Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

I can't understand why the sum of two real waves result in a time dependent wave, but not so for the complex waves. In details, I can't get this passage on p.38-39 in A.C. Phillips, Introduction to Quantum Mechanics:

$$\tag{3.9}\Psi ~=~ A\cos (kR_1 - \omega t) + A\cos (kR_2 - \omega t)$$


$$\tag{3.10}\Psi^2 ~=~ 2A^2\cos^2\left(\frac{k(R_1-R_2)}{2}\right)\cos^2\omega t.$$

This is driving me crazy!

share|cite|improve this question
What part of the equations don't you understand? – Mew Feb 13 '13 at 23:23
Hi Chris, I can't get why you get this function if you square Fi. I've tried to get this with basic sine\cosine transformation but I failed.I tried to ask wolframalpha and it gave me the same formula but with additional terms in the argument of cos^2(wt). It seems easy, but I can't see how the second follows the first, and it's a pity because it's a fundamental concept I guess. Thank you for your reply! – Lorenzo Feb 14 '13 at 8:56

1 Answer 1

Yes, OP is right. The book contains an error. The formula (3.10) for the classical real wave$^1$ should have been

$$\tag{3.10'}\Psi^2 ~=~ 4A^2\cos^2\left(\frac{k(R_1-R_2)}{2}\right)\cos^2\left(\omega t-\frac{k(R_1+R_2)}{2}\right).$$

Note however, that the qualitative physical conclusions that Phillips draws from eq. (3.10) in subsequent paragraphs remain valid.


$^1$ Note that Phillips is here (on p.38-39) for pedagogical reasons considering a real classical wave (3.9) rather than the correct quantum mechanical wave function.

share|cite|improve this answer
A, thank you very much! I'm such a pedant. I think I've fount two other errors while reading: pg40 (3.13) should be 4A^2 and not 2A^2 pg50 should be e^-(x^2/2a) otherwise you can't normalize it – Lorenzo Feb 14 '13 at 16:07

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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