Schrodinger equation of particle in box

[R-ckay]

hi all please help me... I'm learning schrodinger equation of a particle in a 1-dimensional box. I read a quantum mechanics book written by A. C Phillips. the wavefuction is

ψ (x,t)= N sin (kx) e^-iEt/hbar

but when I compared to what I read from a modern physics book written by Beisser. the wavefunction is:

d²ψ/dx² + 2m/hbar E ψ
=0

and the solution is:

ψ=A sin ((√2mE)/hbar)x + B cos ((√2mE)/hbar)

and we only use:

ψ =N sin (kx) only,

it is different. and I got trouble when I tried to find momentum probability. because it is different. which one is true?

my question is:
1.where "e^-iEt/hbar" comes from?
2. how to get the solution "
ψ=A sin ((√2mE)/hbar)x + B cos ((√2mE)/hbar)" from
d²ψ/dx² + 2m/hbar E ψ
= 0

thank you very much for help.. :)

 

[Chopin]

#1: The sin(kx) is a solution to the time-independent Schrodinger Equation. This means that it only tells you the spatial behavior at the wave at one point in time. To find the time-dependence of the equation, you use the time-dependent version of the Schrodinger Equation, which says:
i\hbar\frac{d}{dt}\Psi(t) = \hat{H}\Psi(t)
If \Psi(x) is an eigenstate of \hat{H}, then \hat{H}\Psi(x)=E\Psi(x), so
i\hbar\frac{d}{dt}\Psi(t,x) = \hat{H}\Psi(t,x) = E\Psi(t,x)
The solution to this differential equation (which you can check by substitution) is
\Psi(t,x) = e^{-iEt/\hbar}\Psi(0,x)

#2: It's a differential equation, so the answer to how to find the solutions is basically "any way you can". In this case, you can make a guess about the form that solutions must take (sinusoidal), and then substitute them into the original equation to figure out whether your guess is right.

The second equation is correct that both sin and cos terms appear in the general solution to the free Schrodinger Equation. However, the first equation is correct that solutions to the equation in an infinite potential well only contain some of those terms. The reason that only sin terms appear in the solution to the particle in a box is that you also have to remember the boundary conditions. Since the box is infinite at x=0 and x=L, the wavefunction must be zero outside of that range, i.e. \Psi(0) = 0 and \Psi(L) = 0. This means that none of the cos terms will work (they're all nonzero at x=0), and of the sin terms, the only ones which work are the ones which are also zero at x=L. This means you're limited to \Psi(x)=sin(2\pi n x/L), where n is an integer.

In this way, the continuous space of general solutions gets reduced to a discrete (or "quantized") set of particular solutions which obey the boundary conditions. This happens any time you have a potential well like this, and is the reason that there are discrete energy levels in systems such as atoms. It's also where "quantum mechanics" gets its name.