Quantum mechanically, is superposition a sum or a product?

Question

This question may sound like a no-brainer, but I'm getting confused after watching this lecture (cf. the slide at minute 5:07).

The context is to motivate the quantization of a field which, for the sake of simplicity, is taken to be a classical, one-dimensional string of length $L$ that is approximated by infinitesimally many oscillating springs. (This is incidentally inspired by Born, Heinsenberg, and Jordan's seminal paper of 1925.) After a lengthy derivation, one gets that the "trajectory" of the classical string is \begin{equation} u(x,t) = \sum\limits_{k=1}^{\infty} q_k (t) \sin\frac{k\pi x}{L} \end{equation} where $q_k$ denotes the harmonic mode $k$. This superposition can thus immediately be recognized as nothing more than a Fourier decomposition.

The lecturer then goes on to argue that the quantum mechanical analog is given by \begin{equation} \Psi = \prod\limits_{k=1}^{\infty} \psi_k (q_k). \end{equation}

While $u$ is a sum, $\Psi$ is instead expressed as product of the various modes. Yet, I've always thought of superposition as a sum---or is that valid only within a mode, not between modes? What am I not understanding here?

EDIT: The lecture in question is titled "Quantum Field Theory 2b - Field Quantization II" by the YouTube channel named ViaScience.

Answer 1

$\Psi$ is not $u$, it is the quantum state that has no classical analogon. In quantum mechanics, or let's say quantum field theory, the observable $u(x, t)$ becomes an operator at each point (that does not necessarily commute with other variables) that acts on a state $\Psi$.

Now, we don't know how to quantize this system of continuously labelled operators, our way out is to rewrite the system as a linear superposition of independent harmonic oscillators (namely, in terms of the classical eigenmodes of the string).

You will see, that $q_k$ is governed by the equation for the harmonic oscillator with a $k$ dependent frequency, but completely independent from the other modes. So we allow ourselves the analogy of quantizing this equation like the harmonic oscillator. (This step is kind of a leap of faith – we do this and study the consequences, we don't know a priori that this yields the correct quantum field theory.)

So our quantization procedure takes $q_k(t)$ and $u(x, t)$ from being real functions to being indexed sets of operators (in the Heisenberg picture). Now each of these oscillators is described by a state $\psi_k(q_k)$ (this state has no time dependence since we are in the Heisenberg picture, where the operators carry the time dependence), and since they are independent the total state is the product of the states of the independent sub-systems (and a function(al) of all the $q_k$): $$ \Psi = \prod \psi_k(q_k). $$

So the formula for $u$ does not represent quantum mechanical superposition, which we talk about if we have a state that is linearly combined from a basis of states. The analogy is rather that of the momentum space and position space representation of elementary quantum mechanics, that are related by a linear transformation as well.