Kramers' Degeneracy and Time-Reversal Symmetry

Hello! Welcome to the twelfth module of our course. In the previous module, we explored topological phases that arise from breaking time-reversal symmetry, culminating in our analysis of the Haldane model. We saw how a cleverly designed, zero-net-flux magnetic field could lead to a Chern insulator phase, exhibiting a quantized Hall conductivity without any external magnetic field. This was a direct consequence of creating a non-zero topological invariant—the Chern number—by breaking time-reversal symmetry ($\mathcal{T}$).

A natural and profound question follows: what happens if we preserve time-reversal symmetry? Does topology become irrelevant? The answer, which has revolutionized condensed matter physics in the last two decades, is a firm no. $\mathcal{T}$-invariant systems can host their own unique and robust topological phases.

To understand these new phases, we must first understand the deep consequences of time-reversal symmetry itself in quantum mechanics. This lesson is dedicated to a cornerstone result known as Kramers' theorem. Our learning outcome is to explain how time-reversal symmetry leads to Kramers' degeneracy in systems with half-integer spin. This principle is the fundamental reason why topological insulators protected by time-reversal symmetry can exist.

1. The Time-Reversal Operator in Quantum Mechanics

In classical mechanics, time reversal is straightforward: we let $t\to -t$. This reverses momentum ($\mathbf{p}\to -\mathbf{p}$) and angular momentum ($\mathbf{L}\to -\mathbf{L}$) but leaves position untouched ($\mathbf{x}\to \mathbf{x}$). In quantum mechanics, we seek an operator, which we'll denote $\mathcal{T}$, that implements this transformation on the corresponding quantum operators.

We therefore postulate that the time-reversal operator $\mathcal{T}$ must satisfy:

• $\mathcal{T}\hat{\mathbf{x}}{\mathcal{T}}^{-1}=\hat{\mathbf{x}}$
• $\mathcal{T}\hat{\mathbf{p}}{\mathcal{T}}^{-1}=-\hat{\mathbf{p}}$
• $\mathcal{T}\hat{\mathbf{S}}{\mathcal{T}}^{-1}=-\hat{\mathbf{S}}$ (Spin is a form of angular momentum, so it must also be reversed).

A Hamiltonian $H$ is said to be time-reversal invariant if it commutes with this operator:

$$[H,\mathcal{T}]=0\text{or}H\mathcal{T}=\mathcal{T}H$$

This means that if $|\psi (t)⟩$ is a solution to the Schrödinger equation, then its time-reversed counterpart, $\mathcal{T}|\psi (-t)⟩$, must also be a solution.

2. The Antiunitary Nature of Time Reversal

What kind of operator is $\mathcal{T}$? Our first guess might be a unitary operator, as is the case for most symmetries (like translations and rotations). However, this leads to a fundamental contradiction. Consider the canonical commutation relation for position and momentum:

$$[{\hat{x}}_i,{\hat{p}}_j]=i\hslash {\delta }_{ij}$$

Let's see what happens when we transform this equation with $\mathcal{T}$. If $\mathcal{T}$ were unitary (${\mathcal{T}}^{-1}={\mathcal{T}}^{†}$), the transformation would be a similarity transformation:

$$\mathcal{T}[{\hat{x}}_i,{\hat{p}}_j]{\mathcal{T}}^{-1}=[\mathcal{T}{\hat{x}}_i{\mathcal{T}}^{-1},\mathcal{T}{\hat{p}}_j{\mathcal{T}}^{-1}]=[{\hat{x}}_i,-{\hat{p}}_j]=-[{\hat{x}}_i,{\hat{p}}_j]=-i\hslash {\delta }_{ij}$$

On the other hand, transforming the right side gives:

$$\mathcal{T}(i\hslash {\delta }_{ij}){\mathcal{T}}^{-1}=i\hslash {\delta }_{ij}$$

We have a contradiction: $-i\hslash {\delta }_{ij}=i\hslash {\delta }_{ij}$, which is impossible.

The only way to resolve this is if the operator $\mathcal{T}$ also acts on the complex numbers in the Hilbert space. Specifically, $\mathcal{T}$ must be antiunitary. An antiunitary operator has two defining properties:

1. It is antilinear: $\mathcal{T}(a|\psi ⟩+b|\varphi ⟩)=a^{\ast }\mathcal{T}|\psi ⟩+b^{\ast }\mathcal{T}|\varphi ⟩$ for any complex scalars $a,b$. Notice the complex conjugation.
2. It preserves the norm: $⟨\mathcal{T}\psi |\mathcal{T}\varphi ⟩=⟨\varphi |\psi ⟩=⟨\psi |\varphi {⟩}^{\ast }$.

With an antiunitary $\mathcal{T}$, the contradiction is beautifully resolved. When we transform the right side of the commutator, we now get $\mathcal{T}(i\hslash {\delta }_{ij})=-i\hslash {\delta }_{ij}$, which perfectly matches the result from transforming the left side.

For a deeper dive into the mathematical properties of antiunitary operators and Wigner's theorem, which states that all symmetries in quantum mechanics must be represented by either unitary or antiunitary operators, the following text is an excellent resource.

3. Time Reversal for Spin and ${\mathcal{T}}^2$

Now, let's consider the action of $\mathcal{T}$ on a spin-$s$ particle. The most common representation of the time-reversal operator for a system with spin is

$$\mathcal{T}=e^{-i\pi S_y/\hslash }K$$

where $S_y$ is the y-component of the total spin operator and $K$ is the complex conjugation operator in the basis of $S_z$ eigenstates. For a single spin-1/2 particle, $S_y=(\hslash /2){\sigma }_y$, so this becomes $\mathcal{T}=e^{-i\pi {\sigma }_y/2}K=-i{\sigma }_{y}K$.

Let's apply the time-reversal operator twice. This seemingly simple operation reveals a profound dichotomy.

$${\mathcal{T}}^2=(e^{-i\pi S_y/\hslash }K)(e^{-i\pi S_y/\hslash }K)=e^{-i\pi S_y/\hslash }K{e}^{-i\pi S_y/\hslash }{K}^{-1}{K}^2$$

Since $K$ is complex conjugation, $K^2=1$. The operator $e^{-i\pi S_y/\hslash }$ is a rotation matrix with real entries in the standard $S_z$ basis, so it is unaffected by complex conjugation ($K{e}^{-i\pi S_y/\hslash }{K}^{-1}=e^{-i\pi S_y/\hslash }$). Therefore:

$${\mathcal{T}}^2=e^{-2i\pi S_y/\hslash }$$

This operator corresponds to a rotation by $2\pi$ around the y-axis. As you know from your study of angular momentum, a $2\pi$ rotation behaves differently for integer and half-integer spin systems:

• Integer Spin (Bosons): A rotation by $2\pi$ is equivalent to the identity. So, ${\mathcal{T}}^2=+1$.
• Half-Integer Spin (Fermions): A rotation by $2\pi$ introduces a minus sign. So, ${\mathcal{T}}^2=-1$.

This distinction is fundamental. It applies to any system based on its total spin. A system containing an odd number of fermions (like a single electron) has half-integer total spin, so ${\mathcal{T}}^2=-1$. A system with an even number of fermions or any number of bosons has integer total spin, so ${\mathcal{T}}^2=+1$.

Test your understanding!

Verify explicitly that ${\mathcal{T}}^2=-1$ for a single spin-1/2 particle using the form $\mathcal{T}=-i{\sigma }_{y}K$. Remember that $K$ complex conjugates everything to its right, so $K(i)=-i$, and that $K^2=1$.

Show answer

Let's apply ${\mathcal{T}}^2$ to an arbitrary two-component spinor state $|\psi ⟩=\left(\begin{array}{c}a\\ b\end{array}\right)$.
First, apply $\mathcal{T}$ once:

$$\mathcal{T}|\psi ⟩=(-i{\sigma }_y)K\left(\begin{array}{c}a\\ b\end{array}\right)=-i\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)\left(\begin{array}{c}{a}^{\ast }\\ b^{\ast }\end{array}\right)=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}{a}^{\ast }\\ b^{\ast }\end{array}\right)=\left(\begin{array}{c}-b^{\ast }\\ a^{\ast }\end{array}\right)$$

Now, apply $\mathcal{T}$ a second time to this new state:

$${\mathcal{T}}^2|\psi ⟩=\mathcal{T}\left(\begin{array}{c}-b^{\ast }\\ a^{\ast }\end{array}\right)=(-i{\sigma }_y)K\left(\begin{array}{c}-b^{\ast }\\ a^{\ast }\end{array}\right)=-i\left(\begin{array}{cc}0& -i\\ i& 0\end{array}\right)\left(\begin{array}{c}-b\\ a\end{array}\right)=\left(\begin{array}{cc}0& -1\\ 1& 0\end{array}\right)\left(\begin{array}{c}-b\\ a\end{array}\right)=\left(\begin{array}{c}-a\\ -b\end{array}\right)=-\left(\begin{array}{c}a\\ b\end{array}\right)=-|\psi ⟩$$

Since this holds for any state $|\psi ⟩$, we have shown that the operator ${\mathcal{T}}^2=-1$.

4. Kramers' Theorem

We are now equipped to prove Kramers' theorem.

Theorem: For any system with time-reversal symmetry ([$H,\mathcal{T}$] = 0) and half-integer total spin (${\mathcal{T}}^2=-1$), every energy eigenstate must be at least two-fold degenerate.

Proof:
The proof is a beautiful argument by contradiction. Let's assume the opposite: that there exists a non-degenerate energy eigenstate $|\psi ⟩$ with energy $E$.

$$H|\psi ⟩=E|\psi ⟩$$

Now, let's act on this equation with the time-reversal operator $\mathcal{T}$.

$$\mathcal{T}H|\psi ⟩=\mathcal{T}E|\psi ⟩$$

Since $H$ is T-invariant, $\mathcal{T}H=H\mathcal{T}$. Since $E$ is a real number and $\mathcal{T}$ is antilinear, $\mathcal{T}E=E\mathcal{T}$. So we get:

$$H(\mathcal{T}|\psi ⟩)=E(\mathcal{T}|\psi ⟩)$$

This shows that the state $|\varphi ⟩\equiv \mathcal{T}|\psi ⟩$ is also an eigenstate with the same energy $E$.

Because we assumed that $|\psi ⟩$ is non-degenerate, its time-reversed partner $|\varphi ⟩$ must be the same state, up to a phase factor $c$.

$$|\varphi ⟩=\mathcal{T}|\psi ⟩=c|\psi ⟩$$

Let's apply $\mathcal{T}$ one more time:

$${\mathcal{T}}^2|\psi ⟩=\mathcal{T}(c|\psi ⟩)=c^{\ast }\mathcal{T}|\psi ⟩=c^{\ast }(c|\psi ⟩)=|c{|}^2|\psi ⟩$$

But for a system with half-integer spin, we know ${\mathcal{T}}^2=-1$. So we must have:

$$-|\psi ⟩=|c{|}^2|\psi ⟩$$

This implies $|c{|}^2=-1$, which is impossible for any complex number $c$. Our initial assumption of a non-degenerate eigenstate must be false.

Therefore, every energy level must be degenerate. The two states $|\psi ⟩$ and $\mathcal{T}|\psi ⟩$ are linearly independent and share the same energy. This guaranteed two-fold degeneracy is called Kramers' degeneracy, and the pair of states $(|\psi ⟩,\mathcal{T}|\psi ⟩)$ is called a Kramers pair.

The following resource provides a concise explanation of this proof.

This degeneracy is not accidental; it is fundamentally protected by time-reversal symmetry. Any perturbation added to the Hamiltonian that preserves T-symmetry (e.g., non-magnetic impurities, electric fields) cannot lift this degeneracy. The only way to break a Kramers pair is to break time-reversal symmetry itself, for example, by applying a magnetic field.

This concept is central to the physics of topological insulators. In a crystal, Kramers' theorem guarantees that at special points in the Brillouin zone that are mapped to themselves by time reversal (e.g., $\mathbf{k}=0$), the electronic bands must come in degenerate pairs: $E_n(\mathbf{k})=E_{n+1}(\mathbf{k})$. The behavior of these Kramers pairs as a function of momentum is what distinguishes a trivial insulator from a topological one.

Conclusion

In this lesson, we established a crucial pillar for understanding $\mathcal{T}$-invariant topological phases. We demonstrated that the quantum mechanical time-reversal operator $\mathcal{T}$ must be antiunitary and that it squares to $+1$ for integer-spin systems but $-1$ for half-integer-spin systems.

Key Takeaways:

• The time-reversal operator $\mathcal{T}$ is antiunitary to be consistent with the canonical commutation relations.
• For systems with integer spin (bosons), ${\mathcal{T}}^2=+1$.
• For systems with half-integer spin (fermions), ${\mathcal{T}}^2=-1$.
• Kramers' theorem states that for any $\mathcal{T}$-invariant Hamiltonian with ${\mathcal{T}}^2=-1$, all energy eigenstates are at least two-fold degenerate.
• These degenerate states, called Kramers pairs, are protected by time-reversal symmetry and cannot be split unless $\mathcal{T}$ is broken.

Preview of the Next Lesson

Now that we have established Kramers' theorem, we can build upon it. In the next lesson, we will introduce a new topological invariant, the ${\mathbb{Z}}_2$ invariant, which is specifically defined for time-reversal invariant systems. We will see how this invariant distinguishes a conventional band insulator from a new type of topological state: the Quantum Spin Hall insulator.