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  Representation on Hilbert space of the product of two symmetry transformations

+ 6 like - 0 dislike

We know by Wigner's theorem that the representation of a symmetry transformation on the Hilbert space is either unitary and linear, or anti-unitary and anti-linear.

Let T and S be two symmetry transformations. Let U(T) and U(S) be the representations of these transformations. What can we say about unitarity or anti-unitarity of U(TS) if we know the unitarity or anti-unitarity of U(T) and U(S)? Why?

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asked Feb 26, 2012 in Theoretical Physics by user15291 (75 points) [ no revision ]
It seems pretty straightforward to me. For an anti-unitary operator $U$ we have $(U \psi, U \chi) = (\psi, \chi)^*$. Then for a product $U V$ of two unitary operators we have $(U V \psi, U V \chi) = (\psi, \chi)$, which means the product $U V$ is unitary. The other cases can be worked out similarly.

This post has been migrated from (A51.SE)
But U(TS) is in general not equal to U(T)U(S). If it would have been, your argument would have worked.

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$U(TS) = U(T)U(S)$, that's part of the definition of a representation. Do you have a counterexample where this is not true?

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@Sidious Lord: Yes, e.g. [projective representations](http://en.wikipedia.org/wiki/Projective_representation).

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@Qmechanic OK, but when we have projective representations, don't we pass to the covering group or add central extensions such that we always have an honest to God representation? I thought there are no obstructions to doing that in relevant physical applications, but I would be interested to know if this is not true.

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@Sidious Lord: In my interpretation of OP's question(v1), in order to be as general as possible, the word _representation_ should not necessarily be understood as implying any group structure or group representation (projective or not), cf. my answer.

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@Qmechanic The word representation in math means isomorphism between one object and another. If the group structure is not preserved, it is not a representation. I agree that projective representations should be also taken into account, but due to slightly different reasoning. Formally, by the way, $U(TS)=U(T)U(S)$ holds for projective representations. Equality in projective spaces and the way they usually explained may confuse, but the group structure is preserved.

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1 Answer

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I) Wigner's Theorem states that a symmetry operation $S: H \to H$ is a unitary or anti-unitary$^{1}$ operator $U(S)$ up to a phase factor $\varphi(S,x)$,

$$ S(x)~=~ \varphi(S,x)\cdot U(S)(x), \qquad x~\in~H,\qquad \varphi(S,x)~\in~\mathbb{C} ,\qquad |\varphi(S,x)|~=~1 .$$

In this context, a symmetry operation $S$ is by definition a surjective (not necessarily linear!) map $S: H \to H$ such that

$$|\langle S(x),S(y)\rangle|~=~|\langle x,y\rangle|,\qquad\qquad x,y~\in~H.$$

Let us introduce the terminology that a symmetry operation $S$ is of unitary (anti-unitary) type if there exists a unitary (an anti-unitary) $U(S)$, respectively.

Moreover, if ${\rm dim}_{\mathbb{C}} H \geq 2$, then one may show that

  1. $U(S)$ is unique up to a constant phase factor, and
  2. $S$ cannot have both a unitary and an antiunitary $U(S)$. In other words, $S$ cannot both be of unitary and anti-unitary type.

II) It follows by straightforwardly applying the definitions, that the composition $S \circ T$ of two symmetry operations $S$ and $T$ is again a symmetry operation, and it is even possible to choose

$$ U(S \circ T)~:=~U(S) \circ U(T).$$ Finally, in the case ${\rm dim}_{\mathbb{C}} H \geq 2$,

  1. $S \circ T$ is of anti-unitary type, if precisely one of $S$ and $T$ are of anti-unitary type, and
  2. $S \circ T$ is of unitary type, if zero or two of $S$ and $T$ are of anti-unitary type.


  1. V. Bargmann, Note on Wigner's Theorem on Symmetry Operations, J. Math. Phys. 5 (1964) 862. Here is a link to the pdf file.


$^{1}$ We use for convenience a terminology where linearity (anti-linearity) of $U(S)$ are implicitly implied by the definition of $U(S)$ being unitary (anti-unitary), respectively.

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answered Feb 26, 2012 by Qmechanic (3,120 points) [ no revision ]
Hi, thanks for your answer. How do you get it straightforwardly without considering U(ST)=U(S)U(T)?

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I updated the answer.

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