# Is application of a Unitary operator equivalent to a phase change of a wave function?

From my understanding, the application of a unitary operator does not change the physics of the system. It is just a more convenient representation in a new eigenbasis.

Now, I have also read today that the eigenvalues of a Unitary operator are complex numbers of moduli one. It is known that multiplication by a complex number is just a rotation in complex plane.

So, can the fact that unitary operator does not change the system be understood in the way that after the application of the unitary operator the eigenstate just changes a phase (rotates in the complex plane without scaling)?

Is this understanding faulty or incomplete or is this a correct way of looking at the application of the unitary operator?

• The best counter example I can think of to this is that time evolution is unitary. This of course does more than simply change the phase. Jul 7, 2021 at 17:54
• It change the system the same way a canonical transformation does, but like a canonical transfo, it doesn't change the physics. See for example part 3.3 in the book: The schwinger action principle and effectif action (and these articles: Canonical transformations in quantum mechanics by Maciej Blaszak and Relations of canonical and unitary transformations for a general time-dependent quadratic Hamiltonian system if you want more informations ). 1/2 Jul 8, 2021 at 12:11
• As Sergei Shmakov and charlie pointed out, the time evolution doesn't just change a phase. If in quantum mechanics, the time evolution is dictated by an unitary operator, classically it can be seen as a canonical transformation... 2/2 Jul 8, 2021 at 12:12
• @Syrocco Thanks a lot for the reference! So my understanding is incorrect here. Could this be because the state vector is an infinite-dimensional vector? This might seem repetitious but I hope you understand where I am coming from. From what I've read in complex analysis multiplication of a complex number with a vector (a complex number in the complex plane) is always a rotation in the complex plane. Now, a ket is also a vector but it doesn't work here so maybe that's cause a ket is infinite-dimensional vector that's why?
– Lost
Jul 8, 2021 at 16:00

It is not equivalent for a general vector (state). You are correct that the eigenvalues of a unitary operator always have modulus one. But think about what that means. It means that if $$|\psi \rangle$$ is an eigenvector of a unitary operator U, then: $$U|\psi \rangle = e^{i\theta}|\psi \rangle$$

So this is true for all eigenvectors, but not necessarily for a general vector. Consider, for example, a vector $$|\psi\rangle = |\psi_1\rangle+|\psi_2\rangle$$, where $$|\psi_{1/2}\rangle$$ are eigenvectors of $$U$$ with different eigenvalues $$e^{i\theta_{1/2}}$$. Then: $$U|\psi \rangle = U|\psi_1 \rangle+U|\psi_2 \rangle= e^{i\theta_1}|\psi_1 \rangle+e^{i\theta_2}|\psi_2 \rangle \neq e^{i\theta}|\psi \rangle$$

So a unitary operator does preserve the magnitudes of vectors, by definition (so the probabilities are conserved - the "physics" is unchanged), but can still rotate the vector in your vector space.

• So what I guessed is only correct for eigen vectors ryt? I mean the fact that an eigenvector gets multiplied by a complex number is equivalent to a rotation in a complex plane which is nothing but a phase change ryt? Its clear from your example that generally this is not true but lets say for an eigen vector is this thinking correct then?
– Lost
Jul 7, 2021 at 20:49
• Yes, you are right, it's true only for eigenvectors. But the other part is not quite true. Multiplication by a complex number is not the same as a rotation. It's actually very difficult to visualize it with complex vector spaces, but think about a real vector space, like vectors on a plane. Multiplying by a number might change the magnitude of the vector, but it doesn't rotate it. So a unitary (orthogonal in the case of real vector spaces) operator doesn't rotate the eigenvectors, that's why they are eigenvectors in the first place. But it does rotate general vectors. Jul 8, 2021 at 4:39
• I would disagree with this line" Multiplication by a complex number is not the same as a rotation. " solely on my knowledge of complex analysis. I am sure of this. Multiplication by a complex number does represent rotation in the complex plane since every complex number can be written in the exponential form which coupled with Euler's formula makes it like rotation.
– Lost
Jul 8, 2021 at 8:36
• Yes, when you multiply a complex number by a complex number of modulus 1, that is a rotation. But you are multiplying vectors (or states) by complex numbers. Physically, $|\psi>$ and $e^{i\theta}|\psi>$ represent the same state in the Hilbert space, whereas rotation implies transforming a vector into a different one (if you use the same basis). Sometimes people say "we only care about directions in Hilbert space". That's why I said it's probably impossible to visualize it, because the multiplication of vectors as numbers on a complex plane by a complex number does not carry the same meaning. Jul 8, 2021 at 17:19
• "the multiplication of vectors as numbers on a complex plane by a complex number does not carry the same meaning" Why? What prevents me from considering a ket as a point in complex plane with infinite dimensions (that is an infinite dimensional vector)?
– Lost
Jul 8, 2021 at 19:51

There are a few concepts intertwined here that we can try to tease apart. Unitary transformations can indeed be considered changes of basis, unitary operators indeed cause changes of phase to their eigenstates, and multiplication by a complex number is indeed a rotation in the complex plane. However, lots of different things can happen when we put all of these concepts together, so let's work through a few ideas.

## When does changing basis change the physics?

Consider the action of walking on a grid defined by two unit vectors $$\mathbf{x}$$ and $$\mathbf{y}$$. We can plot this on a graph, where the horizontal coordinate tells you how far you've moved in the $$\mathbf{x}$$ direction and the vertical coordinate tells you how far you've moved in the $$\mathbf{y}$$ direction. This sets the stage for different notions of changing basis.

If you are moving in the $$\mathbf{x}$$ direction and decide to apply a unitary operation to your velocity vector, all of a sudden you might find yourself moving in the $$\mathbf{y}$$ direction. The unitary has rotated your direction of walking, so now you are doing something physically distinct from what you were doing before.

If you are moving in the $$\mathbf{x}$$ direction and decide to rotate the graph while still moving horizontally, you have applied a unitary operation only on the underlying coordinate system and so you now seem to be moving in the $$\mathbf{y}$$ direction, which again has changed the physics.

If you are moving in the $$\mathbf{x}$$ direction and decide to rotate how you look at the graph, it will look $$\mathbf{y}$$ is the horizontal direction and $$\mathbf{x}$$ is the vertical direction, but it will look like you are moving in the vertical direction, so you are still moving in the $$\mathbf{x}$$ direction such that the underlying physics is unchanged! The distinction here is that we have applied the unitary to both the coordinate system and to the motion on the graph, which amounts to looking at the identical physical situation from a different perspective.

This distinction differentiates between when unitary operations do and do not change the physics. When a unitary operation changes the basis of everything, it is as if we are looking at the same physics from a different perspective, but when a unitary operation only changes the basis of the participants or of the coordinate system or of something else, the physics will indeed change.

## Quantum description of unitaries changing basis vs. physics

Take a quantum state $$|\psi\rangle$$ that undergoes some transformation $$|\psi\rangle\to A|\psi\rangle$$. If we "change the basis" of the state $$|\psi\rangle$$ or the operator $$A$$, that means we are transforming them using a unitary $$U$$: $$|\psi\rangle\to |\psi^\prime\rangle=U|\psi\rangle$$ or $$A\to A^\prime=U A U^\dagger$$. Depending on the combinations of things that we change, the physics will or will not change.

If we only "change the basis" of $$|\psi\rangle$$, we see immediately that $$A|\psi\rangle\neq A|\psi^\prime\rangle=AU|\psi\rangle.$$ Similarly, if we change the basis of $$A$$, we see that $$A|\psi\rangle\neq A^\prime|\psi\rangle=UAU^\dagger|\psi\rangle.$$ Thirdly, if we change the basis of both $$|\psi\rangle$$ and $$A$$, we still change the physics $$A|\psi\rangle\neq A^\prime |\psi^\prime\rangle= UA U^\dagger U|\psi\rangle=UA|\psi\rangle!$$ The only way for us to not change the physics when we change the basis is if we also "look at the problem from a counter-rotated perspective" by changing the basis back at the end: $$A|\psi\rangle=U^\dagger A^\prime |\psi^\prime\rangle.$$

What happens when we want to compare two things in quantum mechanics? We take an inner product $$\langle\phi|\psi\rangle$$. If we change the basis in which we compare the two things, nothing physically changes, because $$\langle \phi|\psi\rangle=\langle\phi^\prime|\psi^\prime\rangle=\langle\phi|U^\dagger U|\psi\rangle.$$ The key is to change the basis for both things being compared. We can similarly see that $$\langle \phi|A|\psi\rangle=\langle \phi^\prime|A^\prime|\psi^\prime\rangle,$$ but always have to remember that everything must have its basis changed for us to observe identical physics.

## Unitaries only cause phase changes to their eigenstates

We can denote the eigenstates of our unitary $$U$$ by $$|u\rangle$$ and their eigenvalues by $$\lambda_u=e^{i\theta_u}$$. A system initially prepared in an eigenstate of $$U$$ will indeed observe no change in its physical properties after undergoing the unitary transformation: $$U|u\rangle=e^{i\theta_u}|u\rangle,$$ and $$e^{i\theta_u}|u\rangle$$ only differs from $$|u\rangle$$ by a global phase.

All states can be formed from eigenstates of $$U$$, so does that mean that no state will change after undergoing the operation $$U$$? We write a pure state as $$|\psi\rangle=\sum_u \psi_u|u\rangle$$ for some set of normalized coefficients $$\{\psi_u\}$$, which will evolve as $$U|\psi\rangle=\sum_u \psi_u e^{i\theta_u}|u\rangle.$$ Now, if not all of the phases $$\theta_u$$ are identical, $$|\psi\rangle$$ has been changed by more than just a global phase! This means it will have physically different properties after undergoing the evolution. It is the fact that $$|\psi\rangle$$ is made from a superposition of different eigenstates $$|u\rangle$$ that gives rise to interesting dynamics.

We can extend this to the above considerations as well. A general operator $$A$$ can also be expressed in the basis comprised from the eigenstates $$|u\rangle$$: $$A=\sum_{u_1,u_2}a_{u_1,u_2}|u_1\rangle\langle u_2|.$$ Changing the basis of $$A$$ changes the operator: $$UAU^\dagger=\sum_{u_1,u_2}a_{u_1,u_2}e^{i(\theta_{u_1}-\theta_{u_2})}|u_1\rangle\langle u_2|.$$ This is a different operator from the original $$A$$ and so the unitary has changed the physics here.

Finally, if we take two states $$|\psi\rangle$$ and $$|\phi\rangle=\sum_u\phi_u|u\rangle,$$ operate on them with $$U$$, and look at the various overlaps, we see that the physics is unchanged as before: \begin{align} \langle \phi^\prime|A^\prime|\psi^\prime\rangle&=\left(\sum_{u_3}\langle u_3|\phi_{u_3}^*e^{-i \theta_{u_3}}\right) \left(\sum_{u_1,u_2}a_{u_1,u_2}e^{i(\theta_{u_1}-\theta_{u_2})}|u_1\rangle\langle u_2|\right) \left(\sum_{u_4}\psi_{u_4}^*e^{i \theta_{u_4}}|u_4\rangle\right) \\ &=\sum_{u_1,u_2,u_3,u_4}\phi_{u_3}^*e^{-i \theta_{u_3}} a_{u_1,u_2}e^{i(\theta_{u_1}-\theta_{u_2})} \psi_{u_4}^*e^{i \theta_{u_4}}\delta_{u_3,u_1}\delta_{u_2,u_4}\\ &=\sum_{u_1,u_2}\phi_{u_1}^*a_{u_1,u_2} \psi_{u_2}=\langle \phi|A|\psi\rangle. \end{align} Again, changing the basis of everything leads to unchanged physics, but changing the basis of only one state, some states, one operator, or some other subset of everything will lead to different physics and different physical predictions.