0
$\begingroup$

There is a section in my lecturer notes for Group Theory that I am finding difficult to understand. We let the row vector $\hat{\phi}=(\phi_1, \phi_2,...,\phi_l)$. For a normal $l$-fold degeneracy, we can write $\hat{S_a}\hat{\phi}=\hat{\phi}\Gamma (\hat{S_a})$, where $\hat{S_a}$ is the similarity transformation of the Hamiltonian and $\Gamma (\hat{S_a})$ is an $l$-dimensional matrix representation of the Group of the Hamiltonian.

The goal now, and the part I do not understand, is to show that this representation in terms of $\Gamma (\hat{S_a})$ matrices is irreducible. The notes start by replacing the row vector $\hat{\phi}$ by a linear combination $\hat{\phi}U$, where $U$ is an $l$-dimensional square matrix. Then we operate on $\hat{\psi}$ with $\hat{S_a}$:

$$\hat{S_a}\hat{\psi}=...=\hat{\psi}U^{-1}\Gamma{(\hat{S_a})}U$$

where I have skipped some of the intermediate steps. Clearly, the representation based on $\hat{\psi}$ and $\hat{\phi}$ are equivalent, since they are related by a similarity transformation.

Suppose now that the representation based on $\hat{\psi}$ is reducible. Then there would be a unitary transformation of the $\phi_j$ such that there are two or more subsets of the $\psi_j$ that transform only among one another under the symmetry operations of the Hamiltonian. This implies that $\hat{S_a}$ applied to any eigenfunction generates eigenfunctions only in the same subset. The degeneracy with the eigenfunctions in the other subset is therefore accidental, in contrast to our assertion that the degeneracy is normal. Hence the representation obtained for a normal degeneracy is irreducible.

The sentence in bold is what I do not understand. I did not know that reducible representations had this property. Please help me understand this step.

$\endgroup$
8
  • $\begingroup$ You have a typo in the sentence starting with "Clearly": one of the two vectors should be $\hat\psi$. The bold sentence explains what reducibility means, namely separate vector subspaces; what do you think it does? $\endgroup$ Commented Nov 28, 2022 at 16:18
  • 1
    $\begingroup$ Thank you for your answer, I have corrected the typo. As for the second part, I am only familiar with 2 equivalent definitions of reducibility: 1) A representation is reducible if all the matrices in it can be brought to block-diagonal form with the same similarity transformation, 2) A representation is reducible if there is a matrix, other than a multiple of the unit matrix, that commutes with all the matrices of the representation (from Schur's 1st lemma). I genuinely do not see how either of these definitions relates to the sentence in bold. $\endgroup$
    – NX37B
    Commented Nov 28, 2022 at 16:37
  • $\begingroup$ So you assumed $U^{-1}\Gamma{(\hat{S_a})}U$ is a concatenation of block matrices not connecting the disjoint subspaces, no? $\endgroup$ Commented Nov 28, 2022 at 16:44
  • $\begingroup$ I think that this is correct, although I must confess that I am not very familiar with this language (disjoint subspace). $\endgroup$
    – NX37B
    Commented Nov 28, 2022 at 17:33
  • $\begingroup$ The subsets of $\hat \psi$ mentioned... $\endgroup$ Commented Nov 28, 2022 at 17:35

1 Answer 1

1
$\begingroup$

The text is an immediate consequence from the definition of the irreducible representation. The mathematical definition of an irreducible representation is a group homomorphism $\Gamma :G \rightarrow \text{GL}(V)$ such that there is no invariant subspace in $V$. That is to say, other than the conditions making $\Gamma$ a group homomorphism, given any $v \in V$ where $v \neq 0$, we can always find some set $S_v=\{g_1,g_2,g_3,\cdots|g_j \in G\}$, which is allowed to depend on $v$, such that $\{\Gamma(g_1)v,\Gamma(g_2)v,\Gamma(g_3)v,\cdots\}$ spans $V$.

A simple example to explain what your lecture note means is the Hamiltonian of a non-relativistic free electron with an applied magnetic field in $z$ direction $$\hat{H}=-{\hbar^2 \over 2m}\nabla^2+h\sigma_z \tag{1}$$ and the Hilbert space for the electron is $$\bigg\{\int{c(k)|k \rangle dk}\bigg|\int{|c(k)|^2dk}=1\bigg\} \times \bigg\{\alpha_1|\uparrow \rangle+\alpha_2|\downarrow \rangle \bigg| |\alpha_1|^2+|\alpha_2|^2=1\bigg\} \tag{2}$$ where $|k \rangle$ is the momentum ket. The first set in Eq. (2) accounts for the spatial part of the electron wavefunction and the second for the spin part. Due to the presence of the magnetic field, the Hamiltonian does not have complete $3$ dimensional rotational symmetry, but it still possesses rotational invariance for rotation around $z$ axis. This means, under the unitary transformation $$\Gamma(\theta)=e^{-i \theta L_z} \otimes e^{-i \theta S_z}, \theta \in \mathbb{R} \tag{3}$$ we have $$\Gamma(\theta)\hat{H}\big(\Gamma(\theta)\big)^{-1}=\hat{H} \tag{4}$$ However, the Lie Group $\Gamma$ is no longer irreducible in the Hilbert space of Eq. (2). As we can see, both $\{\alpha|\uparrow \rangle|\alpha \in \mathbb{C}\}$ and $\{\alpha|\downarrow \rangle|\alpha \in \mathbb{C}\}$ are the subspaces of the spin part wavefunctions, and they are invariant under the action of $\Gamma$. Hence, the degeneracy in spin is not guaranteed in Eq. (1), and this is almost always true except the condition when $h=0$, where the Hamiltonian accidentally reduces to the form which has a higher $3$ dimensional rotational symmetry. The idea of reducible representations can be quite useful in crystals, where continuous rotation symmetries further reduce to discrete point groups, and consideration of the interplay between spins including the nuclei spin and the point group can give us predictions about degeneracies in energy levels.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.