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In my QFT book the author (Schwartz) mentioned, that we found two unitarily inequivalent representations of the Lorentz group. However, he never really introduced the idea of unitarily equivalent representations. So my question is: What is the relevance of unitarily in-/ equivalent representations in physics? Why do we talk about it/ what does it mean physically, if two representations are unitarily in-/ equivalent?

PS: The mathematical definition: https://encyclopediaofmath.org/wiki/Unitarily-equivalent_representations

Update: The statement in from the book 'Quantum Field Theory and the Standard Model' by Schwartz. On page 169 (Chapter 10.3) he compares the 4-vector representation of the Lorentz group to the Dirac representation (Bispinor representation). He states that the representations are inequivalent, which, I guess, should have meant unitarily inequivalent.

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    $\begingroup$ What makes you suspect it is not the same as the mathematical definition? Where is your text saying this and why does it trouble you? $\endgroup$ Jul 6, 2021 at 18:24
  • $\begingroup$ I think you misunderstood what I was trying to say. I don't doubt that my book uses the same definition. I am just wondering, why we care about the representations being unitarily inequivalent. After all, there must be some reason, why physics textbooks mention it. $\endgroup$
    – sika_98
    Jul 6, 2021 at 19:24
  • $\begingroup$ I updated the post and added some information on where my question first came from. I am looking for the general reason on why we talk about unitary in-/equivalence in QFT/QM. I am quite new to the whole concept of representations, so maybe I am missing something obvious. Regarding your question on what I am looking for: I would prefer a conceptual answer, that gives me intuition on what unitary in-/equivalence means in physics. I would rather not read an essay on the subject right now, mainly because I probably still lack quite a bit of maths to understand a specialized text. $\endgroup$
    – sika_98
    Jul 6, 2021 at 20:00

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There's not really one right answer to this question, but here is one way to look at it.

Time evolution in quantum mechanics is unitary (ignoring measurements -- which I will ignore throughout this answer). In the Schrodinger picture, (and ignoring measurements), the state at an initial time $t=t_i$, which we will call $|\psi,0\rangle$ is always related to the state at a final time $t=t_f$ $|\psi,t\rangle$ by some unitary operator $U(t_f,t_i)$ \begin{equation} |\psi,t_f\rangle = U(t_f,t_i)|\psi,t_i\rangle \end{equation} So if two states are related by a unitary transformation, then it is in principle possible for the time evolution of quantum mechanics to take us from one state to the second state. Said another way, starting from one member of a unitary representation, there's nothing to stop unitary evolution from taking the state into any other members of the representation. On the other hand, if two states are not related by a unitary transformation, then it is not possible for the state to evolve from one to the other, so these two states describe two different physical systems that are not interchangeable. (There's a subtlety here, since in an interacting quantum theory you can get from one type of particle to another by unitarily evolving through multi-particle states; but often "unitary representations of the Poincaire group" are used to classify single particle states)

Representations of a symmetry are useful, because the symmetry allows us to classify possible states. For example, if we have rotational symmetry, then finding representations of this rotational symmetry (angular momentum eigenstates) tell us the possible sets of states in Hilbert space. In the case of angular momentum, a given representation tells us the total angular momentum and the possible values of the $z$ component of the angular momentum.

Combining these, unitary representations are important because time evolution can take us from one state in the representation to another. For example, within a unitary representation of the rotation group, we can transform from one value of the $z$ component of angular momentum to another, within some representation labeled by the total angular momentum.

Finally, we want to consider unitarily equivalent representations as being really the same thing (physically). Representations that differ by a unitary transformation of the basis vectors are just related by a change of basis. Since physical results are independent of the choice of basis, we want to "mod out" by choosing to work from one standard representative from a set of unitarily equivalent representations.

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  • $\begingroup$ Thank you very much for your nice answer! Now I finally understand what we mean by a "physical system" in this context. $\endgroup$
    – sika_98
    Jul 7, 2021 at 8:49
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The point your text is making is that when you find bunches of 4×4 matrices satisfying the Lorentz algebra (10.21) & (10.69), you don't actually have the "same" physics structure. A representation may be characterized by more labels than its dimensionality!

Over and above the matrices written down, there is an infinity of such matrix sets satisfying the same algebra, but we classify them in equivalence classes consisting of unitary transforms of each set: so changing vector bases operated upon by the Vs, yields unitarily (orthogonally, if you wish to preserve real vectors) equivalent sets of Vs, all examined in one breath, by a standard representative, as given. They all give the same physics, and have the same (Lorentz) invariant Casimirs, etc, so studying one is enough; which everybody picks.

But Mat points out there is yet another 4D representation, for which there is no unitary equivalence transformation s. t. $US_{\mu\nu}U^{-1}=V_{\mu\nu}$, and so it need not give you the same invariants, and physics in general. There is no basis change from 4-vector space to 4-spinor space! The Casimirs are different, for instance the square of the Pauli-Lubanski pseudovector specifying the spin, etc., and so it is a very different representation than the vector one above.

In fact, he tells you it is reducible, in contrast to the vector one, and demonstrates that. Its equivalence class rotates you among different bases, Dirac, Weyl, Majorana, unitarily equivalent, so these describe the same physics, since basis changes are not meaningful for scalar products etc...

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  • $\begingroup$ That clears up the text quite a bit, thank you. One follow-up question: When you say, that the Casimirs are different, you mean that their eigenvalues are different, while the Casimir itself is the same element of the algebra, right? Because, afaik the definition of a Casimir Operator itself is independent of the representation. $\endgroup$
    – sika_98
    Jul 7, 2021 at 8:34
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    $\begingroup$ Yes, of course! The P-L eigenvalue is s(s+1) times a constant. $\endgroup$ Jul 7, 2021 at 9:37
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    $\begingroup$ Then I got it right, thank you! $\endgroup$
    – sika_98
    Jul 7, 2021 at 9:48
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    $\begingroup$ For example, the eigenvalues of the rotation group Casimir label the representation. $\endgroup$ Jul 7, 2021 at 10:09
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In physics unitary equivalence basically just means “the same up to a (unitary) change of basis.”

Consider as an example the standard Pauli matrices. They are usually written so that $\sigma_z$ is diagonal, but there’s nothing to prevent you from choosing a basis where $\sigma_y$ is diagonal. In fact the two choices are unitarily equivalent since there is a single matrix $U$ so that $U\sigma_zU^\dagger=\sigma_y$, where $U$ takes you from the basis where $\sigma_z$ is diagonal to the basis where $\sigma_y$ is diagonal. The same $U$ will take $\sigma_i$ in one basis to $\sigma_k$ in the other.

Of course, the physics (or more precisely the symmetries baked in the problem) doesn’t change because you choose one basis over another. Thus, you don’t have to rework all the group theory of spin-1/2 particles just because you chose a basis where $\sigma_y$ is diagonal: all the structural results - v.g. what are the possible resulting total spins when you couple to spin-1/2 particles - carry over between unitarity equivalent representations. All you need (if you actually need it), is to find the change of basis and you can confidently continue as before.

When representations are NOT equivalent, you can’t make such a change of basis. There are “many ways” in which to reps can’t be equivalent.

Clearly two representations of different dimension can’t be connected by a unitary $U$: the states in the $S=1$ triplet are fundamentally different from the state in the $S=0$ singlet, even if the both arise as a result of coupling two $s=1/2$ particles. The triplet states are symmetric under permutation, and singlet is antisymmetric, and a simple change of basis cannot take you from 3 symmetric state to one antisymmetric state.

In some cases (as in your example), even if you have a pair of 4-dim. reps, they are not equivalent, meaning that some properties (eigenvalues of some generators, or something else) will be different in the two cases.

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  • $\begingroup$ Clear answer, thanks! $\endgroup$
    – sika_98
    Jul 7, 2021 at 8:56

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