-3
$\begingroup$

The following 10 numbers:

enter image description here

completely specifies a self-propagating Schrodinger wavefunction:

enter image description here

If one is able to produce/induce these 10 excitation numbers directly onto a pre-existing wavefunction, is this equivalent to have created a particle, since the wavefunction is the complete description of a particle?

Improve this question
$\endgroup$
14
  • $\begingroup$ There's a lot of other information about the particle that's not in the wavefunction. E.g., the mass $\endgroup$
    – DanDan0101
    Dec 9, 2022 at 2:22
  • $\begingroup$ @DanDan0101 thank you. I am not fully clear what is inherent and what is not inherent in the wavefunction. Suppose we have the wavefunction of the hydrogen atom. Since this wavefunction is able to fully produce the orbital shapes of hydrogen, then the masses of proton and electron, the charges of proton and electron, etc... must have been encoded into the wavefunction for it to produce the orbital shapes correctly? $\endgroup$
    – James
    Dec 9, 2022 at 2:25
  • $\begingroup$ Please do not post images of texts you want to quote, but type it out instead so it is readable for all users and so that it can be indexed by search engines. For formulae, use MathJax instead. $\endgroup$
    – Níckolas Alves
    Dec 9, 2022 at 2:53
  • $\begingroup$ Why do you think that a wavefunction can be “completely specified” by its values at a few discrete locations? There are an infinite number of wavefunctions with these values. $\endgroup$
    – Ghoster
    Dec 9, 2022 at 6:40
  • 1
    $\begingroup$ Discretization is a reasonable approach to simulating continuous systems. But you seem to be jumping to unreasonable conclusions. The Schrodinger equation, whether discretized or not, doesn’t describe particle creation. $\endgroup$
    – Ghoster
    Dec 9, 2022 at 7:01

1 Answer 1

Reset to default
2
$\begingroup$

You mentioned a "pre-existing wavefunction". This means the particle was already there. What you did was simply collapse the particle's wavefunction to a new value.

It is possible to create and annihilate particles and these sorts of things happen all the time, but ordinary QM can't handle these phenomena. Instead, we describe them in terms of quantum field theory, which is suitable to discuss systems with variable particle number. Ordinary QM only describes a unique particle that can't be created or annihilated.

Improve this answer
$\endgroup$
7
  • $\begingroup$ thank you. Suppose we have the wavefunction of a hydrogen atom. This wavefunction exists in theory all over 3D space, is this right? Suppose now, some distance from where the main action is, we induce this 10 excitation numbers, does this mean a new particle has entered the equation? There is no measurement, because this is presupposing we can access the wavefunction directly, which normally we can't. $\endgroup$
    – James
    Dec 9, 2022 at 2:58
  • $\begingroup$ @James The wavefunction models mathematically the hydrogen atom. It is not the hydrogen atom . $\endgroup$
    – anna v
    Dec 9, 2022 at 4:46
  • $\begingroup$ @annav I guess it's the particle creation that is causing much stir in the comments... My main question is actually only this: does this 10 excitation numbers above (without any formula, just 10 numbers) completely describe a particle and how it will behave in QM (because no new particles can enter the picture), and the time evolution is completely & deterministically encoded into the wavefunction? $\endgroup$
    – James
    Dec 9, 2022 at 4:55
  • 1
    $\begingroup$ @James The answer is , no, these numbers cannot model a hjydrogen atom, they are not enough, in the same way that ten points of a pencil on paper cannot draw a building, even if they are points that belong to a complete drawing of the building. $\endgroup$
    – anna v
    Dec 9, 2022 at 4:58
  • 1
    $\begingroup$ @James that is what mathematics is used for, to enable us to pick up those numbers from the infinity of real and imaginary numbers that can model specific observations or data. Think of the straight line of a point particle in space . It is by far more smart to model it by picking coordinates and writing with them the function the line follows, than to write down an infinite number of space points. $\endgroup$
    – anna v
    Dec 9, 2022 at 5:11

Not the answer you're looking for? Browse other questions tagged or ask your own question.