Applying particle in a box boundaries to an eigenfunction

[ReidMerrill]

Homework Statement

Use the eigenfunction Ψ(x) =A'e^ikx + B'e^-ikx rather than Ψ(x)=Asinkx + Bcoskx to apply the boundary conditions for the particle in a box. A. How do the boundary conditions restrict the acceptable choices for A’ and B’ and for k? B. Do these two functions give different probability density if each is normalized?

Homework Equations

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The Attempt at a Solution

I know that the limits are Ψ(0)=Ψ(a)=0, Ψ(x)=0

When I apply that to the equation
Ψ(0) =A'e^ik0 + B'e^-ik0 = A'+B'= 0
and
Ψ(a) =A'e^ika + B'e^-ika

And I don't know what to do from here. Any help would be appreciated

 

[TSny]

I know that the limits are Ψ(0)=Ψ(a)=0, Ψ(x)=0

Ψ(0)=Ψ(a)=0 is good. But why Ψ(x)=0?

When I apply that to the equation
Ψ(0) =A'e^ik0 + B'e^-ik0 = A'+B'= 0
and
Ψ(a) =A'e^ika + B'e^-ika

And I don't know what to do from here. Any help would be appreciated

Use what you found from Ψ(0) = 0 to simplify Ψ(a)=0.

 

[ReidMerrill]

Ψ(0)=Ψ(a)=0 is good. But why Ψ(x)=0?Use what you found from Ψ(0) = 0 to simplify Ψ(a)=0.

I've got further since I posted this.
From A'+B'=0
B'=-A'
Ψ(a)=A'(e^ika-e^-ika
e^ika=e^-ika

But I'm stuck again.

 

[TSny]

I've got further since I posted this.
From A'+B'=0
B'=-A'
Ψ(a)=A'(e^ika-e^-ika)
e^ika=e^-ika

But I'm stuck again.

If you know how to plot the complex numbers e^ika and e^-ika in the complex plane, then you should be able to see the condition for them to be equal.

However, a better way is to recall how to write e^ika and e^-ika in terms of the trig functions sin(ka) and cos(ka), or vice versa.