Schrodringer's Equation (Quantum Phys)

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Homework Statement


In Eq. 38-18 Keep both terms, putting A=B=Ψo. The equation then describes the superposition of two matter waves of equal amplitude, traveling in opposite directions.(Recall that this is the condition for a standing wave.)
(a) Show that |Ψ(x,t)|^2=2Ψo^2[1+Cos(2kx)]
(b) Plot tihs function, and demonstrate that it describes the square of the amplitude of a standing matter wave.
(c) Show that the nodes of this standing wave are located at
x=(2n+1)1/4(λ) where n=1,2,3,. . .
and λ is the de Broglie wavelength of the particle.
(d) Write a similar expression for the most probable locations of the particle.


Homework Equations


(1) Ψ(x,t)=Ae^[i(kx-wt)]+Be^[-i(kx-wt)] (this is Eq. 38-18)
(2) λ=h/p (de Broglie wavelength)
(3) e^(iθ)=cosθ+isinθ
(4) e^(-iθ)=cosθ-isinθ

The Attempt at a Solution



What i did was took equation 1 above and substituted B in for A on the first part. That gave me
Ψ(x,t)=B(e^[i(kx-wt)]+e^[-i(kx-wt)])

I then substituted (kx-wt) for θ so that I could use equations 3 and 4. This yielded(after some algebra):
Ψ(x,t)=B(2cos(kx-wt))

Next i used the angle difference formula to separate the KX and WT and then I squared both sides of the equations. After about 4 more lines of work and some degree reducing I ended with and have hit a roadblock:

4B^2=cos(kx)cos(wt)Cos(kx-wt)+ (1-cos2kx)/2 + (1-cos2wt)/2

The last two terms were sin^2(kx) sin^2(wt) but i reduced the degrees because the final equation I'm looking for is all cosines.
 
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Are you sure you copied equation 38-18 correctly?
 
Yes it is a solution to Schrodinger's equation for a free particle traveling in an arbitrary "x" direction.
 
Well, the sign of ω is incorrect in the second term.
 
Oops, I do have it copied correctly in my notes, just not in my post.
 
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