# Probability amplitude in basic quantum mechanics

I came across this proportionality statement in my quantum mechanics notebook: $\psi(x,t)$ is proportional to

\begin{align} \cos(kx - wt) &= \exp(i(kx-wt)) + \exp(-i(kx-wt)) \\ &= \exp (i(kx-wt)) \end{align} I looked through most commonly used textbooks for quantum mechanics and I couldn't find this in them. Can you help me figure this out? Thanks.

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The equality $\exp(i(kx-wt)) + \exp(-i(kx-wt)) = \exp (i(kx-wt))$ is incorrect, perhaps you recorded something else incorrectly too?

Generally, wavefunctions that go as $e^{i(kx - \omega t)}$ represent travelling waves and are used in calculating transmission/reflection coefficients for potentials, while a wavefunction that behaved like $\sin(kx)$ represent a standing wave. I'm not sure which one you wrote down.

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Well actually it's a proportionality sign. But I didn't know how to enter it. This comes somewhere near Born's rule and I noticed some factors missing. From Euler's formula, it is understood that there has to be a factor of 2. But the exp(i(kx-wt)) appeared out of nowhere. – Artemisia Sep 30 '13 at 4:22
Hmm it's incorrect even with a proportionality sign. Maybe you're thinking of $e^{i\theta} + e^{-i\theta} = 2 Re(e^{i\theta}) = 2\cos(\theta)$. – xuanji Sep 30 '13 at 4:25
Yes that's what I had in mind. But this was blindly copied down during the lecture without much thought. – Artemisia Sep 30 '13 at 4:27
If it's near Born's rule (en.wikipedia.org/wiki/Born_rule), perhaps it's just a wavefunction given as an example to demonstrate applying Born's rule on. BTW, you can use latex on physics.SE (meta.math.stackexchange.com/questions/5020/…) to enter things like the proportionality sign. – xuanji Sep 30 '13 at 4:32
So this is just an example? Can you please explain the steps as to how this works? With the exponent function, I mean. – Artemisia Sep 30 '13 at 4:40

For a free particle (no external potential) the Schroedinger equation reads $$i \hbar \partial_t \Psi (x,t) = -\dfrac{\hbar^2}{2m} \partial_x^2 \Psi (x,t)$$ which has plane wave solutions that will have the functional form $$\Psi(x,t) \sim \exp(i(kx - \omega t)) + \exp(-i(kx - \omega t)) .$$ Of course, you know that this is equal to twice the cosine of $kx-\omega t$. You determine the constant of proportionality by requiring that the wave function be normalized. The exponential with positive argument (assuming $kx - \omega t > 0$) is the solution of a plane wave travelling to the right (positive x- direction) and the other is that which is travelling to the left. It may have been that your professor was talking about plane waves travelling to the right? Were there any boundary conditions? If so, I bet that it was that there is either a delta function or an infinite wall to the left of the particle. Schroedinger's equation must be supplemented with boundary conditions (as any other B.V.P must) if it is to be soluble.

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Hmmm... ok I understand this. But this particular section came before Schroedinger's rule. He hasn't started that yet. This comes just before Fourier transforms. Just to provide an expression for $$\psi(x,t)$$ in terms of the exponent function. – Artemisia Oct 1 '13 at 5:23