# Representation on Hilbert space of the product of two symmetry transformations

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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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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.

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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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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.

Reference:

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 (3,110 points)
Hi, thanks for your answer. How do you get it straightforwardly without considering U(ST)=U(S)U(T)?

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