1
$\begingroup$

Is there any need to prove the existence of an operator $U$ which represents the action of symmetry transformation on rays in Hilbert space? Or is it enough just to prove that it is unitary and linear or antiunitary and antilinear?

I do not see how we should go about finding this operator. I got the answer that this operator is given, but I don't see how. The only thing that is given is the symmetry transformation which takes the whole ray $R_1$ into, let's say, $R_2$. But what then is the action of the operator on this ray? Which vector in $R_1$ will be taken into which in $R_2$?

My guess is that we can show that coefficients of the old and a new transformed vector are the same, so we can just say that in this way the transformation is translated into an operator. Every orthonormal vector of the old basis has the same coefficient as the one in the new basis.

$\endgroup$
  • 2
    $\begingroup$ I'm not sure if this is quite a duplicate, but it's very closely related to this question. It also follows up on this one but is not a duplicate of that. $\endgroup$ – David Z Mar 12 '16 at 11:03
  • 1
    $\begingroup$ The statement of Wigner's theorem is: "Given a ray transformation that is a symmetry, there is an (anti-)unitary operator that represents it on the Hilbert space". The proofs should construct said operator from the ray transformation, else they are not proofs. I do still not understand what your actual issue is. $\endgroup$ – ACuriousMind Mar 12 '16 at 11:05
  • $\begingroup$ Might be a duplacate, yes, I am sorry. And my apologies to ACuriousMind. I would love to be able to explain it better, but it seems that I am not. My problem is obviously that I dont see how is this operator constructed. In Steven Weinbergs book there is a proof but I do not see where construction of the operator happen. Could you please, in the form of an answer, point out where do we construct the operator? Of course, it might be that the entire proof IS the construction.... $\endgroup$ – Žarko Tomičić Mar 13 '16 at 12:53
0
$\begingroup$

Source of confusion was obviously that I did not understand what it means to prove the existance of such a thing as an operator. Details of the proof aside, in order to prove this, you simply construct the operator and in proofs of Wigners theorem that is exactly what is done. On the basis of the fundamental assumption about our transformation which is probability conservation, we made a connection betwen the old and a new set of vectors by defining a legal maping that is, defining a maping that associates exatly one vector from the new set to the vectors from the old one. After all, new vectors in a given ray differ from the old ones by a multiplicative constant in the first place, so just by seeing that we can choose a phase of the new vectors to be this or that, ve can construct the operator. Of course there is much more to this story but I believe, in the end, that this was my confusion. To add, the operator is purely mathematical object as opposed to transformation. So we have a certain freedom in its construction.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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