My questions are related with everything discussed in this post already. It's about the section on Active and Passive transformation in the 1st chapter (Linear Vector Spaces) of R. Shankar's Principles of Quantum Mechanics. I'll paste the relevant part here:

Suppose we subject all the vectors $|V\rangle$ in a space to a unitary transformation

$$|V\rangle \rightarrow U|V\rangle$$ Under this transformation, the matrix elements of any operator $\Omega$ are modified as follows: $$\langle V'|\Omega|V\rangle \rightarrow \langle UV'|\Omega|UV\rangle = \langle V'|U^\dagger\Omega U|V\rangle$$ It is clear that the same change would be effected if we left the vectors alone and subjected all operators to the change $$\Omega\rightarrow U^\dagger\Omega U$$

What I find hard to understand (or rather, interpret) is the expression $\langle V'|\Omega|V\rangle$. I know that the elements of the matrix representing $\Omega$ in an orthonormal basis are given by $\langle i|\Omega |j\rangle = \Omega_{ij}$ where $i$, $j$, ... are the orthonormal basis vectors. But what could $\langle V'|\Omega|V\rangle$ possibly mean, if $V$ and $V'$ are two arbitrary vectors in the vector space? It seems the result of this expression would just be some arbitrary complex number with no apparent meaning. I think this was already answered in the other post, but I'm not entirely satisfied with that answer and would like a bit more clarity.

Another question I have is how do I interpret the expression $\langle V'|U^\dagger\Omega U|V\rangle$? I already understand at a conceptual level what active and passive transformations are -- that active transformation transforms or changes all the vectors in the space while keeping the basis intact and passive transformation is a change of basis while keeping all the vectors intact -- but how do I connect this with the above expression?

The last question I have is why does Shankar deal only with Unitary operators when it comes to these transformations? Surely, the statements he makes apply to any operator, don't they? Doesn't $\langle UV'|\Omega|UV\rangle = \langle V'|U^\dagger\Omega U|V\rangle$ hold for any operator $U$, not necessarily unitary?

Note that I have looked for answers to these, but I wasn't able to find anything absolutely clear.


1 Answer 1

  1. $|V\rangle$ and $|V'\rangle$ here are any two arbitrary states, they could be two elements of the orthonormal basis. You can completely ignore the fact that they are called 'matrix elements', what the author wants to show here is how the number $\langle V'|\Omega |V\rangle$ changes when the the basis is modified $| i\rangle \to U|i \rangle $

  2. The difference is more in the lines of: for a given unitary transformation do I change the states? or the operators? The conclusion is: it is the same result. Changing the states according to $|V \rangle \to U |V \rangle$ and leaving the operators unchanged is the same as leaving the states unchanged and transforming the operators as $\Omega \to U^\dagger\Omega U$

  3. You are right, this applies even when the transformation is not unitary. The advantage of using an unitary transformation is that it preservers the orthonormality of the states $\delta_{ij} = \langle i | j \rangle \to \langle U i | U j \rangle = \langle i | U^\dagger U |j \rangle = \langle i | j \rangle = \delta_{ij}$. And this actually links back to your first question

  • $\begingroup$ On your third answer, I think you mean the unitary transformation preserves the "orthonormality of the basis vectors"? Also, I don't understand how it's connected to my first question but I'm guessing it has something to do with assuming $|V\rangle$ and $|V'\rangle$ are part of an orthonormal basis. Could you elaborate on that? Thank you very much for your answers. I have been thinking about this for the last two days! $\endgroup$ Nov 21, 2016 at 18:27
  • $\begingroup$ @KillerBunny What I mean is: if the states are orthonormal, then after the transformation, they will remain orthonormal. And about my last statement, you're right, it is the fact that under a unitary transformation, the new states will also form a basis for which you can define the matrix elements as you did in your posting $\endgroup$
    – caverac
    Nov 21, 2016 at 18:31
  • $\begingroup$ All we have talked here is about $\Omega$. What about $U$? Is it an active or a passive transformation in itself? And how does the definition of $\Omega$ being active or passive connect to as given here on this Wikipedia page? en.wikipedia.org/wiki/Active_and_passive_transformation $\endgroup$ Jan 19, 2018 at 9:38
  • $\begingroup$ But why are we concerned about that quantity <v'|A|v>? For example, I could see at A|Uv> and say that AU is the desired new operator for older bases. And ofcourse, I am wrong somewhere, but where? $\endgroup$
    – Prabhat
    Jul 19, 2021 at 18:29

Your Answer

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

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