Superposition of quantum states when states are rays: well-definedness question

Question

I have some confusion (uncertainty? :)) about the compatibility between states and superposition in quantum mechanics. I give a bit of background context and then ask my question at the end of the 4th paragraph.

In the mathematical description of quantum mechanics, here is what I understand states of a system to be. First, they may be described by a unit vector (wave function) $\psi$ in a Hilbert space, with the proviso that all unit vectors of the form $e^{i\theta}\psi$ describe the same state as $\psi$. Second, we could say states of the quantum system are described by nonzero vectors $\psi$ not necessarily of unit length, with the proviso that all of its scalar multiples $c\psi$ for nonzero $c \in \mathbf C$ describe the same state as $\psi$. (The standard convention to focus on states as unit vectors is made to simplify the probabilistic interpretation given by the Born rule, I suppose.) Third, since nonzero vectors that are not scalar multiples of each other are supposed to describe different states of the system, the states can be described without any redundancy as the $1$-dimensional subspaces of the Hilbert space.

The states of the quantum system can undergo superposition: we are allowed to form linear combinations of states and this leads to new states (with dynamics described by Schroedinger's equation). This is why vector spaces are the framework in which quantum states are described.

However, different $1$-dimensional subspaces of a vector space can't be added to get a single $1$-dimensional subspace as a result: if $\psi_1$ and $\psi_2$ are two linearly independent vectors in the Hilbert space, then $\psi_2$ and $-\psi_2$ are supposed to represent the same state but $\psi_1 + \psi_2$ and $\psi_1 - \psi_2$ (more generally, $\psi_1+\psi_2$ and $\psi_1 + c\psi_2$ where $c \not= 1$) are not scalar multiples of each other so they don't belong to the same $1$-dimensional subspace. The set of all linear combinations $\{a\psi_1 + b\psi_2\}$ is a $2$-dimensional subspace, not a $1$-dimensional subspace. Therefore it seems that superposition is not something that can be done to states, as it would not be well-defined. Yet all accounts of QM that I have looked at refer to "superposition of quantum states". The individual $\psi$'s in the Hilbert space are not supposed to be physically meaningful on their own, while the $1$-dimensional subspaces they each span (the rays of the Hilbert space) are physically meaningful, yet the superposition concept makes sense as an operation on the $\psi$'s but makes no sense on the $1$-dimensional subspaces. How is this conundrum resolved? That is my question here.

In contrast to my confusion over states as $1$-dimensional subspaces not being compatible with superposition, there is "no problem" working with $1$-dimensional subspaces when we form a composite of two quantum systems, since the mathematical description of composite systems uses tensor products and a tensor product of $1$-dimensional subspaces is very naturally a $1$-dimensional subspace again.

Answer 1

Superposition is not exactly a linear operation. It's a projection of one. Or perhaps I should say that not all linear combinations are superpositions.

Solutions to the Schroedinger equation form a Hilbert space. But, only normalised solutions are accepted as states. Normalised solutions do not form a Hilbert space. They form a sphere embedded in a Hilbert space. It can be convenient to use the Hilbert space to talk about the states, but then the operations require renormalisation.

When you superimpose by forming an arbitary non-degenerate linear combination of solutions you potentially get an element not on the sphere - which needs to be rescaled to the sphere for it to be normalised. To make sure that the combination is on the sphere, it is required to limit the choices to coefficents whose sum of squares is unity.

Since only points on the sphere are important, we can work with the line (point in the grassman manifold) through the centre to represent the point on the sphere. But, then the operation to work out which line through the centre corresponds to a superposition of two other lines through the centre is not computed by taking the direct sum of the two subspaces.

The state is the normalised solution to the Schroedinger equation. It is a point on the sphere, which is a unit vector in the Hilbert space. Representing this as a line through the origin is convenient in some ways, and inconvenint in others. But, in any case - the superposition operation is not equivalent to linear combinations in the Hilbert space.

Answer 2

1. The physical states of a quantum system are equivalence classes on the unit complex sphere.

In a mathematical sense, a quantum state is a unit vector in a Hilbert space over the complex numbers. However, two of these states can be physically equivalent. The only physical significance comes from $\left<\psi^*|\Theta|\psi\right>$ where $\psi$ is the quantum state and $\Theta$ is an operator corresponding to some observable. Hence if $a$ is a unit complex number $\left<(a\psi)^*|\Theta|(a\psi)\right> = (a^*a)\left<\psi^*|\Theta|\psi\right>$, and $a^*a=1$, so no observable can distinguish the states $\psi$ and $a\psi$. In a physical sense it could be said that $a\psi$ is the same state as $\psi$, even though they are different actual vectors. Mathematically, the term state can be transferred to the equivalence classes - a state in this sense is a collection of unit vectors that are physically equivalent.

Without a literature survey, I would not be sure, but my instinctive reaction is that most writers use the term state to refer to each unit vector, but accept the concept that two states in this sense can be the same (equivalent). The switch between using the equivalence class and the representative of the class is something that happens commonly in mathematical discussion and is intended to be clear from the context. In terms of physical science - this is related philosophically to Mach's principle that nothing that is not observable should be included in a theory. While this sounds like a nice idea, in practice it is impractical and tends to invoke counter productive linguistic convulsions.

Now, the equivalence classes are generated by the relation that $\psi_1 \equiv \psi_2$ iff there exists $a \in C$ such that $\psi_1 = a\psi_2$. If this was real n-space instead of complex n-space then this would equate points on opposite sides of the sphere. Each equivalence class is a pair of points. Through each such pair is exactly one line through the origin - that is, one 1D subspace. So, the equivalence classes, the physical states, can be put in one to one correspondence with the 1D subspaces. The topological space of all these subspaces is a known mathematical object and so the physical states of a quantum system can be studied mathematically by using this space of 1D subspaces.

2. phase is not physical but relative phase is physical

While $a\psi$ and $\psi$ are the same state for $a$ a unit complex number. $a_1\psi_1 + a_2\psi_2$ can be physically different for different values of $a_1/a_2$, which is (check the exponential expression) the relative phase shift between the two vectors in the superposition.

So, given $\psi$, some state, and $\psi_1$ and $\psi_2$ which are two equivalent states then $\frac{1}{\sqrt{2}}(\psi+\psi_1)$ and $\frac{1}{\sqrt{2}}(\psi+\psi_2)$ might not be physically equivalent.

This is not a contradiction - the operation of superposition in and of itself is not so much a physical operation as a mathematical one. In various applications of the theory it is possible to determine the state of a particle to be the superposition of two other states, and in doing this one also has to work out the relative phase. The absolute phase is not measurable.

The states are physically indistinguishable but not mathematically indistinguishable.

The non-measurable nature of the phase refers, strictly, only to the state of the entire quantum system. Physically you cannot have two quantum systems, you only ever have one. We only approximate the case of two relatively non interacting quantum systems in terms of two systems that are then superposed.

Superposition is, strictly, just a mathematical device used to describe a single quantum system in terms of two other quantum systems. Like describing an integer as the product of primes. In quantum theory, the quantum system is usually decomposed mathematically into a superposition of eigen states. But, the only thing that exists in the physical sense is the entire singular quantum state no matter what it is.