bombarding photon is around twice the rest energy of an electron, this pair creation process can occur. Thus for practical purposes, the maximum photon energy that we can use to detect the electron is finite, which means that the electron’s position can be determined only approximately.

To get an idea of the ‘radius’ of an electron using these ideas (I put ‘radius’ in quotes because an electron doesn’t have a rigid boundary in quantum theory), we can proceed as follows. We’ll work only to orders of magnitude, rather than precise quantities.

From the uncertaintly relation, the photon’s momentum is about

$\displaystyle \Delta P\sim\frac{\hbar}{\Delta X} \ \ \ \ \ (2)$

For a photon, the relativistic energy is related to the momentum by

$\displaystyle \Delta E=\Delta Pc \ \ \ \ \ (3)$

where ${c}$ is the speed of light. Therefore, the energy of the incident photon is

$\displaystyle \Delta E\sim\frac{\hbar c}{\Delta X} \ \ \ \ \ (4)$

We therefore want to restrict this energy to less than twice the electron’s rest energy, so

$\displaystyle \Delta E\lesssim2mc^{2} \ \ \ \ \ (5)$

 $\displaystyle \frac{\hbar c}{\Delta X}$ $\displaystyle \lesssim$ $\displaystyle 2mc^{2}\ \ \ \ \ (6)$ $\displaystyle \Delta X$ $\displaystyle \gtrsim$ $\displaystyle \frac{\hbar}{2mc}\sim\frac{\hbar}{mc} \ \ \ \ \ (7)$

The latter quantity is the Compton wavelength of the electron. [When we originally encountered the Compton wavelength is Carroll & Ostlie’s book on astrophysics, they defined it as ${h/mc}$, so Shankar’s Compton wavelength is ${\frac{1}{2\pi}}$ times that of Carroll & Ostlie. However, since we’re working with orders of magnitude, this won’t matter much.]

Thus the Compton wavelength can be taken as a rough size of the electron. We can write this as a fraction of the Bohr radius ${a_{0}}$ using

$\displaystyle a_{0}\equiv\frac{\hbar^{2}}{me^{2}} \ \ \ \ \ (8)$

so that

$\displaystyle \frac{\hbar/mc}{a_{0}}=\frac{\hbar}{mc}\frac{me^{2}}{\hbar^{2}}=\frac{e^{2}}{\hbar c}=\alpha\approx\frac{1}{137} \ \ \ \ \ (9)$

where ${\alpha}$ is the famous fine structure constant. Since ${a_{0}}$ is roughly the radius of a ground-state hydrogen atom, the electron is about 100 times smaller than this.

We can use similar arguments to do some rough calculations on other particles.

Example 1 For example, the pion has a range of about ${10^{-15}\mbox{ m}}$ as a mediator of the nuclear force, so if we take this as ${\Delta X}$ then

 $\displaystyle 2m_{\pi}c^{2}$ $\displaystyle \sim$ $\displaystyle \frac{\hbar c}{\Delta X} \ \ \ \ \ (10)$

The rest energy of an electron is about ${0.5\mbox{ MeV}}$, so we can get an estimate of the rest energy of the pion as follows.

$\displaystyle \frac{m_{\pi}c^{2}}{m_{e}c^{2}}=\frac{\Delta X_{e}}{\Delta X_{\pi}}=\frac{a_{0}/137}{10^{-15}} \ \ \ \ \ (11)$

$\displaystyle a_{0}\approx5\times10^{-11}\mbox{ m} \ \ \ \ \ (12)$

so

$\displaystyle m_{\pi}c^{2}\approx\left(0.5\mbox{ MeV}\right)\frac{5\times10^{-11}}{137\times10^{-15}}=182\mbox{ MeV} \ \ \ \ \ (13)$

The actual rest mass of a pion is around 140 MeV, so this estimate isn’t too bad.

Example 2 The de Broglie wavelength of a particle is defined by

$\displaystyle \lambda=\frac{h}{p} \ \ \ \ \ (14)$

For an electron with kinetic energy 200 eV, we need to find its momentum to calculate ${\lambda}$. The relativistic kinetic energy is

$\displaystyle K=mc^{2}\left(\gamma-1\right) \ \ \ \ \ (15)$

where

$\displaystyle \gamma=\frac{1}{\sqrt{1-v^{2}/c^{2}}} \ \ \ \ \ (16)$

We have

$\displaystyle \gamma=\frac{K}{mc^{2}}+1=\frac{200\mbox{ eV}}{0.5\times10^{6}\mbox{ eV}}+1=1.0004 \ \ \ \ \ (17)$

Thus the electron is travelling at a non-relativistic speed, so to a good approximation we can use Newtonian formulas. The speed is

 $\displaystyle v$ $\displaystyle =$ $\displaystyle c\sqrt{\frac{2K}{mc^{2}}}=c\sqrt{\frac{2\left(200\right)}{0.5\times10^{6}}}\approx0.03c\ \ \ \ \ (18)$ $\displaystyle p$ $\displaystyle =$ $\displaystyle mv=\left(9.1\times10^{-31}\right)\left(0.03\right)\left(3\times10^{8}\right)=7.7\times10^{-24}\mbox{ kg m s}^{-1}\ \ \ \ \ (19)$ $\displaystyle \lambda$ $\displaystyle =$ $\displaystyle \frac{h}{p}=\frac{6.6\times10^{-34}}{7.7\times10^{-24}}\approx10^{-10}\mbox{ m}=1\AA \ \ \ \ \ (20)$

# Runge-Lenz vector and closed orbits

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 13, Exercise 13.2.1.

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The energy levels of hydrogen, when calculated from the Coulomb potential alone (ignoring various perturbations) depend only on the principal quantum number ${n}$ according to

$\displaystyle E=-\frac{1}{n^{2}}\frac{\mu e^{4}}{2\hbar^{2}} \ \ \ \ \ (1)$

The quantization arises entirely from the requirement that the radial function remain finite for large ${r}$, and makes no mention of the angular quantum numbers ${l}$ and ${m}$. Thus each energy level (each value of ${n}$) has a degeneracy of ${n^{2}}$, with ${2l+1}$ degenerate states for each ${l}$. Each symmetry is associated with the conservation of some dynamical quantity, with the degeneracy in ${m}$ due to conservation of angular momentum.

Shankar points out that, in classical mechanics, any potential with a ${\frac{1}{r}}$ dependence conserves the Runge-Lenz vector, defined for the hydregen atom potential as

$\displaystyle \mathbf{n}=\frac{\mathbf{p}\times\boldsymbol{\ell}}{\mu}-\frac{e^{2}}{r}\mathbf{r} \ \ \ \ \ (2)$

where I’ve used ${\mu}$ for the electron mass to avoid confusion with the ${L_{z}}$ quantum number ${m}$.

Although it doesn’t make sense to talk about the orbit of the electron in quantum mechanics, classically we can see that the conservation of ${\mathbf{n}}$ implies that the orbit is closed. We can see this as follows.

First, using

$\displaystyle \boldsymbol{\ell}=\mathbf{r}\times\mathbf{p} \ \ \ \ \ (3)$

we have

 $\displaystyle \mathbf{n}$ $\displaystyle =$ $\displaystyle \frac{1}{\mu}\mathbf{p}\times\left(\mathbf{r}\times\mathbf{p}\right)-\frac{e^{2}}{r}\mathbf{r}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{\mu}\mathbf{r}\times\left(\mathbf{p}\cdot\mathbf{p}\right)-\mathbf{p}\left(\mathbf{r}\cdot\mathbf{p}\right)-\frac{e^{2}}{r}\mathbf{r}\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(\frac{p^{2}}{\mu}-\frac{e^{2}}{r}\right)\mathbf{r}-\mathbf{p}\left(\mathbf{r}\cdot\mathbf{p}\right) \ \ \ \ \ (6)$

In the second line, we used the vector identity

$\displaystyle \mathbf{A}\times\left(\mathbf{B}\times\mathbf{C}\right)=\mathbf{B}\left(\mathbf{A}\cdot\mathbf{C}\right)-\mathbf{C}\left(\mathbf{A}\cdot\mathbf{B}\right) \ \ \ \ \ (7)$

Since we’re dealing with a bound state, ${r}$ must always remain finite, so it must have a maximum value. At this point ${\frac{dr}{dt}=0}$, which means that there is no radial motion, which in turn means that all motion at that point must be perpendicular to ${\mathbf{r}}$. In other words

$\displaystyle \mathbf{p}\cdot\mathbf{r}_{max}=0 \ \ \ \ \ (8)$

Also, from conservation of energy, we have

$\displaystyle E=\frac{p^{2}}{2\mu}-\frac{e^{2}}{r} \ \ \ \ \ (9)$

so at ${r_{max}}$ we have

$\displaystyle p^{2}=2\mu\left(E+\frac{e^{2}}{r_{max}}\right) \ \ \ \ \ (10)$

Plugging these into 6, we get

 $\displaystyle \mathbf{n}$ $\displaystyle =$ $\displaystyle \left(2E+\frac{2e^{2}}{r_{max}}-\frac{e^{2}}{r_{max}}\right)\mathbf{r}_{max}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(2E+\frac{e^{2}}{r_{max}}\right)\mathbf{r}_{max} \ \ \ \ \ (12)$

Exactly the same argument applies to the case where ${r}$ is a minimum: again ${\frac{dr}{dt}=0}$ so ${\mathbf{r}\cdot\mathbf{p}=0}$ and we end up with

$\displaystyle \mathbf{n}=\left(2E+\frac{e^{2}}{r_{min}}\right)\mathbf{r}_{min} \ \ \ \ \ (13)$

If ${\mathbf{n}}$ is conserved (constant), then it must be parallel or anti-parallel to both ${\mathbf{r}_{max}}$ and ${\mathbf{r}_{min}}$, and the latter two vectors must therefore always have the same direction. In other words, the particle reaches its maximum (and minimum) distance always at the same point in its orbit, meaning that the orbit is closed.

In a general (elliptical) orbit, ${r_{max}>r_{min}}$ so ${\frac{e^{2}}{r_{max}}<\frac{e^{2}}{r_{min}}}$. Since ${E<0}$ for a bound orbit, we therefore must have

 $\displaystyle 2E+\frac{e^{2}}{r_{max}}$ $\displaystyle <$ $\displaystyle 0\ \ \ \ \ (14)$ $\displaystyle 2E+\frac{e^{2}}{r_{min}}$ $\displaystyle >$ $\displaystyle 0 \ \ \ \ \ (15)$

This in turn implies that ${\mathbf{n}}$ is anti-parallel to ${\mathbf{r}_{max}}$ and parallel to ${\mathbf{r}_{min}}$.

For a circular orbit, both ${r}$ and ${p}$ are constant, so both the kinetic and potential energies are also constant. From the virial theorem, we know that, for ${V\propto\frac{1}{r}}$

$\displaystyle \left\langle T\right\rangle =-\frac{1}{2}\left\langle V\right\rangle \ \ \ \ \ (16)$

Thus

 $\displaystyle E$ $\displaystyle =$ $\displaystyle T+V\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{V}{2}\ \ \ \ \ (18)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -\frac{e^{2}}{2r} \ \ \ \ \ (19)$

Thus from 12, we see that ${\mathbf{n}=0}$ for a circular orbit.

# Hydrogen atom – radial function at large r

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 13, Exercise 13.1.4.

[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 solving the 3-d Schrödinger equation for a spherically symmetric potential, the radial function has the asymptotic form for large ${r}$, and for the energy ${E<0}$:

$\displaystyle U_{El}\left(r\right)\underset{r\rightarrow\infty}{\longrightarrow}Ar^{\pm me^{2}/\kappa\hbar^{2}}e^{-\kappa r} \ \ \ \ \ (1)$

where

$\displaystyle \kappa\equiv\sqrt{\frac{2m\left|E\right|}{\hbar^{2}}} \ \ \ \ \ (2)$

For the hydrogen atom, the function ${U_{El}}$ is obtained from a series solution of the differential equation with the result

 $\displaystyle U_{El}$ $\displaystyle =$ $\displaystyle e^{-\rho}v_{El}\ \ \ \ \ (3)$ $\displaystyle v_{El}$ $\displaystyle =$ $\displaystyle \rho^{l+1}\sum_{k=0}^{\infty}C_{k}\rho^{k}\ \ \ \ \ (4)$ $\displaystyle \rho$ $\displaystyle =$ $\displaystyle \sqrt{\frac{-2mE}{\hbar^{2}}}r\ \ \ \ \ (5)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \kappa r \ \ \ \ \ (6)$

To keep the wave function finite at large ${r}$, we require the series to terminate, which leads to the quantized energy levels, given by

$\displaystyle E_{n}=-\frac{me^{4}}{2\hbar^{2}n^{2}} \ \ \ \ \ (7)$

The series in 4 terminates at a value of ${k=n-l-1}$, so the function ${v_{El}}$ is a polynomial in ${\rho}$, and thus in ${r}$, of degree ${n}$. Since the actual radial function is

$\displaystyle R_{nl}=\frac{U_{El}}{r} \ \ \ \ \ (8)$

we have that ${R_{nl}}$ is a polynomial of degree ${n-1}$ in ${r}$ multiplied by the exponential ${e^{-\rho}=e^{-\kappa r}}$. That is, for large ${r}$

$\displaystyle R_{nl}\sim r^{n-1}e^{-\kappa r} \ \ \ \ \ (9)$

To show that this is consistent with 1, we use 7 and 2.

 $\displaystyle n$ $\displaystyle =$ $\displaystyle \sqrt{\frac{me^{4}}{2\hbar^{2}\left|E_{n}\right|}}\ \ \ \ \ (10)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \sqrt{\frac{me^{4}}{2\hbar^{2}}}\sqrt{\frac{2m}{\hbar^{2}}}\frac{1}{\kappa}\ \ \ \ \ (11)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{me^{2}}{\kappa\hbar^{2}} \ \ \ \ \ (12)$

Comparing this with 1, we see that

$\displaystyle r^{n}=r^{me^{2}/\kappa\hbar^{2}} \ \ \ \ \ (13)$

so the condition is satisfied.

# Radial function for large r

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Section 12.6.

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

In solving the Schrödinger equation for spherically symmetric potentials, we found that we could reduce the problem to the equation

$\displaystyle \left[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}\right]U_{El}=EU_{El} \ \ \ \ \ (1)$

where ${U_{El}\left(r\right)}$ is related to the radial function by

$\displaystyle R_{El}\left(r\right)=\frac{U_{El}\left(r\right)}{r} \ \ \ \ \ (2)$

We’ve looked at some properties of ${U_{El}}$ (which Griffiths calls ${u}$) for the hydrogen atom, but we can also try to extract some information about ${U_{El}}$ in the more general case where we don’t need to specify the potential ${V}$ precisely. Here we’ll examine what happens as ${r\rightarrow\infty}$.

By looking at 1, we can see that the centrifugal barrier term (the last term in the square brackets) disappears for large ${r}$, so the behaviour is determined by the nature of the potential ${V}$. We might think that, provided ${V\underset{r\rightarrow\infty}{\longrightarrow}0}$, we can just ignore the potential and solve the reduced equation

$\displaystyle \frac{d^{2}U_{E}}{dr^{2}}=-\frac{2\mu E}{\hbar^{2}}U_{E} \ \ \ \ \ (3)$

where we’ve dropped the subscript ${l}$ since we’re ignoring the centrifugal barrier, which is the only term in which ${l}$ appears. In fact, this assumption proves to be faulty, in that the analysis is valid only if ${V\rightarrow\frac{1}{r^{a}}}$ where ${a>1}$, or in other words, if ${rV\left(r\right)\rightarrow0}$. To see why, we need to consider two cases: ${E>0}$ (so that the particle can escape to infinity, since we’re assuming ${V\le0}$ everywhere, and thus that ${E}$ can take on any positive value); ${E<0}$, so that the particle is bound, and there are discrete energy levels. Shakar treats the ${E>0}$ case so we’ll look at the ${E<0}$ case. In this case, 3 has the general solution

$\displaystyle U_{E}=Ae^{-\kappa r}+Be^{\kappa r} \ \ \ \ \ (4)$

where

$\displaystyle \kappa=\sqrt{-\frac{2\mu E}{\hbar^{2}}}=\sqrt{\frac{2\mu\left|E\right|}{\hbar^{2}}} \ \ \ \ \ (5)$

In the most general case, the constants ${A}$ and ${B}$ can be anything, subject to the usual constraint that the overall wave function is normalized. However, in order for this normalization to occur, we can’t have the ${e^{\kappa r}}$ term, since that terms blows up as ${r\rightarrow\infty}$. As we’ve seen in the specific example of the hydrogen atom, when we express the radial function as a series in powers of ${r}$, the series must terminate after a finite number of terms in order to keep the wave function finite, and it is this that results in the quantized energy levels. Although a direct link between the series solution and the form 4 isn’t obvious, the net effect is that, when the energy has one of the allowed discrete values, the term ${Be^{\kappa r}}$ disappears from the asymptotic solution.

The form 4 is valid only under the restriction that ${rV\left(r\right)\rightarrow0}$ for large ${r}$. To see why, suppose we write

$\displaystyle U_{E}=f\left(r\right)e^{\pm\kappa r} \ \ \ \ \ (6)$

for some function ${f}$. If 4 is valid, then ${f}$ should tend to a constant for large ${r}$. We can plug 6 into 1 and ignore the centrifugal term since we’re looking only at large ${r}$. This gives

$\displaystyle \frac{d^{2}U_{E}}{dr^{2}}-\frac{2\mu}{\hbar^{2}}VU_{E}-\kappa^{2}U_{E}=0 \ \ \ \ \ (7)$

Calculating the derivative, we have

 $\displaystyle \frac{dU_{E}}{dr}$ $\displaystyle =$ $\displaystyle \left(f^{\prime}\pm\kappa f\right)e^{\pm\kappa r}\ \ \ \ \ (8)$ $\displaystyle \frac{d^{2}U_{E}}{dr^{2}}$ $\displaystyle =$ $\displaystyle \left(f^{\prime\prime}\pm\kappa f^{\prime}\pm\kappa f^{\prime}+\kappa^{2}f\right)e^{\pm\kappa r}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left(f^{\prime\prime}\pm2\kappa f^{\prime}+\kappa^{2}f\right)e^{\pm\kappa r} \ \ \ \ \ (10)$

Plugging this into 7 we get

$\displaystyle f^{\prime\prime}\pm2\kappa f^{\prime}-\frac{2\mu}{\hbar^{2}}Vf=0 \ \ \ \ \ (11)$

At this point, Shankar assumes that ${f}$ is slowly varying for large ${r}$, which seems reasonable, so we can disregard the second derivative, to get

$\displaystyle f^{\prime}=\mp\frac{\mu}{\kappa\hbar^{2}}Vf \ \ \ \ \ (12)$

or

$\displaystyle \frac{df}{f}=\mp\frac{\mu}{\kappa\hbar^{2}}V\left(r\right)\;dr \ \ \ \ \ (13)$

If we integrate this from some constant lower value ${r_{0}}$ up to an arbitrary large value ${r}$, we have

$\displaystyle f\left(r\right)=f\left(r_{0}\right)\exp\left[\mp\frac{\mu}{\kappa\hbar^{2}}\int_{r_{0}}^{r}V\left(r^{\prime}\right)dr^{\prime}\right] \ \ \ \ \ (14)$

The point now is that if ${V\left(r\right)\rightarrow\frac{1}{r^{a}}}$ with ${a>1}$, then the integral of ${V}$ will be an inverse power of ${r}$, and thus will go to zero as ${r\rightarrow\infty}$. In that case, the RHS of 14 does indeed tend to a constant as ${r\rightarrow\infty}$, and the asymptotic solution 4 is valid. However, if ${V=-\frac{e^{2}}{r}}$ (the Coulomb potential, as found in the hydrogen atom), then the integral of ${V}$ is a logarithm and does not tend to zero for large ${r}$. In this case, we get

 $\displaystyle f\left(r\right)$ $\displaystyle =$ $\displaystyle f\left(r_{0}\right)\exp\left[\pm\frac{\mu e^{2}}{\kappa\hbar^{2}}\ln\frac{r}{r_{0}}\right]\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \left[r_{0}^{\mp\mu e^{2}/\kappa\hbar^{2}}f\left(r_{0}\right)\right]r^{\pm\mu e^{2}/\kappa\hbar^{2}} \ \ \ \ \ (16)$

The quantity in square brackets is a constant, but the last factor is a power of ${r}$ which, for the positive exponent, continues to grow as ${r\rightarrow\infty}$. Thus the asymptotic solution 4 is valid only for potentials that fall off faster than ${\frac{1}{r}}$ for large ${r}$.

# Radial function for small r

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Section 12.6.

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

In solving the Schrödinger equation for spherically symmetric potentials, we found that we could reduce the problem to the equation

$\displaystyle \left[-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}}\right]U_{El}=EU_{El} \ \ \ \ \ (1)$

where ${U_{El}\left(r\right)}$ is related to the radial function by

$\displaystyle R_{El}\left(r\right)=\frac{U_{El}\left(r\right)}{r} \ \ \ \ \ (2)$

We’ve looked at some properties of ${U_{El}}$ (which Griffiths calls ${u}$) for the hydrogen atom, but we can also try to extract some information about ${U_{El}}$ in the more general case where we don’t need to specify the potential ${V}$ precisely. Here we’ll examine what happens as ${r\rightarrow0}$.

The quantity in the square brackets in 1 is an operator which will call ${D_{l}\left(r\right)}$:

$\displaystyle D_{l}\left(r\right)\equiv-\frac{\hbar^{2}}{2\mu}\frac{d^{2}}{dr^{2}}+V\left(r\right)+\frac{l\left(l+1\right)\hbar^{2}}{2\mu r^{2}} \ \ \ \ \ (3)$

If we require ${D_{L}}$ to be hermitian, this results in the condition that, for two functions ${U_{1}}$ and ${U_{2}}$,

$\displaystyle \left.U_{1}^*\frac{dU_{2}}{dr}\right|_{0}^{\infty}-\left.U_{2}\frac{dU_{1}^*}{dr}\right|_{0}^{\infty}=0 \ \ \ \ \ (4)$

If we require ${U_{El}}$ to be normalizable then it must satisfy either

$\displaystyle U_{El}\underset{r\rightarrow\infty}{\longrightarrow}0 \ \ \ \ \ (5)$

which is valid for bound states where ${E>V}$ as ${r\rightarrow\infty}$, or

$\displaystyle U_{El}\underset{r\rightarrow\infty}{\longrightarrow}e^{ikr} \ \ \ \ \ (6)$

where

$\displaystyle k=\sqrt{\frac{2\mu E}{\hbar^{2}}} \ \ \ \ \ (7)$

if ${E>V}$ as ${r\rightarrow\infty}$. In the latter case, we’re using the definition of normalization for an oscillating function. If ${U_{El}\underset{r\rightarrow\infty}{\longrightarrow}0}$ then 4 is 0 at the upper limit. Using the normalization condition for oscillating functions, if ${U_{El}\underset{r\rightarrow\infty}{\longrightarrow}e^{ikr}}$ then 4 is also zero (on average) at the upper limit. Thus in order for ${D_{l}}$ to be Hermitian, we must have

$\displaystyle \left.U_{1}^*\frac{dU_{2}}{dr}\right|_{0}-\left.U_{2}\frac{dU_{1}^*}{dr}\right|_{0}=0 \ \ \ \ \ (8)$

at the lower limit as well.

One way of satisfying this condition is if

$\displaystyle U_{El}\underset{r\rightarrow0}{\longrightarrow}c \ \ \ \ \ (9)$

Because the actual radial function is given by 2, a value of ${c\ne0}$ would give

$\displaystyle R\sim\frac{U}{r}\sim\frac{c}{r} \ \ \ \ \ (10)$

Such a function is still square integrable because an integral over all space introduces a factor of ${r^{2}}$ in the volume element:

$\displaystyle \int R^{2}r^{2}\sin\theta dr\;d\theta\;d\phi \ \ \ \ \ (11)$

Thus the integrand is still finite at ${r=0}$ so the integral itself can be finite.

The problem with ${c\ne0}$ is that the Laplacian of ${\frac{1}{r}}$ gives a delta function:

$\displaystyle \nabla^{2}\frac{1}{r}=-4\pi\delta^{3}\left(\mathbf{r}\right) \ \ \ \ \ (12)$

Unless the potential ${V}$ has a delta function at the origin (which would be quite unusual), the ${\nabla^{2}}$ in the Schrödinger equation can’t be allowed to generate a delta function there, so we must have ${c=0}$.

So far, everything is true for any potential. If we now assume that ${V}$ is less singular than ${\frac{1}{r^{2}}}$ (that is, ${V\underset{r\rightarrow0}{\longrightarrow}\frac{1}{r^{a}}}$ where ${a<2}$), the centrifugal barrier term in 1 will dominate as ${r\rightarrow0}$, so for small ${r}$, 1 reduces to the differential equation

$\displaystyle U_{l}^{\prime\prime}\simeq\frac{l\left(l+1\right)}{r^{2}}U_{l} \ \ \ \ \ (13)$

The ${E}$ in the suffix of ${U}$ has been dropped because the term involving ${E}$ in 1 is negligible compared to the centrifugal barrier for ${r\rightarrow0}$. This equation has solutions

$\displaystyle U_{l}\sim r^{\alpha} \ \ \ \ \ (14)$

Plugging this into 13 we get

$\displaystyle \alpha\left(\alpha-1\right)=l\left(l+1\right) \ \ \ \ \ (15)$

This is a quadratic equation in ${\alpha}$ which has the two solutions

$\displaystyle \alpha=-l,l+1 \ \ \ \ \ (16)$

If we are to have ${U_{El}\rightarrow0}$ as ${r\rightarrow0}$, then we must discard the solution ${U_{l}\sim r^{-l}}$, so we have that

$\displaystyle U_{El}\sim r^{l+1} \ \ \ \ \ (17)$

All of this works only if ${l\ne0}$ since in the case where ${l=0}$ (zero angular momentum), there is no centrifugal barrier and we must look at the form of the potential. Shankar notes that the problems he considers in his book are such that 17 is also valid for ${l=0}$.

# Hydrogen atom – a sample wave function

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 13, Exercise 13.1.3.

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

The wave function for the hydrogen atom can be obtained by a series solution of the differential equation, leading to the result (which I’ve rewritten in Shankar’s notation, although my original post used Griffiths’s notation):

$\displaystyle \psi_{nlm}\left(r,\theta,\phi\right)=\frac{U_{El}\left(r\right)}{r}Y_{l}^{m}\left(\theta,\phi\right) \ \ \ \ \ (1)$

Here, we have

 $\displaystyle U_{El}$ $\displaystyle =$ $\displaystyle e^{-\rho}v_{El}\ \ \ \ \ (2)$ $\displaystyle v_{El}$ $\displaystyle =$ $\displaystyle \rho^{l+1}\sum_{k=0}^{\infty}C_{k}\rho^{k}\ \ \ \ \ (3)$ $\displaystyle \rho$ $\displaystyle =$ $\displaystyle \sqrt{\frac{-2mE}{\hbar^{2}}}r \ \ \ \ \ (4)$

The energy levels of the hydrogen atom are

$\displaystyle E=-\frac{me^{4}}{2\hbar^{2}n^{2}} \ \ \ \ \ (5)$

where ${n=1,2,3,\ldots}$. The coefficients ${C_{k}}$ in 3 are given by a recursion relation

 $\displaystyle C_{k+1}$ $\displaystyle =$ $\displaystyle \frac{-e^{2}\lambda+2\left(k+l+1\right)}{\left(k+l+2\right)\left(k+l+1\right)-l\left(l+1\right)}C_{k}\ \ \ \ \ (6)$ $\displaystyle \lambda$ $\displaystyle =$ $\displaystyle \sqrt{-\frac{2m}{\hbar^{2}E}} \ \ \ \ \ (7)$

Combining ${\lambda}$ and ${E}$, the formula becomes, for a given ${n}$

$\displaystyle C_{k+1}=\frac{2\left(k+l+1\right)-2n}{\left(k+l+2\right)\left(k+l+1\right)-l\left(l+1\right)}C_{k}$

The coefficient ${C_{0}}$ which starts everything off is determined by normalization.

As an example, we can find the wave function ${\psi_{210}}$. In this case ${n=2}$ and ${l=1}$ so the first term in the recursion, with ${k=0}$ gives ${k+l+1=2}$ and ${C_{1}=0}$. The full wave function is then

 $\displaystyle \psi_{210}$ $\displaystyle =$ $\displaystyle \frac{1}{r}\rho^{2}e^{-\rho}C_{0}Y_{1}^{0} \ \ \ \ \ (8)$

To evaluate ${\rho}$ we use the energy for ${n=2}$:

$\displaystyle E_{2}=-\frac{me^{4}}{8\hbar^{2}} \ \ \ \ \ (9)$

This gives

$\displaystyle \rho=\sqrt{\frac{2m^{2}e^{4}}{8\hbar^{4}}}r=\frac{me^{2}}{2\hbar^{2}}r=\frac{r}{2a_{0}} \ \ \ \ \ (10)$

where ${a_{0}}$ is the Bohr radius

$\displaystyle a_{0}\equiv\frac{\hbar^{2}}{me^{2}} \ \ \ \ \ (11)$

Plugging everything into 8, using ${Y_{1}^{0}=\sqrt{\frac{3}{4\pi}}\cos\theta}$, we have

$\displaystyle \psi_{210}=\sqrt{\frac{3}{4\pi}}\frac{C_{0}}{4a_{0}^{2}}re^{-r/2a_{0}}\cos\theta \ \ \ \ \ (12)$

Normalizing gives the condition

$\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}\psi_{210}^{2}r^{2}\sin\theta dr\;d\theta\;d\phi=1 \ \ \ \ \ (13)$

Working out the integral (using software or tables) gives

 $\displaystyle \frac{3}{2}a_{0}C_{0}^{2}$ $\displaystyle =$ $\displaystyle 1\ \ \ \ \ (14)$ $\displaystyle C_{0}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2}{3a_{0}}} \ \ \ \ \ (15)$

So the final wave function is

$\displaystyle \psi_{210}=\frac{1}{\sqrt{32\pi a_{0}^{3}}}\frac{r}{a_{0}}e^{-r/2a_{0}}\cos\theta \ \ \ \ \ (16)$

which agrees with Shankar’s equation 13.1.27.

# Isotropic harmonic oscillator in 3-d – use of spherical harmonics

Shankar, R. (1994), Principles of Quantum Mechanics, Plenum Press. Chapter 12, Exercise 12.6.11.

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

We’ve solved the 3-d isotropic harmonic oscillator before, so we’ve already solved most of Shankar’s exercise 12.6.11. We can quote the results here. The solution has the form

$\displaystyle \psi_{Elm}=\frac{U_{El}\left(r\right)}{r}Y_{l}^{m}\left(\theta,\phi\right) \ \ \ \ \ (1)$

The earlier solution uses notation from Griffiths’s book, but as the end result is the same, it’s not worth going through the derivation again using Shankar’s notation.

The potential is

$\displaystyle V(r)=\frac{1}{2}m\omega^{2}r^{2} \ \ \ \ \ (2)$

The radial equation to be solved is

$\displaystyle \frac{d^{2}u}{d\rho^{2}}=\left(-1+\frac{l(l+1)}{\rho^{2}}+\rho_{0}^{2}\rho^{2}\right)u \ \ \ \ \ (3)$

If we define

 $\displaystyle \kappa^{2}$ $\displaystyle \equiv$ $\displaystyle \frac{2\mu E}{\hbar^{2}}\ \ \ \ \ (4)$ $\displaystyle \rho$ $\displaystyle \equiv$ $\displaystyle \kappa r\ \ \ \ \ (5)$ $\displaystyle \rho_{0}$ $\displaystyle \equiv$ $\displaystyle \frac{\mu\omega}{\hbar\kappa^{2}}=\frac{\hbar\omega}{2E} \ \ \ \ \ (6)$

Taking the asymptotic behaviour of the radial function for small and large ${r}$ into account leads us to a solution of form

$\displaystyle u(\rho)=\rho^{l+1}e^{-\rho_{0}\rho^{2}/2}v(\rho) \ \ \ \ \ (7)$

Note that Griffiths’s ${v}$ is not the same as Shankar’s ${v}$, the latter of which is defined by Shankar’s equation 12.6.49.

This gives a differential equation for Griffiths’s ${v}$

$\displaystyle \rho\frac{d^{2}v}{d\rho^{2}}+2\left(l+1-\rho_{0}\rho^{2}\right)\frac{dv}{d\rho}+\rho(1-\rho_{0}(2l+3))v=0 \ \ \ \ \ (8)$

The function ${v}$ can be solved as a power series, giving

$\displaystyle v(\rho)=\sum c_{j}\rho^{j} \ \ \ \ \ (9)$

Substituting into 8 leads to the recursion relation

$\displaystyle c_{q+2}=\frac{\rho_{0}(2q+2l+3)-1}{(q+2)(q+2l+3)}c_{q} \ \ \ \ \ (10)$

with ${c_{1}=0}$, so that ${c_{q}=0}$ for all odd ${q}$. The requirement that the series terminates at some finite value of ${j}$ leads to the quantization condition on ${E}$:

 $\displaystyle E$ $\displaystyle =$ $\displaystyle \hbar\omega\left(q_{max}+l+\frac{3}{2}\right) \ \ \ \ \ (11)$

or, defining ${n=q_{max}+l}$,

$\displaystyle E_{n}=\hbar\omega\left(n+\frac{3}{2}\right) \ \ \ \ \ (12)$

We worked out the degeneracies in the earlier post as well, so that the degeneracy of ${E_{n}}$ is

$\displaystyle d\left(n\right)=\frac{1}{2}(n+1)(n+2) \ \ \ \ \ (13)$

To complete Shankar’s exercise, we need to work out the eigenfunctions for ${n=0}$ and ${n=1}$. For ${n=0}$, ${q_{max}=l=0}$, so only ${c_{0}\ne0}$ and we have

 $\displaystyle v\left(\rho\right)$ $\displaystyle =$ $\displaystyle c_{0}\ \ \ \ \ (14)$ $\displaystyle u\left(\rho\right)$ $\displaystyle =$ $\displaystyle c_{0}\rho e^{-\rho_{0}\rho^{2}/2}\ \ \ \ \ (15)$ $\displaystyle \psi_{000}$ $\displaystyle =$ $\displaystyle \frac{u}{r}Y_{0}^{0}\ \ \ \ \ (16)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}\kappa e^{-\rho_{0}\rho^{2}/2}Y_{0}^{0}\ \ \ \ \ (17)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}\sqrt{\frac{2\mu3\omega}{4\pi\hbar}}e^{-\mu\omega r^{2}/2\hbar} \ \ \ \ \ (18)$

where in the fourth line we used

 $\displaystyle \kappa$ $\displaystyle =$ $\displaystyle \frac{\sqrt{2\mu E}}{\hbar}=\frac{\sqrt{2\mu\frac{3}{2}\hbar\omega}}{\hbar}=\sqrt{\frac{3\mu\omega}{\hbar}}\ \ \ \ \ (19)$ $\displaystyle Y_{0}^{0}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{4\pi}} \ \ \ \ \ (20)$

Normalizing this requires that

 $\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}\psi_{000}^{2}r^{2}\sin\theta dr\;d\theta\;d\phi$ $\displaystyle =$ $\displaystyle c_{0}^{2}\frac{6\mu\omega}{\hbar}\int_{0}^{\infty}e^{-\mu\omega r^{2}/\hbar}r^{2}dr\ \ \ \ \ (21)$ $\displaystyle$ $\displaystyle =$ $\displaystyle 1 \ \ \ \ \ (22)$

This is a standard Gaussian integral and can be done using software or tables so we get

$\displaystyle c_{0}=\frac{\sqrt{6}}{3}\left(\frac{\mu\omega}{\pi\hbar}\right)^{1/4} \ \ \ \ \ (23)$

This gives a wave function of

$\displaystyle \psi_{000}=\left(\frac{\mu\omega}{\pi\hbar}\right)^{3/4}e^{-\mu\omega r^{2}/2\hbar} \ \ \ \ \ (24)$

which agrees with the earlier result.

For ${n=1}$, the degeneracy is, from 13

$\displaystyle d\left(1\right)=3 \ \ \ \ \ (25)$

The three possibilities are ${m=0,\pm1}$ which are reflected in the three spherical harmonics ${Y_{1}^{0,\pm1}}$. The radial function is the same in all cases, and is obtained from ${q_{max}=0}$, ${l=1}$. From 7, this gives

 $\displaystyle v\left(\rho\right)$ $\displaystyle =$ $\displaystyle c_{0}\ \ \ \ \ (26)$ $\displaystyle u\left(\rho\right)$ $\displaystyle =$ $\displaystyle c_{0}\rho^{2}e^{-\rho_{0}\rho^{2}/2}\ \ \ \ \ (27)$ $\displaystyle \psi_{11m}$ $\displaystyle =$ $\displaystyle \frac{u}{r}Y_{1}^{m}\ \ \ \ \ (28)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}\kappa^{2}re^{-\rho_{0}\rho^{2}/2}Y_{1}^{m}\ \ \ \ \ (29)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}\frac{5\mu\omega}{\hbar}re^{-\mu\omega r^{2}/2\hbar}Y_{1}^{m} \ \ \ \ \ (30)$

Again, we work out ${c_{0}}$ by imposing normalization. For example

 $\displaystyle \psi_{111}$ $\displaystyle =$ $\displaystyle c_{0}\frac{5\mu\omega}{\hbar}re^{-\mu\omega r^{2}/2\hbar}Y_{1}^{1}\ \ \ \ \ (31)$ $\displaystyle$ $\displaystyle =$ $\displaystyle -c_{0}\frac{5\mu\omega}{\hbar}re^{-\mu\omega r^{2}/2\hbar}\sqrt{\frac{3}{8\pi}}\sin\theta e^{i\phi} \ \ \ \ \ (32)$

The normalization integral is

 $\displaystyle \int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{\infty}\psi_{111}^{2}r^{2}\sin\theta dr\;d\theta\;d\phi$ $\displaystyle =$ $\displaystyle c_{0}^{2}\left(\frac{5\mu\omega}{\hbar}\right)^{2}\frac{3}{8\pi}2\pi\int_{0}^{\pi}\int_{0}^{\infty}e^{-\mu\omega r^{2}/\hbar}r^{4}\sin^{3}\theta dr\;d\theta\ \ \ \ \ (33)$ $\displaystyle$ $\displaystyle =$ $\displaystyle c_{0}^{2}\frac{75}{8}\sqrt{\frac{\pi\hbar}{\mu\omega}}=1\ \ \ \ \ (34)$ $\displaystyle c_{0}$ $\displaystyle =$ $\displaystyle \frac{2\sqrt{6}}{15}\left(\frac{\mu\omega}{\pi\hbar}\right)^{1/4} \ \ \ \ \ (35)$

I used Maple to do the integrals. This gives a wave function of

$\displaystyle \psi_{111}=-\sqrt{\frac{\mu\omega}{\hbar}}\left(\frac{\mu\omega}{\pi\hbar}\right)^{3/4}re^{-\mu\omega r^{2}/2\hbar}\sin\theta e^{i\phi} \ \ \ \ \ (36)$

We can work out the other two wave functions the same way (I used Maple, so I won’t go into the details):

 $\displaystyle \psi_{11-1}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{\mu\omega}{\hbar}}\left(\frac{\mu\omega}{\pi\hbar}\right)^{3/4}re^{-\mu\omega r^{2}/2\hbar}\sin\theta e^{-i\phi}\ \ \ \ \ (37)$ $\displaystyle \psi_{110}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2\mu\omega}{\hbar}}\left(\frac{\mu\omega}{\pi\hbar}\right)^{3/4}re^{-\mu\omega r^{2}/2\hbar}\cos\theta \ \ \ \ \ (38)$

The ${\psi_{110}}$ here is the same as ${\psi_{001}}$ in our rectangular solution set. The other two are linear combinations of ${\psi_{100}}$ and ${\psi_{010}}$ from our rectangular set, which were (the suffixes in these 2 equations stand for ${x}$, ${y}$ and ${z}$, and not ${n}$, ${l}$ and ${m}$):

 $\displaystyle \psi_{100}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}r\sin\theta\cos\phi\ \ \ \ \ (39)$ $\displaystyle \psi_{010}$ $\displaystyle =$ $\displaystyle \sqrt{\frac{2m\omega}{\hbar}}\left(\frac{m\omega}{\pi\hbar}\right)^{3/4}e^{-m\omega r^{2}/2\hbar}r\sin\theta\sin\phi \ \ \ \ \ (40)$

We have

 $\displaystyle \psi_{111}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\psi_{100}+i\psi_{010}\right)\ \ \ \ \ (41)$ $\displaystyle \psi_{11-1}$ $\displaystyle =$ $\displaystyle \frac{1}{\sqrt{2}}\left(\psi_{100}-i\psi_{010}\right) \ \ \ \ \ (42)$