Homework solutions#
Homework 1 solutions (TA: Renée Goodman)
Homework 2 solutions (TA: Ivan Martinez)
Homework 3 solutions (TA: Clément Fortin)
Homework 4 solutions (TA: Renée Goodman)
Homework 5 solutions (TA: Ivan Martinez)
Homework 6 solutions (TA: Clément Fortin)
Homework 7 solutions (TA: Renée Goodman)
Homework 8 solutions (TA: Ivan Martinez)
Homework 9#
Question 1: the singlet and triplet states as spin eigenstates
The \(S_z\) operator is straightforward because we can write \(\hat{S}_z = \hat{S}_{1,z} + \hat{S}_{2,z}\). Applying this to the triplet states you will find that they are eigenstates with eigenvalues \(m_s\hbar=(1,0,-1)\times \hbar\), while the singlet state has eigenvalue 0. For \(\hat{S}^2 = \hat{S}_1^2 + \hat{S}_2^2 + 2\hat{\mathbf{S}}_1\cdot \hat{\mathbf{S}}_2\), we can use Townsend equation (5.10) which gives
Then
as expected for total spin \(s=1\). The state \(\ket{\downarrow\downarrow}\) gives the same answer.
For the \(m_s=0\) states, we need
and the same for \(\ket{\downarrow\uparrow}\). This means that the \(m_s=0\) triplet state (symmetric) has \(s=1\) whereas the \(m_s=0\) singlet state (antisymmetric) has \(s=0\).
Question 2: variational principle estimate of the ground state energy
Townsend 12.7. We have a potential \(V(x) = a|x|\) and want to estimate the ground state energy with the trial wavefunction \(\psi(x) = (b/\pi)^{1/4} e^{-bx^2/2}\). The contribution of the potential term in the Hamiltonian is
The kinetic energy term is
The estimate for the ground state energy is therefore
Minimize the right hand side with respect to \(b\):
Substituting in this value of \(b\) gives
(Note that this has the correct units since according to the definition of \(V\), \(a\) is an energy per unit length).
Question 3: spin-spin interaction between electrons
Townsend 12.3. Estimate the singlet-triplet splitting of the two electrons in helium due to a direct spin-spin interaction.
The magnetic moment of the electron due to its spin is \(\mu= (e/m_e)\times \hat{S}\) (see spin-orbit coupling in lecture 17). Since the magnetic moment is \(\propto 1/m\) it should be 2000 times larger for an electron compared to a proton. So we can take the hyperfine splitting discussed in section 5.2 of Townsend and multiply by 2000.
We can get the size of the hyperfine splitting from the energy of a 21cm photon (this is the photon involved in the hyperfine transition), \(hc/\lambda \approx 6\mu\mathrm{eV}\). Multiplying by 2000 gives an estimate for the energy of the electron-electron interaction of \(\boxed{\approx 10^{-3}\ \mathrm{eV}}\).
We could also estimate the energy from first principles by writing down the energy associated with the magnetic dipole of one of the electrons in the magnetic field of the other, \(\mu B\). The magnetic field of a dipole is \(B\sim \mu_0 \mu / 4\pi r^3\), and we could take a typical distance between the electrons of \(r\sim a_0\) (the Bohr radius). The energy is then
Then simplify using \(a_0 = \alpha^{-1}(\hbar/m_ec)\) to get
the same order of magnitude as the previous estimate.
Homework 10#
Townsend 13.3: Validity of the Born approximation
Start with Townsend (13.48) which is the condition that the Born approximation is valid:
(ie. the wavefunction near the origin is dominated by the incident plane wave, so our approximation for solving the integral equation (5) is a good one).
Now assume the potential has some range \(a\), so we only have to consider values of \(r\) in the in the integral up to \(r\approx a\). At low energy where \(ka\ll 1\), we can approximate \(e^{ikr}\approx 1\) and \(\sin kr\approx kr\), giving
Then treating \(V(r)\approx V_0\approx \) constant for \(r<a\), we can do the integral to get
So we have a condition for the Born approximation to be valid at low energies \(ka\ll 1\)
The condition for the Born approximation to be valid at high energies \(ka\gg 1\) is (Townsend 13.49),
so we see that if the low energy condition is already satisfied then at high energies we’re dividing by the large number \((ka)^2\) and so the high energy condition will automatically be satisfied as well.
Note the physical interpretations of these criteria: the high energy condition is that the particle energy should be much larger than the magnitude of the scattering potential \(V_0\); the low energy condition is that the potential should be small enough that there are no bound states.
Townsend 13.7: Born approximation example
Start with the scattering amplitude for a spherically-symmetric potential,
With \(V(r) = V_0 e^{-r/a}\), this becomes
Therefore
or in terms of the scattering angle
Notice that this has the correct units which we can see from the \(a^2\) factor. The first term on the right hand side is dimensionless.
Townsend 13.10: p-wave scattering from a hard sphere
In the region \(r>a\) where \(V=0\), the general solution for \(\ell=1\) is
where \(\hbar^2k^2/2\mu=E\) (e.g. see section 10.4 of Townsend). We include the Neumann function here because we exclude the origin so there’s no problem with it blowing up. The boundary condition at \(r=a\), \(R(a)=0\) can be used to determine one of the coefficients:
so therefore
At large distance \(kr\rightarrow\infty\), this becomes
(using Townsend 13.63).
We are looking for the asymptotic form
where \(\delta_1\) is the phase shift. For \(\delta_1\ll 1\), this is
where we assume \(\cos\delta_1\approx 1\) and \(\sin\delta_1\approx \delta_1\).
Therefore we identify
where in the last step we expand for \(ka\ll 1\) using Townsend (10.69) to get the forms of \(j_1\) and \(\eta_1\):
Townsend 13.11: comparing Born approximation and partial wave expansion
The potential we are considering is the finite well \(V=-V_0\) for \(r<a\) and \(V=0\) for \(r>a\). The s-wave cross-section is given by equation (13.93) of Townsend:
where \((k_0a)^2 = (2\mu a^2/\hbar^2)(V_0+E)\) (Townsend 13.88a). We’re going to be doing a comparison at low energy, so let’s take \(E\ll V_0\), and let’s also take \(k_0a\ll 1\) which is the condition from Townsend problem 13.3 for the validity of the Born approximation:
Now we can compute the Born approximation cross-section and compare the answers. Use equation (8) for the scattering amplitude given a spherically-symmetric potential. This gives
To compute the angular integral exactly, we would need to express \(q\) in terms of the scattering angle \(\theta\). But we are interested here in low energy, so let’s first expand the integrand for \(qa\ll 1\):
which has no \(q\)-dependence. As we would expect at low energy the cross-section is isotropic (\(\ell=0\) dominates).
This gives
Both approaches indeed give the same expression for the cross-section in this limit \(k_0a\ll 1\) where the Born approximation should be valid.
Chapter 14 practise problems#
Townsend 14.10: exciting a charged particle in a 1D harmonic oscillator
In this question, an oscillating electric field is applied to a charged particle in a 1D harmonic oscillator. The perturbation to the Hamiltonian is
We start off in the ground state and we need to use time-dependent perturbation theory to calculate the probability that the particle is found in state \(\ket{n}\) after a time \(t\). Apply equation (9)
and therefore
By now you will have seen this kind of matrix element before: use the ladder operators to evaluate it \(\hat{x} = (\hbar/2 m\omega_0)^{1/2}(\hat{a}+\hat{a}^\dagger)\) which implies that the only possible \(\ket{n}\) is the first excited state \(\ket{1}\):
Using Mathematica to evaluate the time integral (setting \(n=1\) since we know now that \(n=1\) is the only option), I get
Therefore
(If you check you’ll see this does have the correct dimensions).
We can evaluate it for the case \(\Delta\omega = \omega-\omega_0\rightarrow 0\), just have to be careful with the limit. The numerator is
and the denominator is
Therefore
Townsend 14.11: a time-dependent electric field applied to the hydrogen atom
Aligning our \(z\)-axis with the applied electric field, the perturbation to the Hamiltonian is
We start off in the ground state and we need to use time-dependent perturbation theory to calculate the probability that the particle will be found in a 2p state as \(t\rightarrow\infty\). Apply equation (9):
where \(\omega_{21} = |E_0-E_1|/\hbar\) is the photon frequency involved in the transition. The matrix element is
(using the integrals given in lecture 25) (we need \(m=0\) on the left-hand side to get a non-vanishing matrix element).
Therefore
Townsend 14.14: radiative transition in the harmonic oscillator
We’re going from the initial state \(\ket{i} = \ket{n=1}\ket{0}\) (first excited state of the oscillator, no photon) to final state \(\ket{i} = \ket{n=0}\ket{1}\) (ground state of the oscillator, one photon). The calculation is similar to the decay of the hydrogen atom from 2p to 1s (lecture 25), but with different spatial integrals. The perturbing Hamiltonian is
The density of states we need for the outgoing photon is
Then applying Fermi’s Golden rule gives
where similarly to lecture 25 we’ve assumed \(e^{ikr}\approx 1}\) (photon wavelength \(\gg\) size of the oscillator).
Simplifying the prefactor gives
Notice that this has the correct units (\(\mathrm{s}^{-1}\)).
We actually have three different choices for the \(\ket{n=1}\) state. In Cartesian coordinates, this corresponds to one of the three choices \(\ket{n_x,n_y,n_z}=\ket{1,0,0}\), \(\ket{0,1,0}\), or \(\ket{0,0,1}\). Or in spherical coordinates we have the three states \(\ket{\ell, m} = \ket{1,1}\), \(\ket{1,0}\) or \(\ket{1,-1}\). As with the hydrogen atom, the different states couple to different components of the polarization vector \(\epsilon\) of the outgoing photon.
The simplest way to evaluate the matrix element is to use Cartesian coordinates, i.e. write
Then we can write the momentum operator in terms of the ladder operators. For example, if we pick the starting state in Cartesian coordinates \(\ket{1,0,0}\) (so \(n_x=1\)), then the only operator that couples this to the ground state \(\ket{0,0,0}\) will involve the ladder operator in the \(x\)-direction, or the \(\hat{p}_x\) term:
(using Townsend eq 7.13). The rate of decay for this state is then
We would get a similar answer for the other two states \(n_y=1\) or \(n_z=1\) but with the \(\epsilon_x\) replaced by the other components of the polarization vector, \(\epsilon_y\) or \(\epsilon_z\) respectively. If we add up these three rates and divide by 3 we can get the average decay rate assuming there is no preferred initial state:
We used the fact that the polarization vector is a unit vector (has unit length). This doesn’t depend on the outgoing photon direction, so we can do the integral \(\int d\Omega\rightarrow 4\pi\), and also multiply by 2 because each outgoing direction has two photon polarizations, to get the total decay rate
The other way to do this would be to use spherical coordinates instead and look at the \(\ket{\ell,m}\) states. Then you need to convert the \(\hat{p}\) in the matrix element to \(\hat{r}\) (lecture 25), expand \(x\), \(y\) and \(z\) in terms of \(Y_{\ell,m}\)’s and do the angular and radial integrals. This is done in section 14.7 of Townsend (equations 14.157 to 14.166). In spherical coordinates, the decay of the \(m=1\), \(m=0\) or \(m=-1\) states involves the polarization components \(\epsilon_x+i\epsilon_y\), \(\epsilon_z\) and \(\epsilon_x-i\epsilon_y\) respectively, i.e. if you start in a state with \(m\neq 0\), then the non-zero angular momentum in the initial state is carried away by a circularly-polarized photon. In the Cartesian treatment, we just have motion in one Cartesian direction so the states connect to linear-polarization in that direction.
Townsend 14.16: density of states for an electron
Start with the phase space density in momentum space and transform into energy:
and with \(E = p^2/2m\), \(dE = p dp/m\) we get the answer
Including spin would give an extra factor of 2 for the two spin states.
Townsend 14.17: Photoelectric effect
In the photoelectric effect, we go from the electron being in the ground state of hydrogen with an incoming photon, write this initial state as \(\ket{i} = \ket{1s}\ket{1}\), to the electron in an outgoing plane wave state and no photon, final state \(\ket{f}=\ket{\mathbf{k}}\ket{0}\). The perturbing Hamiltonian that couples the electron and photon states is (lecture 25)
(assume the proton is infinitely heavy and use \(\mu=m_e\)).
The matrix element to plug into Fermi’s Golden Rule is therefore
where \(\epsilon\) is the photon polarization direction. The momentum operator is easiest if we operate to the left and convert it to the electron momentum in the final state \(\mathbf{p}_e\):
where \(k_f=p_e/\hbar\).
Finally, putting in \(\psi_{1s}\) gives
where \(\mathbf{q}=\mathbf{k}-\mathbf{k}_f\) is the momentum transfer.
We can use the same trick as in the Born approximation (lecture 22, see derivation of equation (8)) to write the integral
Now using Fermi’s Golden Rule we can assemble an expression for the rate
Now use the hint in the question to divide by \(c/V\) for the incident photon flux, to convert to a differential cross-section. Then simplifying factors and using \(\alpha = e^2/4\pi\epsilon_0\hbar c\), I find
which agrees with the answer given in the question.