Quantum - Postulate of getting specific momentum p = |A_p|^2

[laser]

Homework Statement

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Relevant Equations

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The first equation states that every wavefunction can be written as a sum of wavefunctions of definite momentum, with A_p being defined as the coefficients in the expansion such that when you take the |wavefunction|^2 it equals 1 - fine.

We then multiply by the wavefunction conjugate and integrate, and using some Kronecker delta trickery we get an expression for A_p. - I'm also okay with this.

But then we say that P(p) = |A_p|^2 is a postulate... From my experience in maths/physics, postulates should be "facts". But here we see that A_p came from somewhere... why are the coefficients in the expansion related to the probability of finding a certain momentum? Is this just experimentally observed and we take it to be true?

Please let me know how you think about this, as I feel like I am thinking of the notion of a "postulate" wrongly.

 

[kuruman]

But here we see that A_p came from somewhere... why are the coefficients in the expansion related to the probability of finding a certain momentum?

You have
$$\psi(x)=\sum_l A_l ~\psi_l(x)$$ It follows that $$\psi^*(x)=\sum_m A_m^* ~\psi_m^*(x)$$Now suppose you consider $$\int_0^L \psi(x)\psi^*(x)dx~,$$ substitute from above and use some more Kronecker delta trickery. What do you get?

 

[laser]

Firstly, I am not sure why you changed the subscripts when going to the wavefunction conjugate.

Also, the equation you asked me to consider is the same one in my image! So I will get A_m. But I don't see how that answers my question.

 

[kuruman]

Firstly, I am not sure why you changed the subscripts when going to the wavefunction conjugate.

Because they are two different summations and anyway the indices are dummy.

Also, the equation you asked me to consider is the same one in my image! So I will get A_m. But I don't see how that answers my question.

It is not the same. Take a second careful look. Like I said, what happens when you substitute the summations with two different indices in the equation that I asked you to consider.

 

[laser]

Because they are two different summations and anyway the indices are dummy.

It is not the same. Take a second careful look. Like I said, what happens when you substitute the summations with two different indices in the equation that I asked you to consider.

Ah okay I see. You get the sum over m of |A_m|^2

which is equal to the above image.

As it turns out, the above image is also equal to the probability density over all the space = 1! (from another postulate)

This leads to another question: why is |A_p|^2 = probability density a postulate as well as the |wavefunction|^2 = probability density? Surely just one of them being a postulate is enough!

 

[kuruman]

This leads to another question: why is |A_p|^2 = probability density a postulate as well as the |wavefunction|^2 = probability density? Surely just one of them being a postulate is enough!

The postulate is
$$\int_0^L \psi(x)\psi^*(x)dx=1.\tag{1}$$
Please provide the exact statement of the problem as was given to you and how "postulate" is mentioned in it. Note that $$\sum_p |A_p|^2=1$$ is another way of saying the same thing as equation (1).

 

[laser]

The postulate is
$$\int_0^L \psi(x)\psi^*(x)dx=1.\tag{1}$$
Please provide the exact statement of the problem as was given to you and how "postulate" is mentioned in it. Note that $$\sum_p |A_p|^2=1$$ is another way of saying the same thing as equation (1).

I'm just reading some notes, it says that equation (1) is a postulate, but it also said that the sum of |A_p|^2 = 1 is also a postulate. Probably just a mistake in the notes, as either postulate can be used to derive the other!

Source: https://oyc.yale.edu/sites/default/files/notes_quantum_9.pdf page 21 equation (34)

 

[kuruman]

I wouldn't worry about it.