Superposition - basic question

1. Sep 20, 2009

Nick R

Hello, I am brand new to this stuff and am trying to get my head around it all. I've spent considerable time trying to understand this from Landau's book on the subject (chapter 1 of course).

I bet I'd get more answers by being more brief but I always find that asking the problem carefully sometimes helps me understand the problem better.

A wavefunction, which completely describes the states of a quantum object, can be decomposed in terms of its eigenfunctions,

$$\psi = \sum_{n} a_{n}\psi_{n}$$

Eigenvalues (maybe a physical quantity) correspond to the eigenfunctions by

$$\widehat{f}\psi_{n} = f_{n}\psi_{n}$$

Where $$\widehat{f}$$ is the operator that corresponds to the quantity in question.

From this, we see that the value of $$a_{n}$$ for a given eigenfunction in the decomposition is (somehow) related to the "probability" that the physical quantity $$f$$ has the value $$f_{n}$$.

Given, is

$$\int |\psi_{n}(q)|^{2}dq = 1$$

and

$$\int |\psi(q)|^{2}dq = 1$$

How does it follow that $$|a_{n}|^{2}$$ is the probability of the physical quantity $$f$$ having the value $$f_{n}$$? The reasoning presented in the book is not clear to me - it is a sort of deductive reasoning that seems like guesswork.

Of course if this is a probability then,

$$\sum |a_{n}|^{2} = 1$$

I don't understand how this follows from the other things.

Here is why I am having a problem with this:

I can see it all works if the following is true:

$$\psi = a_{0}\psi_{0} + a_{1}\psi_{1} + ... + a_{n}\psi_{n}$$

$$|\psi| = \sqrt{|a_{0}\psi_{0}|^{2} + |a_{1}\psi_{1}|^{2} + ... + |a_{n}\psi_{n}|^{2}}$$

$$\int |\psi|^{2}dq = \int |a_{0}\psi_{0}|^{2}dq + \int |a_{1}\psi_{1}|^{2}dq + ... + \int |a_{n}\psi_{n}|^{2}dq$$

$$= |a_{0}|^{2}\int |\psi_{0}|^{2}dq + |a_{1}|^{2}\int |\psi_{1}|^{2}dq + ... + |a_{n}|^{2}\int |\psi_{n}|^{2}dq$$

Truth of this rests on the truth of two identities for complex numbers.

$$|(a+bi)(c+di)|^{2} = |a+bi|^{2}|c+di|^{2} IDENTITY ONE$$
According to my calculations this is true.

$$|(a+c) + (b+d)i|^{2} = |a+bi|^{2} + |c+di|^{2} IDENTITY TWO$$
According to my calculations this is false, unless there is a constraint $$2ac = -2bd$$.

What is going on here? Is there some sort of constraint?

2. Sep 20, 2009

Staff: Mentor

I think your missing piece of information is that the $\psi_n$ are orthogonal, that is,

$$\int \psi^*_m(q) \psi_n(q) dq = 0$$

for $m \ne n$.

This is what lets you go from

$$\psi = a_{0}\psi_{0} + a_{1}\psi_{1} + ... + a_{n}\psi_{n}$$

to

$$\int |\psi|^{2}dq = \int |a_{0}\psi_{0}|^{2}dq + \int |a_{1}\psi_{1}|^{2}dq + ... + \int |a_{n}\psi_{n}|^{2}dq$$

3. Sep 20, 2009

Nick R

Thanks I think that makes a lot of sense.

Basically "cross terms" looking similar to

$$\psi^*_m(q) \psi_n(q)$$

arise in the expression for $$|\psi|^{2}$$, and are eliminated when they are integrated, leaving only the terms looking like

$$\psi^*_n(q) \psi_n(q)$$

Thanks a bunch.