Diamagnetic Shift of Excitons: Weak-Field Perturbation Theory

Hello! Welcome back to our study of excitons.

In our last lesson, we undertook the crucial task of deriving the exciton Hamiltonian in a magnetic field. We found that the center-of-mass and relative motions are coupled, but that by introducing the conserved pseudomomentum $K$ , we could separate the problem. This left us with an effective Hamiltonian for the relative motion that includes the familiar Coulomb interaction, an orbital Zeeman-like term, a diamagnetic term, and a motional Stark term.

Today, we will analyze this Hamiltonian in a physically important limit: the weak magnetic field regime. For most optical experiments, excitons are created with nearly zero center-of-mass momentum ( $K \approx 0$ ), which simplifies our starting point considerably by eliminating the motional Stark term. Our goal is to apply time-independent perturbation theory to this simplified Hamiltonian to achieve this lesson's learning outcome: to calculate the diamagnetic shift of exciton energy levels in the weak-field regime. This quadratic shift is often the first and most prominent signature of a magnetic field in magneto-spectroscopy experiments.

1. The Perturbation Problem

Let's begin with the relative Hamiltonian we derived in the previous lesson, setting the pseudomomentum $K = 0$ and assuming the magnetic field $B$ is along the z-axis ( $B = B z^$ ):

H^_{r e l} = H^_{0} (\frac{p ^ ^{2}}{2 μ} - \frac{e ^{2}}{4 π ϵr}) + H^_{Z}^{'} (\frac{m _{h}^{*} - m _{e}^{*}}{m _{h}^{*} + m _{e}^{*}}) \frac{e B}{2 μ} l^_{z} + H^_{D}^{'} \frac{e ^{2} B ^{2}}{8 μ} (x^{2} + y^{2})

We can divide this into an unperturbed part and a perturbation:

Unperturbed Hamiltonian $H^_{0}$ : This is the familiar Hamiltonian for a hydrogen-like atom with reduced mass $μ$ and a screened Coulomb potential. We have already solved this in Module 2. Its eigenfunctions are the hydrogenic wavefunctions $∣ n, l, m ⟩$ with corresponding energies $E_{n}^{(0)} = - R_{y}^{*} / n^{2}$ , where $R_{y}^{*}$ is the effective Rydberg energy.
Perturbation $H^^{'} = H^_{Z}^{'} + H^_{D}^{'}$ : This contains all the magnetic field terms. In the weak-field limit, the energy shifts induced by $H^^{'}$ are much smaller than the exciton binding energy, making perturbation theory an appropriate tool. The perturbation consists of:
- $H^_{Z}^{'}$ : A term linear in $B$ that looks like a Zeeman term. It splits states with different orbital angular momentum $m$ .
- $H^_{D}^{'}$ : A term quadratic in $B$ . This is the diamagnetic term.

2. A Quick Refresher on Perturbation Theory

Your quantum mechanics courses have equipped you well with perturbation theory. As a brief refresher on the core result we'll be using, you might find this video helpful. It rigorously derives the first-order corrections to both energy and wavefunctions.

Deriving 1st Order Perturbation Theory (Energy and Wavefunction Corrections)

This video from the channel Andrew Dotson provides a clear, step-by-step derivation of first-order non-degenerate perturbation theory.

Please watch the segment from 10:23 to 17:35. It covers the derivation of the first-order energy correction, arriving at the well-known result that it is the expectation value of the perturbing Hamiltonian in the unperturbed state.

The key result for our purposes is the first-order correction to the energy of a non-degenerate state $∣ n^{(0)} ⟩$ :

Δ E_{n}^{(1)} = ⟨ n^{(0)} ∣ H^^{'} ∣ n^{(0)} ⟩

3. Deriving the Diamagnetic Shift

Let's apply this to the exciton ground state, which is the 1s state $∣1, 0, 0 ⟩$ . This state is non-degenerate. The first-order energy correction is:

Δ E_{1 s}^{(1)} = ⟨ 1, 0, 0∣ H^_{Z}^{'} + H^_{D}^{'} ∣1, 0, 0 ⟩ = ⟨ 1, 0, 0∣ H^_{Z}^{'} ∣1, 0, 0 ⟩ + ⟨ 1, 0, 0∣ H^_{D}^{'} ∣1, 0, 0 ⟩

First, consider the Zeeman term $H^_{Z}^{'}$ . Since it is proportional to $l^_{z}$ , and the 1s state is an s-orbital with $m = 0$ , the operator $l^_{z}$ acting on it gives zero. Therefore, the first-order Zeeman correction for the 1s state is zero:

⟨ 1, 0, 0∣ H^_{Z}^{'} ∣1, 0, 0 ⟩ = 0

This means the lowest-order energy shift for the ground state comes entirely from the diamagnetic term.

Now, let's calculate the diamagnetic contribution:

Δ E_{1 s, dia} = ⟨ 1, 0, 0∣ H^_{D}^{'} ∣1, 0, 0 ⟩ = ⟨ 1, 0, 0 \frac{e ^{2} B ^{2}}{8 μ} (x^{2} + y^{2}) 1, 0, 0 ⟩

We can pull the constants out of the expectation value:

Δ E_{1 s, dia} = \frac{e ^{2} B ^{2}}{8 μ} ⟨ 1, 0, 0∣ x^{2} + y^{2} ∣1, 0, 0 ⟩

The calculation boils down to finding the expectation value of $(x^{2} + y^{2})$ for the 1s hydrogenic state. The 1s wavefunction is spherically symmetric, so the expectation values of $x^{2}$ , $y^{2}$ , and $z^{2}$ must be equal:

⟨ x^{2} ⟩ = ⟨ y^{2} ⟩ = ⟨ z^{2} ⟩ = \frac{1}{3} ⟨ r^{2} ⟩

Therefore, $⟨ x^{2} + y^{2} ⟩ = \frac{2}{3} ⟨ r^{2} ⟩_{1 s}$ .

For a hydrogenic 1s state, the expectation value of $r^{2}$ is a standard result, $⟨ r^{2} ⟩_{1 s} = 3 (a_{B}^{*})^{2}$ , where $a_{B}^{*}$ is the exciton's effective Bohr radius.

Substituting this in, we get:

⟨ x^{2} + y^{2} ⟩_{1 s} = \frac{2}{3} (3 (a_{B}^{*})^{2}) = 2 (a_{B}^{*})^{2}

Finally, we arrive at the diamagnetic shift for the 1s exciton:

Δ E_{1 s, dia} = \frac{e ^{2} B ^{2}}{8 μ} (2 (a_{B}^{*})^{2}) = \frac{e ^{2} ( a _{B}^{*} ) ^{2}}{4 μ} B^{2}

This result is often written as $Δ E_{dia} = σ B^{2}$ , where we define the diamagnetic shift coefficient $σ$ :

σ_{3 D} = \frac{e ^{2} ( a _{B}^{*} ) ^{2}}{4 μ} = \frac{e ^{2}}{4 μ} (\frac{4 π ϵ ℏ ^{2}}{μ e ^{2}})^{2} = \frac{4 π ^{2} ϵ ^{2} ℏ ^{4}}{μ ^{3} e ^{2}}

The key features of this result are that the energy shift is positive (a "blue shift") and that it is quadratic in the magnetic field strength $B$ .

Let's confirm our derivation with a key text on this topic.

Excitons in Magnetic Fields

The paper 'Excitons in Magnetic Fields' provides a concise and clear summary of the theory we are discussing. It is an excellent resource for seeing how these fundamental concepts are presented in the literature.

Please read the subsection within 'Three-Dimensional Case' that begins with 'To define the low-B and high-B limits...'. Follow the argument leading to Eq. (14), which calculates the first-order correction for the 1s state. Note how the result is expressed in terms of a 'diamagnetic-shift coefficient' in Eq. (15). You will see that their result matches ours exactly.

Test your understanding!

The diamagnetic shift coefficient $σ_{3 D}$ is proportional to $1/ μ^{3}$ . Two different semiconductors, A and B, have the same dielectric constant, but the exciton reduced mass in A is half that in B ( $μ_{A} = 0.5 μ_{B}$ ). Which material would you expect to show a larger diamagnetic shift for its 1s exciton state, and by what factor?

Show answer

Since $σ \propto 1/ μ^{3}$ , the material with the smaller reduced mass will have a significantly larger diamagnetic shift.
The ratio of the coefficients is $\frac{σ _{A}}{σ _{B}} = (\frac{μ _{B}}{μ _{A}})^{3} = (\frac{μ _{B}}{0.5 μ _{B}})^{3} = 2^{3} = 8$ .
Semiconductor A will show a diamagnetic shift that is 8 times larger than that of semiconductor B. This illustrates the high sensitivity of this measurement to the effective mass.

4. Physical Interpretation and Connection to Experiments

The diamagnetic shift represents the energy cost of confining the electron and hole's relative motion. The Lorentz force tries to make the particles execute cyclotron orbits, "squeezing" the exciton wavefunction in the plane perpendicular to the field. This increased confinement leads to a higher kinetic energy, which manifests as the positive, quadratic energy shift.

The magnitude of the shift is proportional to $⟨ x^{2} + y^{2} ⟩$ , which is a measure of the exciton's spatial extent (its cross-sectional area) perpendicular to the field. This provides a direct and powerful experimental tool: by measuring the diamagnetic shift coefficient $σ$ , one can determine the exciton's root-mean-square radius if the reduced mass is known, or vice versa.

This quadratic $B^{2}$ dependence is a hallmark feature of magneto-photoluminescence spectra in the low-field regime.

Magnetophotoluminescence of exciton Rydberg states...

To see how this theoretical result is applied in practice, let's look at a research paper that measures the diamagnetic shift for exciton Rydberg states.

First, read the paragraph 'We next consider the diamagnetic shift...' in the main text (section on page 205420-3). This section introduces the weak-field limit and Eq. (2), which is the formula for the diamagnetic shift. Note how they connect the coefficient \sigma directly to the expectation value (\langle r^2 angle). (This paper deals with a 2D material, where r^2 = x^2+y^2, simplifying the prefactor compared to our 3D case). Then, look at Figure 3(c). You can see the experimental data (points) for the 1s and 2s states being fit by quadratic curves (solid lines), confirming the B^2 dependence we derived. Finally, for the direct derivation in the 2D case, refer to the Appendix, specifically Eqs. (A5) through (A10), which lay out the perturbative calculation very clearly.

This connection between a clean theoretical derivation and tangible experimental data is precisely what makes magneto-optics a powerful probe of semiconductor properties. Your own work on magneto-PL of CuO will involve analyzing just these kinds of shifts.

Conclusion

In this lesson, we have taken a significant step in understanding the interaction of excitons with magnetic fields. By applying first-order perturbation theory to the simplified $K = 0$ Hamiltonian, we derived one of the most fundamental results in magneto-exciton physics.

Key Takeaways:

In the weak-field limit and for $K = 0$ , the exciton Hamiltonian can be treated as an unperturbed hydrogenic problem plus a small magnetic perturbation.
For s-states (like the 1s ground state), the first-order energy correction from the linear Zeeman term is zero.
The lowest-order energy shift is the diamagnetic shift, which arises from the $B^{2}$ term in the Hamiltonian.
Using perturbation theory, we calculated this shift to be $Δ E_{dia} = σ B^{2}$ , where the diamagnetic shift coefficient $σ$ is proportional to the square of the exciton's effective Bohr radius and inversely proportional to the cube of its reduced mass.
This quadratic energy increase with the magnetic field is a key experimental signature used to determine fundamental exciton parameters like size and effective mass.

Preview of the Next Lesson

We have now calculated the quadratic diamagnetic shift. But what about the term in the Hamiltonian that is linear in $B$ ? In our next lesson, we will focus on analyzing the linear Zeeman splitting of exciton states, which requires considering states with non-zero angular momentum and incorporating the intrinsic spin of the electron and hole. This will allow us to determine effective g-factors and complete our picture of the exciton in a weak magnetic field.

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