In Ballentine's book on QM in Chapter 3 he states, that if a state vector is transformed there is a corresponding transformation on the operators associated with observables. In math, if $$ |\mathbf{\psi'}\rangle = U|\mathbf{\psi}\rangle$$


$$ A |\mathbf{\psi} \rangle = a_n |\mathbf{\psi} \rangle $$

Then there must be a transformed observable with.

$$ A' | \mathbf{\psi'} \rangle = a_n |\mathbf{\psi'} \rangle $$

Which implies that

$$ A' = UAU^{-1} $$

However, I am not quite following the justification for the statement that the eigenvalues must be the same above. I get confused by the active and the passive view of transformations and how it is applied here. Ballentine says he is using the active point of view. So a state vector, $|\mathbf{\psi} \rangle$, gets moved to a new location in the same coordinate system.

Why do we also transform the observables such that the following is true? $$ A' | \mathbf{\psi'} \rangle = a_n |\mathbf{\psi'} \rangle $$


Based on the comments below, there is some confusion about my exact questions. I understand that if you transform $A$ such as

$$ A' = UAU^{-1} $$

That you will get,

$$ A' | \mathbf{\psi'} \rangle = a_n |\mathbf{\psi'} \rangle $$

My question is why are we not just using the same operator A on the transformed state vector, $|\psi \rangle$?

  • $\begingroup$ The following might be useful: $A'|\psi'\rangle=(UAU^{-1})(U|\psi\rangle)=UA(U^{-1}U)|\psi\rangle=U(A|\psi\rangle)=a_n(U|\psi\rangle)=a_n|\psi'\rangle.$ $\endgroup$
    – Charlie
    Nov 14, 2020 at 19:02
  • $\begingroup$ @Charlie, that wasn't my question. I am asking why we even transform the observables as $A'=UAU^{-1}$? Why not just keep the observables the same and measure a new position say with the position operator after a translation? $\endgroup$
    – Jeff
    Nov 14, 2020 at 19:27
  • $\begingroup$ @Jeff Ah! Then I misunderstood your question as well, I would suggest slightly editing it to be a little clearer. :) $\endgroup$
    – Philip
    Nov 14, 2020 at 19:30
  • $\begingroup$ I guess this is a related question. physics.stackexchange.com/questions/580968/… $\endgroup$
    – Jeff
    Nov 14, 2020 at 19:34

2 Answers 2


Let me first show you what's really going on here, your problem will short out automatically. I'll talk in terms of 2D transformation so that It will become easy to visualize.


The following says, that when you apply a transformation on some vector, you get a new vector (for visual view). The next one is $$A|\psi\rangle=a_n|\psi\rangle$$ This suggests that a vector $|\psi\rangle$ is an eigenvector of $A$ that mean after the transformation it's just scale by a factor.

Now the question we want to ask is this If I apply a transformation $U$ on whole space so that every vector gets transformed and so is $|\psi\rangle$ which is a eigen vector of $A$, What will be a new transformation $A'$ for which $|\psi\rangle$ will be eigen vector?

My question is why are we not just using the same operator A on the transformed state vector, $|ψ⟩$?

The reason is simple because transformed $|\psi\rangle$ is no longer eigenvector for $A$. But I want to give a little more sense.. so get along with me.

Let's go back to out question of interest. Now for simplicity, We can think of $U$ as rotation. The idea is that after the tranformation (rotation ) every vector which is in direction of $|\psi\rangle$ should lie on the same line after the tranforamation. For example after 90 degree rotation $\hat{i}$ and $2\hat{i}$ will remain parallel.

So we suspect that every eigenvector of $A$ should be on the same line after the tranformation. Now to find such matrix : We first reverse the effect of rotation by an inverse matrix. So that we back to our original state so $U^{-1}U|\psi\rangle$. The next step is to use the fact that we know the tranformation for which this is eigen vector and so $AU^{-1}U|\psi\rangle$. Now we will again apply our tranformation to reverse the effect $UAU^{-1}U|\psi\rangle$.

In Active picture this said nothing but that after the tranformation the eigenvector lie on the same line.

But there is something very nice is going on Passive picture. That's you to find As short note:


This suggest a short of mathematical empathy. That's is to change in perspective. The matrix $U$ is what change is prespective.If you didn't get, here.

  • $\begingroup$ It still isn't wholly satisfying for me. The position observable, $X$, has eigenvectors for every point in space. So if you do a space translation by $\mathbf{a}$, of $|x\rangle$, $|x'\rangle = |x+a\rangle$. But, Ballentine is insisting that one also translates the position operator $X$. $\endgroup$
    – Jeff
    Nov 15, 2020 at 0:22
  • $\begingroup$ I think the key thing here is that Ballentine wants to perform a transformation on a state vector but not change any of the observables. I think that is what you were alluding to with keeping the eigenvectors on the same line under the rotation. $\endgroup$
    – Jeff
    Nov 15, 2020 at 4:19
  • $\begingroup$ When you didn't change the operator that's what you call the active transformation that's the case for which I have given explanation. But you can do the same with passive transformation where the transformation will change. I hinted on last paragraph. $\endgroup$ Nov 15, 2020 at 5:11

I think I understand now. I'll say that I think the book's description is confusing.

Essentially, what Ballentine is saying is that the laws of phyiscs are invariant under Galilean transformations.

In other words, if we have some state vector, $|\psi\rangle$, and we perform a space time transformation to another frame of reference then the laws of physics should be the same.

So, for example if we perform a translation,

$$|\psi'\rangle = e^{-i\mathbf{a}\cdot \mathbf{P}/\hbar}|\psi\rangle $$

Then, for each Observable, there should be a way to translate them also, so that if an observer was also translated they would observe the same things as in the untranslated system. If you look at at eigenvector of an observable this means that for,

$$ A|\phi_n\rangle = a_n |\phi_n\rangle$$ $$ A'|\phi'_n\rangle = a_n |\phi'_n\rangle$$

In other words there exists an A' such that for that translated system is observed the same as the untranslated system, with

$$ A' = U A U^{-1} $$

For the position case this ends up with, $$Q' = Q-\mathbf{a}\cdot I$$

Or in other words, the translated observer will end up subtracting $\mathbf{a}$ from the positions they record.

The presentation of all of this feels a bit muddled to me. Really, we have two observers in two different coordinate systems. We are essentially finding an operator $A'$ such that for the translated state vector, $\psi'\rangle = U \psi\rangle$, in our coordinate system, that tells us what the observables would look like to the observer in a translated coordinate system.

  • 1
    $\begingroup$ No, it's not only referring to Galilean transformations, but to more general symmetries. See this answer to another related question on symmetries in quantum mechanics. $\endgroup$ Nov 15, 2020 at 17:54
  • $\begingroup$ Thanks, Massimo. Can you provide an example? Ballentine, says "The laws are invariant under certain space-time symmetry operations, including displacements, rotations, and transformations between frames of reference in uniform relative motion. Corresponding to each such space-time transformation there must be a transformation of observables." I believe what you are saying is correct, but I am wondering what a concrete example is and whether it renders my answer incorrect if I try an extend it to all symmetries. $\endgroup$
    – Jeff
    Nov 15, 2020 at 19:00

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