Quantum states transform like vectors?

Question

One may define vectors as tuples satisfying the vector transformation law, or as elements of a vector space. On the latter, the superposition of two quantum states is another state, and the product of two states is non-state scalar. The vacuum state $|\psi\rangle=0$ exists and, for a scalar $a$ and a state $|\psi\rangle$, the product $a|\psi\rangle$ is still a state vector. Briefly, the axioms of a vector space are satisfied. My question regards the idea that state vectors transform like vectors.

Since states may not be position states, or states with any sort of directionality, I am not sure how to apply the concept of coordiante transformation in an abstract vector space. For instance, if I say,

"Let $\hat R(\hat n,\phi)$ be a rotation operator and $|\psi\rangle=|a\rangle+|b\rangle$ so that $\hat R|\psi\rangle=\hat R|a\rangle+\hat R|b\rangle$ obviously preserves the `angle' between $a$ and $b$, meaning that $\hat R|a\rangle$ and $\hat R|b\rangle$ are still orthogonal,"

then I am not satisfied that I have shown the proper vector transformation. Since the states themselves are the spanning basis of the state space, I am not quite getting the gist of what a coordinate transformation would be such that I could make a demonstration like the one below copied from Wolfram. What is the general proof that quantum states transform like vectors even when they are not states directly related to $\mathbb{R}^3$ coordinates? Following the form below, if I start with

$$|\psi\rangle=\sum_k a_k|\psi_k\rangle\quad\implies\quad\psi_i=a_{ik}\psi_k$$

then how should I proceed? Namely, what are my $x_i$ and $x_j$ in the vector space of quantum states?

Answer 1

I wonder if you are confusing several things here. Vectors are defined by the axioms of the vector space.

When you talk about vectors transforming in a certain way, I think you are talking about the vectors in the tangent space of some differentiable manifold. This an additional concept. Not all vectors must have a tangent space in which they 'live'. For example, differential one-forms (e.g. $dx,\,dy,\,dz$) are vectors, dual to vectors you are more used to, and they certainly do not transform in the way you describe. By definition, one-forms will do the opposite. A more life-like example is magnetic field. It is an axial vector, and as a result it transforms under reflection in a rather curious way.

As far as state-vectors transforming in a certain way. I feel like the natural way to go here is Wigner's theorem

Answer 2

States live in a Hilbert space which is typically infinite-dimensional where standard coordinate transformations don't apply. However, unitary representations of the rotational group on elements of the Hilbert space can be applied. These representations follow the following rules ($R$ -- $R_1$ and $R_2$ too -- is a rotation from the rotation group SO(3)):

$$U(R) U(R)^\dagger =1 \quad\text{and}\quad U(R_1)U(R_2) = U(R_1 R_2)$$

Trivially one has also (no rotation) : $$U(1) = 1$$

However, it depends strongly on the state what effect they might have. If for instance it acts on a momentum state $|\mathbf{p} \rangle$ one gets:

$$U(R)|\mathbf{p} \rangle = |\mathbf{ R p} \rangle$$

This means that if

$$\hat{p}_z |\mathbf{p} \rangle = p_z |\mathbf{p} \rangle$$

then the rotated state will an eigenvector to a different eigenvalue:

$$\hat{p}_z |R\mathbf{p} \rangle = Rp_z |R\mathbf{p} \rangle$$

Of course the action of the rotation $R$ on $p_z$ is one we are used to in classical physics.

A Hilbert space is a vector space which moreover has a scalar product and is complete. So states indeed live in a vector space, but not the one we used to from classical physics. In order to explain what the vector space means I give a simple example:

One can decompose a member $f$ -- we assume for this very simple example $f$ as periodic with a period of $2\pi$ -- of the Hilbert space (typically a function $\in\, L_2$, but other mathematical spaces are possible (with differently defined scalar product)) in a Fourier series:

$$f(\phi) = \sum_{n=0,\ldots,\infty} a_n E_n \equiv \sum_{n=0,\ldots,\infty} a_n e^{in \phi}$$

Then the $e^{in \phi}$ are the base vectors $E_n$ and the $(a_n)$ are the components of the Hilbert space vector $f$. (I denoted the base vectors with a capital $E$ in order to avoid confusion with the $e$ from the exponential function). The components can, as it is well known, computed by projecting the vector $f$ on the different base vectors by means of the scalar product:

$$a_n = <f, E_n> = < f, e^{in\phi}>$$

which in a Hilbert space is an integral:

$$ <f,g> = \int f(\phi) \overline{g(\phi)} d\phi$$

Therefore:

$$ a_n = \int_0^{2\pi} f(\phi) e^{-in\phi} d\phi$$