Wavefunction squared is Prob. Density?

In summary: Electric field in an electromagnetic wave gives the intensity of the wave at that point. In the modern quantum sense, this also means the probability density of finding the photon at that point.Comparing with matter waves, we don't know what is actually waving. The wave itself is matter, so we can't speak of matter waving in a matter wave. That is why we've got a mathematical entity called wave-function which does the same function as amplitude of the wave. Squaring this we get the intensity i.e probability of finding the material particle per unit volume at that point. This probability/vol is nothing but probability density.Is there any empirical evidence to prove this directly? No it's not
  • #1
shredder666
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First time I learned this, I just memorized it, but now when I think about it, it seems so arbitrary. Is there a reason on why it is?

And isn't it true that wavefunction cannot be...observed? In that case how can we be sure that this statement (the title) is true?
 
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  • #2
Squared amplitude of the Electric field in an electromagnetic wave gives the intensity of the wave at that point i.e a measure of concentration of the substance which is waving in the wave. In the modern quantum sense, this also means the probability density of finding the photon at that point (Resnick-Halliday)

Comparing with matter waves, we don't know what is actually waving. The wave itself is matter, so we can't speak of matter waving in a matter wave. That is why we've got a mathematical entity called wave-function which does the same function as amplitude of the wave. Squaring this we get the intensity i.e probability of finding the material particle per unit volume at that point. This probability/vol is nothing but probability density.
 
  • #3
is there any empirical evidence to prove this directly? or will there be any "observable" evidence to prove this?
 
  • #4
shredder666 said:
is there any empirical evidence to prove this directly? or will there be any "observable" evidence to prove this?

Every directly observable property is related to the square of the wave function. The phase (which the wave function has but the square does not) comes into play when you have "interference" or superpositions. I.e. the wave function is the sum of two states. The square of a sum is not the same as the sum of the squared terms, so this is observable.
 
  • #5
Can we observe the wave function directly? No it's not a physical observable. However, the density which follows directly from the wave function is absolutely observable, for example in x-ray diffraction you are actually probing the electron cloud and then working backward to get the positions of the nuclei.
 
  • #6
This concept is often called the Born Postulate.
 
  • #7
This is one reason QM is not exactly an "elegant" theory (imo). QM can pretty much be thought of as a bunch of rules invented by people so that the answers come out correctly. When Planck invented his constant (h), he had no idea why it should exist, he just noticed that by quantizing the allowed energy levels in a black body, he could resolve the ultraviolet catastrophe. When Schroedinger came up with his equation, he had no idea what the wave-function was! He though it was some electron cloud density distribution and only later did Born come up with his probabilistic take on it. Schroedinger himself never accepted Born's postulate. So, in this sense, there is no good reason exactly why the wave-function absolute squared IS the probability distribution, but it's just that the rules and math work out that way.

This is NOT to say, though, that QM is somehow inferior. It is a VERY powerful tool that we have, and is one of the most physically accurate ones.
 
  • #8
This is true. None of the the constants were empirically determined or derived from other quantities. I wouldn't say that they invented them so that the answers came out correctly, in Planck's case the constant fell out of his derivation. A hypothesis is basically just a guess that we check against our results. They hypothesized these and it turns out that experiment agrees with them to as many decimal places as we can measure, e.g. speed of light.

Is it intrinsically satisfying? maybe not. Is the theory of gravity intrinsically satisfying? Probably not though it shades our later interpretations of things like QM because it makes us want to describe them in terms of objects we can pick up off the coffee table. Sh*ts bananas.
 
  • #9
However, whereas Einstein derived all of SR through essentially 2 postulates (principle of relativity, and constancy of the speed of light), and then went on to develop GR using the principle of equivalence, no such elegant derivations exist in QM. QM is much more "invented" in some sense.

What I mean is whereas SR and GR have overarching principles from which we derive most things, QM was developed much more ad-hoc, with many different equivalent formulations, and a lot of it is just "oh, if we describe it this way, we get a good match with experiment!". The fact that QM was developed by many many scientists while the foundations of SR and GR were more or less developed by Einstein probably has a good bit to do with this.
 
  • #10
shredder666 said:
First time I learned this, I just memorized it, but now when I think about it, it seems so arbitrary. Is there a reason on why it is?
Yes. The wave function is the description of a special kind of state called a pure state.

A general state is described instead by a density matrix R - a positive semidefinite Hermitian linear operator of trace 1 in the most general case. Its diagonal elements define the probability density. If R is diagonal and remains so in the course of time, the system behaves like a classical stochastic system. Any off-diagonal elements account for quantum behavior.

Pure states are the special case when the density matrix has rank 1. in this case, it can be written as R=psi psi^* with a vector psi, the state vector or wave function of the system. (psi^* is the conjugate transposed vector.) This wave function is not completely determined by the state since multiplying psi by any number of absolute value one (a ''phase'') gives another such vector. (This is the reason why psi cannot be observable.)

By looking at the diagonal elements of R=psi psi^*, you find that the probability density is the square of the absolute values, the precise formulation of the property alluded to in the title of the thread. Since the wave function may be complex, the absolute values matter.
shredder666 said:
And isn't it true that wavefunction cannot be...observed? In that case how can we be sure that this statement (the title) is true?
The wave function cannot be observed, but the probabilities can (to an accuracy determined by how many repetitions are feasible). If the predictions from the wave function match the observed probabilities, one concludes that the wave function seems ok.
 
Last edited:

1. What is a wavefunction squared?

A wavefunction squared is a mathematical representation of a quantum state in quantum mechanics. It is the square of the amplitude of a wavefunction, which gives the probability of finding a particle in a specific location or state.

2. How is a wavefunction squared related to probability density?

In quantum mechanics, the square of the wavefunction, also known as the probability amplitude, is directly related to the probability density. The probability density represents the likelihood of finding a particle in a specific location or state, while the wavefunction squared gives the amplitude of that probability.

3. Why is wavefunction squared used in quantum mechanics?

Wavefunction squared is used in quantum mechanics because it provides a way to describe and calculate the probabilities of a particle's position or state. In the quantum world, particles do not have definite positions, so the wavefunction squared allows us to understand the likelihood of where a particle may be found.

4. How is wavefunction squared different from a traditional wavefunction?

Wavefunction squared is a more useful representation of a quantum state compared to a traditional wavefunction. While a traditional wavefunction gives the amplitude of a particle's probability, the wavefunction squared gives the actual probability of finding the particle in a specific location or state.

5. Can wavefunction squared be negative?

No, wavefunction squared cannot be negative. In quantum mechanics, probabilities must always be positive values, and since wavefunction squared represents the probability density, it must also be positive. If a negative value is obtained, it is interpreted as a phase shift in the wavefunction and does not affect the overall probability.

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