Determine the wave function of the particle

[rmfw]

[STRIKE][/STRIKE]

Homework Statement

A particle coming from +∞ with energy E colides with a potential of the form:

V = ∞ , x<0 (III)
V = -V₀ , 0<x<a (II)
V = 0, x>a (I)

a) Determine the wave function of the particle considering that the amplitude of the incident wave is A. Writting the amplitude of the reflected wave at x=a in the form

$\frac{B}{A} = e^{i\delta}$

determine $\delta$ . What is the value of $\delta$ in the limit where V₀ = 0 ?

b) Determine the probability density current in x>a

The Attempt at a Solution

a)

For region I :

$\Psi_{I} (x) = Ae^{-ikx} + Be^{ikx} , k=\frac{\sqrt{2mE}}{\hbar}$

II:

$\Psi_{II} (x) = Csin(lx) + Dcos(lx) , l=\frac{\sqrt{2(mE+V_{0}}}{\hbar}$

III:

$\Psi_{III} (x) = 0$Boundary conditions:

at x=0:

$0 = Csin(lx) + Dcos(lx) (=) D = 0$

So $\Psi_{II} (x) = Csin(lx)$

at x=a:

$Ae^{-ika} + Be^{ika} = Csin(la)$ (1)

and

$-ikAe^{-ika} + ikBe^{ika} = lCcos(la)$ (2)

Dividing (1) for (2) I got:

$\frac{B}{A} = e^{-2ika} \frac{(-\frac{1}{l}tan(la)ik-1)}{(1-\frac{1}{l}tan(la)ik)}$

How can I get rid of that constant that is multiplying by the exponential? Is this even right?

 

[Simon Bridge]

B/A appears to be a complex number - is that right?

Note: did you try writing $\psi_{II}$ as a sum of complex exponentials?

 

[rmfw]

B/A appears to be a complex number - is that right?

Note: did you try writing $\psi_{II}$ as a sum of complex exponentials?

Hi Simon Bridge, thank you for your reply.

I did write psiII as a sum of complex exponentials and I got the following:

$\frac{B}{A} = \frac{\frac{k}{l}e^(ika-ila)-\frac{k}{l}e^(-ika + ila) - e^(-ika -ila) - e^(-ika + ila)}{\frac{k}{l}e^(ika-ila) - \frac{k}{l}e^(ika+ila) +e^(ika+ila) + e^(ika-ila)}$

is there any way to simplify this ? This problem is killing me, because I'm not even sure what "writing the amplitude of the reflected wave at x=a in the form $\frac{B}{A}$ " means.

I mean, I assumed the amplitude of the reflected wave at x=a is B, if I write B/A it isn't the amplitude of the reflected wave anymore.

 

[Simon Bridge]

B/A is the proportion of the incident amplitude that is reflected.
If you multiplied it by 100, you's be able to say, "100B/A percent got reflected".

You basically have to look for common factors and cancel things off a bit at a time.
It is not going to be easy. But that, I am afraid, is the exercise.

Note: $e^{a-b}+e^{-a-b}=(e^a+e^{-a})e^{-b}$ ... stuff like that. maybe $e^{-b}$ cancels something in the denominator? If not - look for something else.

Of course, since they are all constant terms, it may be possible to add them up geometrically.
Separate the real and imaginary components.

 

[paranormal]

Your approach to solving for the wave function in each region is correct. However, in order to determine the value of \delta, it may be helpful to rewrite the wave function in region II as a combination of exponential functions rather than trigonometric functions. This can be done by using the identities sin(x) = (e^{ix}-e^{-ix})/2i and cos(x) = (e^{ix}+e^{-ix})/2.

Substituting these identities into the wave function in region II, we get:

\Psi_{II} (x) = C\frac{e^{ilx}-e^{-ilx}}{2i}

Using this form for the wave function, the boundary conditions at x=0 and x=a can be used to solve for the coefficients C and \delta.

At x=0, the wave function becomes:

\Psi_{II} (0) = C\frac{e^{il(0)}-e^{-il(0)}}{2i} = C

At x=a, the wave function becomes:

\Psi_{II} (a) = C\frac{e^{ila}-e^{-ila}}{2i} = C\frac{e^{2ila}-1}{2ie^{ila}} = C\frac{e^{2ila}-e^{-2ila}}{4ie^{ila}}

Using the boundary condition equation (1) from your attempt, we can write:

Ae^{-ika} + Be^{ika} = C\frac{e^{2ila}-e^{-2ila}}{4ie^{ila}}

Substituting in the expressions for A and B that you found in your attempt, we get:

Ae^{-ika} + \frac{A}{e^{2ika}} = C\frac{e^{2ila}-e^{-2ila}}{4ie^{ila}}

Dividing both sides by Ae^{-ika}, we get:

1 + \frac{1}{e^{2ika}} = C\frac{e^{2ila}-e^{-2ila}}{4ie^{ila}Ae^{-ika}}

Using the boundary condition equation (2) from your attempt, we can write:

-ikAe^{-ika} + ik\frac{A}{e^{2ika}} = C\frac{l}{e^{ila}}

Substituting in the expressions for A and