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I How is Graphene's Hamiltonian rotationally invariant?

  1. Mar 2, 2017 #1
    Graphene's Hamiltonian contains first order derivatives (from the momentum operators) which aren't invariant under simple spatial rotations. So it initially appears to me that it isn't invariant under rotation. From reading around I see that we also have to perform a rotation on the Pauli matrices that always appear with these operators which would then leave the Hamiltonian invariant. I don't understand how to do this rotation or why this is the case? Any help would be greatly appreciated
     
  2. jcsd
  3. Mar 6, 2017 #2

    DrDu

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    Science Advisor

    I suppose you refer to a hamiltonian of the form ##H=p\cdot \sigma=2(p_+ \sigma_-+p_- \sigma_+)## with ##p_\pm=(p_x\pm i p_y)/2## and ##\sigma_\pm=(\sigma_x\pm i\sigma_y)/2##. Upon rotation, ##p_\pm## gets multiplied with a phase factor. This can be compensated multiplying ##\sigma_\pm## with a phase factor, too. This is an example of a gauge transformation as it corresponds to multiplication of the electronic states with a phase factor, which is not directly observable.
     
  4. Mar 13, 2017 #3
    There're two approaches here.

    First is to notice that the specific presentation of the Pauli matrices is not predetermined. As long as those matrices satisfy the correct commutation and anti-commutation relations, ## [\sigma_i, \sigma_j] = 2i \epsilon_{ijk}\sigma_k ## and ## \{\sigma_i, \sigma_j \} = \delta_{ij} ##, they would do. For example, there's nothing wrong in choosing
    $$
    \sigma_x = \frac{1}{4}
    \begin{pmatrix}
    \sqrt{6} & 2 + i \sqrt{6} \\
    2 - i \sqrt{6} & -\sqrt{6}
    \end{pmatrix},
    \sigma_y = \frac{1}{4}
    \begin{pmatrix}
    -\sqrt{2} & \sqrt{10 - i 4 \sqrt{6}} \\
    \sqrt{10 + i 4 \sqrt{6}} & \sqrt{2}
    \end{pmatrix},
    \sigma_z = \frac{1}{2}
    \begin{pmatrix}
    \sqrt{2} & - i \sqrt{2} \\
    i \sqrt{2} & - \sqrt{2}
    \end{pmatrix}.
    $$
    The question is just what's the meaning of the components of the spinor. When we choose ## \sigma_z = diag(1, -1)## the components correspond to amplitudes to be at different sub-lattices, ##\psi_{1,2}##. But one can orient the quantization axis in any way one wants. What if I want the components to be symmetric and antisymmetric combinations of #\psi_{1,2}#. I can do that but the form of the ##\sigma## matrices will change. The representation above, for instance, corresponds to some specific choice of the quantization axis.

    This has the direct relation to the rotational symmetry of graphene's Hamiltonian ## H = p_i \sigma_i##. After rotation around the ##z## axis, we have
    $$
    \widehat{R}(\theta) H = p_i \widetilde{\sigma}_i.
    $$
    So, we got Hamiltonian with slightly modified ##\sigma## matrices. if we ask ourselves, what do these new ##\widetilde{\sigma}## matrices correspond to, we'll see that they take this form if we rotate the pseudospin coordinate system around the ##z## axis. This makes sense: if we rotate the system, we need to rotate all objects, including spinors.

    The second approach is to notice right away that
    $$
    \widehat{R}(\theta) H = e^{i \sigma_z \theta/2}H e^{-i \sigma_z \theta/2},
    $$
    which formally expresses the argument above.

    And, of course, this is not a gauge transformation.
     
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