# QM Basis Transformation Through Unitary Operator

I've a short question about basis transformations in QM. Suppose I have two bases $$\{|{\phi_n}\rangle\}$$ and $$\{|{\phi_n'}\rangle\}$$. For brevity, we can make them orthonormal. I know that any state vector can be expanded in terms of both bases:

$$|\psi\rangle = \sum_n |\phi_n\rangle \langle \phi_n | \psi \rangle = \sum_n |\phi_n'\rangle \langle \phi_n' | \psi \rangle$$

I also understand that the operator mapping $$|\phi_n\rangle$$ to $$|\phi_n'\rangle$$, $$\hat{U}$$, is a unitary operator.

Now I have always been told that a change of basis does not change the state vector. If I have expressed my state vector in the basis $$\{|{\phi_n}\rangle\}$$ and want to express it in terms of $$\{|{\phi_n'}\rangle\}$$ I should just apply the identity transformation $$\hat{1} = \sum_n |\phi_n'\rangle\langle\phi_n'|$$ and I will have changed the basis.

In this process, the corresponding column matrix will change as if a unitary matrix was applied. From this observation, Zettili's book page 115-116 and the lecture notes of a course I'm following conclude that in order to express a state vector in a new basis, we should actually apply a unitary operator like this:

$$|\psi_\text{new}\rangle = \hat{U}|\psi_\text{old}\rangle$$

I've seen this conclusion in multiple places now, but cannot quite follow. Why do we change the state $$|\psi \rangle$$? Isn't the state invariant of the basis chosen? That is the approach of for instance Sakurai in section 1.5.

Or are we keeping the basis fixed while applying the unitary operator? An analogy I see often is a 2D-rotation. Rotating the $$(x,y)$$ coordinate system through an angle $$\theta$$ to become $$(x', y')$$ while keeping the vector fixed is a transformation of basis (from the $$(x,y)$$ to the $$(x',y')$$-system). But mathematically, I can also rotate the state vector through an angle $$-\theta$$ and treat the old $$(x,y)$$-axes as the new $$(x', y')$$-axes. Is this what the sources that confuse me are doing (in a QM context)?

The state doesn't change, of course, but its representation in the new basis does. This is quite intuitive, let's take the example of rotations in 2D space: consider a unit vector in the $$xy$$ plane that is given by $$V = \begin{pmatrix}1\\0\end{pmatrix}.$$ If I were to describe this vector in a basis $$x'y'$$ which is rotated with respect to $$xy$$ by an angle $$\theta$$ clockwise, it shouldn't be too hard to see that it corresponds to $$V' = \begin{pmatrix}\cos{\theta}\\\sin{\theta}\end{pmatrix}.$$

The vector stays "the same", but its representation changes depending on the choice of the coordinates: what was earlier pointing along the $$x$$ axis will now appear to be point along some line along the angle $$\theta$$. If we wanted to relate these two representations, in 2D we'd use a special orthogonal transformation (a rotation), and we could say that

$$V' = R V,\quad \quad \text{where}\quad R = \begin{pmatrix}\cos\theta &\sin{\theta}\\ -\sin{\theta}&\cos{\theta}\end{pmatrix}.$$

Such a transformation preserves the lengths, orientations, and dot-products of vectors. Similarly, in QM, the transformations that preserve the "length" (inner products) are unitary transformations $$\hat{U}$$, since they keep the dot-product invariant, since $$\langle\phi'|\psi'\rangle = \langle\phi'|\hat{U}^\dagger\hat{U}\psi\rangle =\langle\phi|\psi\rangle.$$

• I'm not sure I follow. In dirac notation, the state I start out with is $|\psi\rangle = |x\rangle$. Then in the new basis we get $|\psi \rangle = \cos \theta |x'\rangle + \sin \theta |y'\rangle$. This is perfectly clear to me: the vector stays identical but the matrix elements change (through a unitary matrix!). But if I were to apply a unitary OPERATOR to my state vector I would get a completely different vector: $\hat{U}|\psi\rangle = \cos \theta |x\rangle - \sin \theta |y\rangle$. That's only the same if I start "pretending" that the basis vectors after the transformations are "new", right? Aug 30, 2020 at 18:57
• So in that case we have $|V'\rangle \neq |V\rangle$? Since $\hat{R}$ is not the identity operator? Aug 30, 2020 at 19:10
• Yes, they both represent the same physical state (i.e., the stick) but their representations are different in our individual bases. Aug 30, 2020 at 19:11
• For example: consider two people ($A$ and $B$) moving relative to each other, observing a particle in a state of definite momentum. They both measure the particle's momentum: $A$ measures a momentum $p_A$, and $B$ measures a momentum $p_B$. $A$ sees the state of the particle as being $|p_A\rangle$, and $B$ sees it as being in $|p_B\rangle$. Of course, there is nothing special about moving constantly relative to each other, and so their worldviews are equivalent, and all probabilities that $A$ will find $p_A$ and $B$ will find $p_B$ are the same. The transformation between states is unitary. Aug 30, 2020 at 19:21
• Sorry for asking so many questions, but isn't the representation just the associated column matrix? I know that the column matrix, containing the components $\langle x' | V \rangle$ etc. changes with a unitary transformation. But the notion that the state vector changes ($|V\rangle \neq |V'\rangle$) appears to completely contradict the fact that we can express a ket vector in multiple bases through the identity transformation: $|\psi\rangle = \sum_n |\phi_n\rangle \langle \phi_n | \psi \rangle = \sum_n |\phi_n'\rangle \langle \phi_n' | \psi \rangle$ Aug 30, 2020 at 19:22