Find the wave function of a particle bound in a semi-infinite square well

[BPMead]

Homework Statement

Consider the semi-infinite square well given by V(x) = -V₀ < 0 for 0≤ x ≤ a and V(x) = 0 for x > a. There is an infinite barrier at x = 0 (hence the name "semi-infinite"). A particle with mass m is in a bound state in this potential with energy E ≤ 0. Solve the Schrödinger equation to derive ψ(x) for x ≥ 0. Use the appropriate boundary conditions and normalize the wave function so that the final answer does not contain any arbitrary constants.

Homework Equations

[-h_bar²/2m]ψ'' + V(x)ψ = Eψ

The Attempt at a Solution

• Schrödinger Equation for 0 ≤ x ≤ a and x > a:
  [-h_bar²/2m]ψ'' - V₀ψ = Eψ, 0 ≤ x ≤ a
  [-h_bar²/2m]ψ'' = Eψ, x ≥ a
• Rewrite Schrödinger equations:
  ψ'' + 2m(E+V₀)/h_bar² = 0, 0 ≤ x ≤ a
  ψ'' + 2mE/h_bar² = 0, x ≥ a
• Solve Schrödinger equations:
  ψ₁ = A₁e^ik₁x + B₁e^-ik₁x, k₁ = sqrt[2m(E+V₀)]/h_bar, 0 ≤ x ≤ a
  ψ₂ = A₂e^ik₂x + B₂e^-ik₂x, k₂ = sqrt[2mE]/h_bar, x ≥ a
• k₂ is negative, and the wave function must not blow up at x = ∞, so A₂ = 0:
  ψ₁ = A₁e^ik₁x + B₁e^-ik₁x, k₁ = sqrt[2m(E+V₀)]/h_bar, 0 ≤ x ≤ a
  ψ₂ = B₂e^-ik₂x, k₂ = sqrt[2mE]/h_bar, x ≥ a
• Apply boundary conditions:
  ψ₁(0) = 0
  ψ₁(a) = ψ₂(a)
  ψ'₁(a) = ψ'₂(a)
  
  1st condition: A₁ + B₁ = 0
  2nd condition: A₁e^ik₁a + B₁e^-ik₁a = B₂e^-ik₂a
  3rd condition: ik₁A₁e^ik₁a - ik₁B₁e^-ik₁a = -ik₂B₂e^-ik₂a

Now I have 3 equations for 3 unknowns, A₁, B₁, and B₂. But I have been trying to solve this algebraically for quite awhile, and I just can't get it to work. When I solve A₁ and B₁ in terms of B₂ and try to plug them into the third condition, I just get B₂ cancelling on both sides. Maybe I'm being really dumb about basic math but I would really appreciate if someone could help with this. Thanks!

 

[vela]

Homework Statement

Consider the semi-infinite square well given by V(x) = -V₀ < 0 for 0≤ x ≤ a and V(x) = 0 for x > a. There is an infinite barrier at x = 0 (hence the name "semi-infinite"). A particle with mass m is in a bound state in this potential with energy E ≤ 0. Solve the Schrödinger equation to derive ψ(x) for x ≥ 0. Use the appropriate boundary conditions and normalize the wave function so that the final answer does not contain any arbitrary constants.

Homework Equations

[-h_bar²/2m]ψ'' + V(x)ψ = Eψ

The Attempt at a Solution

• Schrödinger Equation for 0 ≤ x ≤ a and x > a:
  [-h_bar²/2m]ψ'' - V₀ψ = Eψ, 0 ≤ x ≤ a
  [-h_bar²/2m]ψ'' = Eψ, x ≥ a
• Rewrite Schrödinger equations:
  ψ'' + 2m(E+V₀)/h_bar² = 0, 0 ≤ x ≤ a
  ψ'' + 2mE/h_bar² = 0, x ≥ a
• Solve Schrödinger equations:
  ψ₁ = A₁e^ik₁x + B₁e^-ik₁x, k₁ = sqrt[2m(E+V₀)]/h_bar, 0 ≤ x ≤ a
  ψ₂ = A₂e^ik₂x + B₂e^-ik₂x, k₂ = sqrt[2mE]/h_bar, x ≥ a
• k₂ is negative, and the wave function must not blow up at x = ∞, so A₂ = 0:

Actually, the way you defined k₂, it's imaginary because E ≤ 0. It's usually better to define $k_2 = \sqrt{-2mE}/\hbar$, so you can avoid those error-inducing factors of i.

ψ₁ = A₁e^ik₁x + B₁e^-ik₁x, k₁ = sqrt[2m(E+V₀)]/h_bar, 0 ≤ x ≤ a
ψ₂ = B₂e^-ik₂x, k₂ = sqrt[2mE]/h_bar, x ≥ a

• Apply boundary conditions:
  ψ₁(0) = 0
  ψ₁(a) = ψ₂(a)
  ψ'₁(a) = ψ'₂(a)
  
  1st condition: A₁ + B₁ = 0
  2nd condition: A₁e^ik₁a + B₁e^-ik₁a = B₂e^-ik₂a
  3rd condition: ik₁A₁e^ik₁a - ik₁B₁e^-ik₁a = -ik₂B₂e^-ik₂a

Now I have 3 equations for 3 unknowns, A₁, B₁, and B₂. But I have been trying to solve this algebraically for quite awhile, and I just can't get it to work. When I solve A₁ and B₁ in terms of B₂ and try to plug them into the third condition, I just get B₂ cancelling on both sides. Maybe I'm being really dumb about basic math but I would really appreciate if someone could help with this. Thanks!

Once you get it down to everything in terms of one constant, you're finished with this part. To determine the constant, you require the wave function to be normalized.

What conditions did you get on k₁ and k₂? A non-trivial solution exists for only certain values.

 

[BPMead]

Thank you so much for your response! I can see why you would define k₂ that way, I guess I kept it imaginary so that ψ₁ and ψ₂ would have a similar format. The way I defined them should still theoretically work right? Or do I have to change it?

So I should solve in terms of one constant, but then how do I normalize when there are two wave functions? Do I integrate ψ₁ from 0 to a and then add that to the integral of ψ₂ from a to ∞, and set that equal to 1?

 

[vela]

Thank you so much for your response! I can see why you would define k₂ that way, I guess I kept it imaginary so that ψ₁ and ψ₂ would have a similar format. The way I defined them should still theoretically work right? Or do I have to change it?

It'll work out either way.

So I should solve in terms of one constant, but then how do I normalize when there are two wave functions? Do I integrate ψ₁ from 0 to a and then add that to the integral of ψ₂ from a to ∞, and set that equal to 1?

Yes, that's what you would do.

 

[BPMead]

Okay, when I solve that system of equations, I get:

A₁ = B₂e^-ik₂a / (e^ik₁a - e^-ik₁a)
B₁ = -B₂e^-ik₂a / (e^ik₁a - e^-ik₁a)

Plugging these into normalize:
Let j = e^-ik₂a / (e^ik₁a - e^-ik₁a)

jB₂₂∫e^ik₁x - e^-ik₁xdx + B₂∫e^-ik₂x = 1

Then I get
jB₂/ik₁ [e^ik₁a-e^ik₂(0)] - jB₂/-ik₁ [e^-ik₁(a) - e^-ik₁(0)] + B₂/-ik₂ (e^-ik₂(∞) - e^-ik₂(0)) = 1

What is e^-ik₂(∞)? Is that still 0? Even if it is, my answer looks unbelievably ugly.

 

[vela]

You can simplify that a bit using
$$\sin x = \frac{e^{ix}-e^{-ix}}{2i}$$ In fact, if you step back a bit, your first boundary condition tells you A₁=-B₁ so that you can write $\psi_1(x) = A \sin k_1x$.

What is e^-ik₂(∞)? Is that still 0?

I just noticed a mistake in your original post. The way you defined k₂, it's purely imaginary and equal to
$$k_2 = i\left(\frac{\sqrt{-2mE}}{\hbar}\right)$$ where the quantity in the parentheses real and positive. Consequently, ik₂ is purely real and negative. This implies that the A₂ term vanishes as x→∞ but the B₂ term blows up.

 

[BPMead]

Okay, if we are going to include trig functions anyway, there is a simpler way that I saw to set this up:

ψ₁ = A₁sin(k₁x) + B₁cos(k₁x)
Since ψ₁(0) = 0, B₁ = 0

So:

ψ₁ = A₁sin(k₁x)
ψ₂ = A₂e^ik₂x

Since ψ₁(a) = ψ₂(a):
A₂ = A₁sin(k₁a) / e^ik₂a

Now normalize to find constants:
∫A₁sin(k₁x)dx + ∫A₂e^ik₂xdx

Solving this, I get C = [1-cos(k₁a)-sin(k₁a)/e^ik₂a]^-1

Which doesn't really look simple enough to be right...

 

[vela]

Remember you have to integrate the square of the modulus to normalize the wave function.

 

[BPMead]

Ah! You're right. So what I got was:

A₁²∫sin²(k₁x)dx + A₁²sin²(k₁a)/e^2ik₂a∫e^ik₂xdx = 1

That works out to:

1 = A₁²[a/2 - (1/4)sin(2k₁a) - sin²(k₁a)/(ik₂e^ik₂a)]

Is that the right way to get A₁?

 

[vela]

Right idea. I really suggest you get rid of the unnecessary i's.