When comparing a classical and a quantum string, the resulting wave function is made up of the sum of the individual modes for the former but the product for the latter. That is at least what I gather from a lecture on quantum field theory, cf. particularly the slide at minute 5:07. The narrator says that

[Unlike the classical modes], the quantum mechanical modes are oscillating independently so the quantum state of the entire system is the product of the quantum states of the individual modes.

Is this one of those dreaded postulates? How can we explain this intuitively from first principles?

PS: Incidentally, does the superposition principle (of quantum fame) only apply within any given mode? Why would that be the case given that EM waves can be made up of infinitely many frequencies?

  • 1
    $\begingroup$ The modes of different frequencies oscillate independently in classical mechanics as well. $\endgroup$ Commented May 16, 2020 at 0:36
  • $\begingroup$ What is a classical wavefunction? Did you mean the classical field configuration, i.e. the embedding of the worldsheet into the target spacetime? $\endgroup$ Commented May 17, 2020 at 10:53
  • $\begingroup$ I'm not familiar with the notion of worldsheet or target spacetime. By classical wave function, I simply referred to the function representing the string, i.e., $u(x, t)$ in the slide in question. $\endgroup$
    – Tfovid
    Commented May 17, 2020 at 11:59

1 Answer 1


There isn't anything arbitrary going on here, it's just a consequence of how we go from a set of classical states to a set of quantum states.

Classically, the position of the string at some moment can be specified by an amplitude for each mode, so it can be written in the form $(c_1, c_2, c_3, \ldots)$. Now, the general rule here is that the quantum state is formed by considering superpositions of all possible classical positions. So for one mode, you would have an amplitude for each value of $c_1$, and hence a wavefunction $\psi(c_1)$. For two modes, you need a two-argument wavefunction $\psi(c_1, c_2)$, which represents the amplitude for having values $(c_1, c_2)$, and this goes on similarly for more modes. This fact that the general wavefunction has this form is no more than the superposition principle.

The reason that the lecturer calls this a "product" is because the "tensor product" is the mathematical tool that does this: the $n$-fold tensor product of the set of one-argument wavefunctions is the set of $n$-argument wavefunctions. But that's just a name. The principle at play is just superposition.

  • $\begingroup$ Something that could be emphasized about the last paragraph: the wavefunction is, in general, not a product of individual wavefunctions for each mode, but rather a general function. There can be entanglement between modes. $\endgroup$
    – Javier
    Commented May 15, 2020 at 23:56
  • $\begingroup$ ... but why wouldn't the quantum wave function be $\sum\limits_{k=1}^{\infty} \psi(c_{k})$, then? That's at least what it looks like to be classically. $\endgroup$
    – Tfovid
    Commented May 16, 2020 at 7:33
  • $\begingroup$ @Tfovid I think the core issue here is a mixup between fields and particles. Individual quantum particles always have wavefunctions with one position argument, describing superpositions of where the particle could be classically. But this question is about fields. The state of a field has many arguments, because it must describe superpositions of all possible classical field states. $\endgroup$
    – knzhou
    Commented May 16, 2020 at 7:36
  • $\begingroup$ @Tfovid It is always the case that when you go from classical to quantum, you need more information to specify the state. In 1D, a classical particle's position is specified by 1 number, but a quantum particle's state is specified by a whole function of $1$ argument. And a classical string's position is specified by a function of $1$ argument, but a quantum string's state is specified by a function of many arguments (formally, infinitely many if you include all the modes, i.e. it would be a function of functions, aka a functional). $\endgroup$
    – knzhou
    Commented May 16, 2020 at 7:38
  • $\begingroup$ @knzhou, I've been grappling with this for the past week but still can't wrap myself around it. I do understand how quantum mechanically the position is described by a probability distribution. My problem is, referring to the slide I mentioned above, that classically, the particle's "trajectory" is a sum of modes $u(x, t) = \sum_k q_k \cdots$ whereas quantum mechanically it is a product $\psi(x) = \prod_k \psi_k(q_k) \cdots$. $\endgroup$
    – Tfovid
    Commented May 27, 2020 at 6:35

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