Action of a symmetry operation on a Bloch state

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  • #1
dRic2
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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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  • #2
Twigg
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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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  • #3
dRic2
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Hi, thanks for the answer.
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".
 
  • #4
Twigg
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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.
 
  • #5
dRic2
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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:
$$\hat H e^{i\phi}\ket{\psi_k} = - \partial^2_x(e^{i\phi}\ket{\psi_k}) + V(x)e^{i\phi}\ket{\psi_k}$$
$$- e^{i\phi}\partial^2_x(\ket{\psi_k}) - k_y^2e^{i\phi}\ket{\psi_k} + e^{i\phi}V(x)\ket{\psi_k}$$
$$e^{i\phi}[- \partial^2_x + V(x)]\ket{\psi_k} - k_y^2e^{i\phi}\ket{\psi_k} = e^{i\phi}(\hat H -k_y^2)\ket{\psi_k} \neq \hat M \hat H \ket{\psi_k} $$

which contradicts the assumption ##[H,M]=0##

Did I make a mistake?
 
  • #6
Twigg
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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 $$\left(e^{i\phi}\right)^2 = e^{ik_x a}$$ wich 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.
 
  • #7
dRic2
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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?
 
  • #8
Twigg
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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 $$[\hat{Q},-\frac{\hbar^2}{2m} \nabla^2 + V(\mathbf{x})]\psi = 0$$
 

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