I have to admit that I don't know much about String Theory (or QFT, for that matter ..), but, if we assume that String Theory is the correct description of fundamental particles, what is the correct way to think about a wavefunction, in the context of String Theory?

For example, take an electron being shot from an emitter to a target. I usually imagine the wavefunction as spreading out from the target in three dimensions, possibly traveling through two or more slits; interfering with itself along the way; and then spontaneously deciding to collapse down onto one of the atoms that makes up the target surface. Where does the string fit into this picture?

Does String Theory say anything about the mysterious wavefunction collapse? (I assume it must do, otherwise can it really be described as a 'theory of everything'?)


It was mentioned in one of the answers that, in string theory, 'point particles' are described as strings. Hang on though .. in QM we were told: "there are no point particles, there are only wavefunctions." But, now in string theory, apparently these point particles are back again, only they're now described as strings. So, where did the wavefunctions go?

  • $\begingroup$ String theory is just a special case of quantum mechanics. You can superpose different string configurations, just like you superpose different particle configurations. The collapse is there too, with the same issues. $\endgroup$ – knzhou Aug 28 '18 at 0:14
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    $\begingroup$ If you think of the SM, string theory, etc. as programs, quantum mechanics is the operating system they all run on. $\endgroup$ – knzhou Aug 28 '18 at 0:14
  • $\begingroup$ @knzhou but if string theory is supposed to be the 'theory of everything', shouldn't that be the operating system that QM is running on? I thought QM was supposed to be a special case of string theory - i.e. where gravity/relativity is not important? $\endgroup$ – Time4Tea Aug 28 '18 at 0:26
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    $\begingroup$ No, "theory of everything" is just a general term for a quantum theory that includes both the Standard Model and gravity. It does not, contrary to the name, explain literally everything. $\endgroup$ – knzhou Aug 28 '18 at 0:38

I think one important thing to mention is that Copenhagen Interpretation of Quantum Mechanics is not the only interpretation out there; in particular there is no "need" for wavefunction collapse if you don't want it. Alternative interpretations include the Many Worlds Interpretation (which makes sense in light of the development of Decoherence in quantum theories; for more on this interpretation I'd recommend this paper by Max Tegmark).

I don't think String Theory should be viewed as something that can resolve between different interpretations of Quantum Mechanics; after all String Theory is still a Quantum Theory. In particular, you still have wavefunctions as before; but instead of describing what you might think of as 'point particles', they describe strings.

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  • $\begingroup$ I think one thing I am confused about is that in QM, we were told: "there are no 'point particles', only 'wavefunctions'". Now, in String Theory, we're being told: "so, those point particles that we did away with in QM are actually strings ..". So, what happened to the wavefunctions then? Perhaps this should be a different question, but it seems like these 'particles' can be whatever we want/need them to be, to suit a given situation. $\endgroup$ – Time4Tea Aug 28 '18 at 0:23
  • $\begingroup$ Who told you that? Wavefunctions aren’t physical; amplitudes are physical, and these give the probability of finding point particles in the relevant states. In string theory wavefunctions merely apply to strings instead of point particles. $\endgroup$ – bapowell Aug 28 '18 at 12:01
  • $\begingroup$ @bapowell surely they must be physical to some extent, if they can interfere with themselves? $\endgroup$ – Time4Tea Aug 28 '18 at 12:30
  • $\begingroup$ What I'm saying is that we don't observe wavefunctions; we do observe particles. $\endgroup$ – bapowell Aug 28 '18 at 14:05
  • $\begingroup$ @bapowell so, basically, the 'strings' are an alternative way to describe particles in the situations where they can be considered to act like discrete 'point particles' (such as when they interact with other particles)?. When they are traveling through space, the strings should be considered as a wavefunction, like in QM? $\endgroup$ – Time4Tea Aug 28 '18 at 14:55

One of the tenets of quantum mechanics is that the wave function characterises fully the state of the system. In other words, all you can know about a particular system is encoded in the wave function.

So in the case of a point particle, the wave function encodes the probabilities of its position, momentum and energy. The electron also has spin, so the wave function encodes information about that too.

In the case of a 1D string, instead of measuring the position, one might ask “what is the string’s vibrational state”? So the wavefunction encodes that too.

Now, note that your way of thinking about the wavefunction of a particle as something spread out in physical space only works in the case of a single point particle! Already for the simple case of a particle with spin 1/2 you would need two such wavefunctions (one for the spin up and one for the spin down). I won’t go too deep into it, but the wavefunction of a system is a function that assigns to each result of (compatible) measurements, a complex number.

So back to the point particle case, the $\psi(x)$ is a complex number assigned to the position $x$. So you can confuse it with, or think of it as, a function on physical space, even though it is a function on the possible results of measuring the position. In the case of a two particle system, the wavefunction is of the form $\psi(x_1, x_2)$ so its a function from the 6 dimensional space of pairs of 3D vectors $(x_1, x_2)$ to the complex numbers. Good luck thinking of this as a function on physical 3D space!

With regards of “what is real” in quantum mechanics, the jury is still out. The maxim “there are no point particles, only wafefunctions” is a bit misleading. While it’s true that you can’t reason correctly about quantum systems as simple billiard-ball-like point particles moving in potentials, not everyone would agree the wavefunction is real, or physically real, whatever that might mean for them. The wavefunctions are essential tools for reasoning and calculations, but you can’t think of them, consistently, as physical objects out there in physical space, like one might think of sound waves or light waves.

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  • $\begingroup$ Thanks for your answer, but your third paragraph seems to be incomplete. $\endgroup$ – Time4Tea Aug 28 '18 at 12:11
  • $\begingroup$ @Time4Tea thanks! I’m in mobile and I had lost that sentence in the editor. I’ll take it out $\endgroup$ – Andrea Aug 28 '18 at 12:31
  • $\begingroup$ Thanks. As I mentioned in a comment to one of the other answers though, surely the wavefunctions must be physical to some extent, if they can interfere with themselves, in response to a particular configuration of physical slits? The case of a wavefunction of two particles is interesting, which I hadn't considered. I don't know much about the technical details of QM, but wouldn't that combined wavefunction be given by: $\psi(x_1,x_2)=\psi_1(x_1).\psi_2(x_2)$ (i.e. a superposition)? In that case, the wavefunction for each particle could be visualized in 3D space. $\endgroup$ – Time4Tea Aug 28 '18 at 12:54
  • $\begingroup$ I guess clearly I need to try to read more into the details of QM in general, which might help me understand better. $\endgroup$ – Time4Tea Aug 28 '18 at 12:55
  • $\begingroup$ @Time4Tea The cases when you can write $\psi(x_1, x_2) = \psi_1(x_1)\psi_2(x_2)$ (this is called a tensor product of single particle states) are quite uncommon in practice. They require that the particles haven’t interacted with each other (i.e. they are not entangled) and even then, that they are of different kind (this is known as the spin-statistics theorem, or Pauli exclusion principle). In general, the state of the combined system will be a sum (also called superposition) of such tensor product states. The simplest case is of two bosons with no spins, where you can write: $\endgroup$ – Andrea Aug 28 '18 at 13:48

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