# Action of a symmetry operation on a Bloch state

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• dRic2
In summary: Q}## is the symmetry operator. There are probably many symmetry operators that commute with the Hamiltonian but don't scale the phase, for example, the identity operator. So I think it's just a matter of finding a useful constraint for your particular problem.
dRic2
Gold Member
Generally speaking, if the Hamiltonian has a specific symmetry defined by an operator M, that is ##[H,M]=0##, when I apply such symmetry operator to a Bloch state I would expect the state to be left unchanged up to a phase:
$$M\ket{ψ_{\mathbf k}}=e^{iϕ(\mathbf k)}\ket{ψ_{\mathbf k}}$$
For the simple case of a translation of a lattice vector R the phase it gets is just ##e^{−i \mathbf k \cdot \mathbf R}##.

What about more complicated symmetries? In my case, I would like to consider a 2D periodic lattice where there is also an additional symmetry called "glide mirror symmetry". This symmetry is composed by a reflection along the x-axis (##y \rightarrow −y##) and a translation in the x-direction of half of a lattice vector (##x \rightarrow x - \frac a 2##), where ##a## a is the lattice vector).

If I apply this symmetry to a Bloch state then what is the resulting phase? I get stuck at the very beginning:
$$M \psi_{\mathbf k}(\mathbf x) = e^{-i[k_x(x-\frac a 2)-k_yy]} u(x-a/2, -y)$$

Thanks

(Let's consider spinless electrons, i.e. scalar Bloch states)

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So you start with a Bloch state $$|\psi \rangle = e^{i\mathbf{k}\cdot \mathbf{x}} u(\mathbf{x})$$ where ##u(\mathbf{x})## is a real valued function. Applying your glide mirror operator M, \begin{align*} M|\psi \rangle &= e^{i (k_x (x-a/2) - k_y y)} u(x-a/2,-y) \\ &= e^{i(k_x a / 2 - 2k_y y)} e^{i\mathbf{k} \cdot \mathbf{x}} u(x-a/2,-y) \end{align*} You know from the fact that ##M## commutes with the Hamiltonian that ##M|\psi \rangle = e^{i\phi} |\psi \rangle##, so comparing that with the above result: $$u(x,y) = u(x-a/2,y)$$ $$\phi = -k_x a / 2 - 2k_y y$$

Edit: This post is just wrong. Skip down to post #6.

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dRic2
I arrived at the same conclusion, but I posted because I felt weird about the phase being position-dependent (## 2k_y y##). But now that you also suggested the same solution I think it does make sense that the phase is position-dependent. In fact, for example, the point on the x-axis (y=0) are not affected by the Mirror-symmetry and their phase would be just ##\phi=-k_xa/2## (as expected by a translation). On the other hand, the more I move far away from the x-axis, the more the mirror symmetry will be "important".

Twigg
I had the same gut reaction, but I felt ok with after realizing the Bloch phase ##\phi = \mathbf{k} \cdot \mathbf{x}## is also position-dependent.

But wait... something is wrong
$$\hat M \hat H \ket{\psi_k} = \hat M E_k \ket{\psi_k} = E_k e^{i\phi}\ket{\psi_k}$$
While
$$\hat H \hat M \ket{\psi_k} = \hat H e^{i\phi}\ket{\psi_k}$$
if ##\phi=-k_xa/2 - 2k_yy## and ##\hat H = - \partial^2_x + V(x)## then:
$$\hat H e^{i\phi}\ket{\psi_k} = - \partial^2_x(e^{i\phi}\ket{\psi_k}) + V(x)e^{i\phi}\ket{\psi_k}$$
$$- e^{i\phi}\partial^2_x(\ket{\psi_k}) - k_y^2e^{i\phi}\ket{\psi_k} + e^{i\phi}V(x)\ket{\psi_k}$$
$$e^{i\phi}[- \partial^2_x + V(x)]\ket{\psi_k} - k_y^2e^{i\phi}\ket{\psi_k} = e^{i\phi}(\hat H -k_y^2)\ket{\psi_k} \neq \hat M \hat H \ket{\psi_k}$$

Did I make a mistake?

You're right. I forgot something silly, and my last post was dead wrong. The glide mirror symmetry behaves exactly like a mirror symmetry in non-periodic situations. The reason is that $$\hat{M}^2 \psi(\mathbf{x}) = \psi(x-a,-(-y)) = e^{ik_x a} \psi(\mathbf{x})$$ since a is the lattice interval and ##\psi(\mathbf{x}) = e^{i \mathbf{k}\cdot\mathbf{x}}u(\mathbf{x})##. This means that the phase ##\phi## defined by $$\hat{M} \psi(\mathbf{x}) = e^{i\phi} \psi(\mathbf{x})$$ must be a root of $$\left(e^{i\phi}\right)^2 = e^{ik_x a}$$ which implies $$\phi = \frac{k_x a}{2} + \pi N$$ for any integer ##N##. In a much prettier form, you can say $$\hat{M} \psi(\mathbf{x}) = \pm e^{ik_x a / 2} \psi(\mathbf{x})$$ I think you'll find it easy to prove that this commutes.

dRic2
This leaves me with one last question: do you think that the phase induce by any symmetry has to be position-independent? I mean the reasoning of post #5 should be quite general for any operator does is supposed to commute with H, right?

dRic2 said:
do you think that the phase induce by any symmetry has to be position-independent?

The phase of the Bloch wavefunction scales with position, so I don't see a problem. I'm sure you could derive a general constraint by solving the equation $$[\hat{Q},-\frac{\hbar^2}{2m} \nabla^2 + V(\mathbf{x})]\psi = 0$$

## 1. What is a symmetry operation in the context of Bloch states?

A symmetry operation refers to a transformation that preserves the overall structure or properties of a system. In the context of Bloch states, it refers to a transformation that leaves the Hamiltonian of a system unchanged, such as a rotation, translation, or inversion.

## 2. How does a symmetry operation affect a Bloch state?

A symmetry operation acts on a Bloch state by transforming it into another Bloch state with the same energy and momentum, but possibly with a different phase or direction. This means that the symmetry operation does not change the physical properties of the Bloch state, but only its mathematical representation.

## 3. What is the significance of symmetry operations in the study of Bloch states?

Symmetry operations play a crucial role in the study of Bloch states because they provide a powerful tool for understanding the behavior and properties of these states. By identifying the symmetries of a system, we can make predictions about the behavior of Bloch states and use this knowledge to design materials with specific properties.

## 4. Can a Bloch state be invariant under all symmetry operations?

No, it is not possible for a Bloch state to be invariant under all symmetry operations. This is because the Hamiltonian of a system must have at least one symmetry, and the Bloch states are eigenstates of the Hamiltonian. Therefore, any symmetry operation that leaves the Hamiltonian unchanged will also change the Bloch state.

## 5. Are there any other factors that can affect the action of a symmetry operation on a Bloch state?

Yes, there are other factors that can affect the action of a symmetry operation on a Bloch state, such as the presence of impurities or defects in the material. These can break the symmetry of the system and lead to changes in the behavior of Bloch states.

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