Finding a General Solution for Eigenfunctions in Quantum Mechanics

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SUMMARY

The discussion focuses on finding the general solution for eigenfunctions in quantum mechanics, specifically for a potential step defined by V(x) = 0 for x < 0 and V(x) = V(0) for x > 0, with E > V(0). The time-independent Schrödinger equation governs the behavior of the particle, leading to two linearly independent solutions in each region. The general solution can be expressed as linear combinations of these solutions, incorporating boundary conditions to determine coefficients. The eigenfunctions can be verified by applying the Hamiltonian operator to confirm they satisfy the eigenvalue equation H φ(x) = E φ(x).

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  • Understanding of the time-independent Schrödinger equation
  • Familiarity with linear differential equations and their solutions
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  • Basic proficiency in complex numbers and wave functions
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Bunting
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Consider a potential step described by the function

V(x) = 0 for x < 0
V(x) = V(0) for x > 0

E > V(0) and is incident from the left.
First it asks a few questions about what if it were a classical particle approaching the barrier. Much of this I understand and am OK with. Then we start treating the particle as a quantum thing so its governed by the TI Schrödinger EQ.

So, what it wants me to do which I am a bit unsure about is find the "general solution for the eigenfunctions" in the 2 regions. I have searched a few sites I found on google, and some lecture notes - but they don't give me so much of a derivation or an answer, as a hint so I am a bit hazy. What I think there looking for is...

phi(sub1)(x) = A(sub1)*e(i*k(sub1)*x) + A'(sub1)*e-(i*k(sub1)*x)
phi(sub2)(x) = A(sub2)*e(i*k(sub2)*x) + A'(sub2)*e-(i*k(sub2)*x)

But I don't really understand why, or how I will achieve a general solution out of this.

Thanks for anything!
 
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Hi Bunting,

In your situation the time indepedent Schrödinger equation is a second order linear differential equation. You probably remember from a course on differential equations that such an equation has two linearly independent solutions. Any linear combination of these two solutions with constant coeffecients is also a solution. In other words, knowing two linearly independent solutions to the equation allows you to write down the general solution to the equation. Clearly then the general solution in a region will contain two undetermined coeffecients that you must fix using boundary conditions, etc.

Physically, the solutions may be chosen to consist of left and right moving waves in both regions because E > V(x) everywhere. Of course, other choices are possible, but the plane waves which you have written down are the most physically transparent choice. As I indicated above, you would determine the unknown coeffecients from boundary conditions. For example, if you were interested in the scattering of particles coming from negative infinity, then you would have an incoming wave in the x < 0 region (incident beam), an outgoing wave in the x < 0 region (reflected beam), and an outgoing wave in the x > 0 region (transmitted wave). You would not have an incoming beam in the x > 0 region, hence one of your coefficients would be zero. Other conditions obtain from considering the boundary at x = 0.

Hope this helps.
 
if I do this in the conventional way (rearrange into an equation with a y'' a y' and a y and come up with two roots of the form...

alpha(1,2) = ± m((4*h(bar)^2*[v(x) - E])/2m) / h(bar)^2

I have read some examples which seem to suggest a better way of going about this is to let phi = A*e^ikx - but if the way I did it works then I don't really see a good reason to do this.

So from here I am assuming your saying plug these roots into a complementary function of the form..

y(x) = Ae(alpha(1)*x) + Be(alpha(2)*x)

But then does this give me a complete solution ? I still have A's and B's in there - I have been given no initial conditions to work work specific to phi(x) = whatever or phi'(x) = whatever, and I have a 0 on the RHS now (as I took the E*phi(x) and substracted it from the other side) like I would be used to being given on a problem such as this.

One other thing that's troubling me is the significance of the incidence from the left and not the right - would this simply be a matter of sign or would it need a reversal of the trigonometric properties also ? (cos->sin and whatnot).

Thanks for the reply, I think its getting there :)
 
Last edited:
ok, scratch that last post - I've been re-reading lecture notes and think I've got a greater understanding - though the "eigenfunctions" are still eluding me. I know what they ARE, just not really if I have foundthem...

First I separated the equation into the regions defined (x>0 and x<0) then ended up with 2 equations...

phi(x) = Ae[i*k(1)*x] + Be[i*k(1)*x] - I don't know why I need i in this one in particular...
phi(x) = Ce[k(2)*x] + De[k(2)*x]

k(1) = sqrt(2*m*E) / h(bar)
k(2) = sqrt(2*m*(V(0) - E)) / h(bar)

so I was wondering how I can "justify" this, and what I could write as the eigenfunctions of this.
 
Hi Bunting,

Knowing what the eigenfunctions are allows you to easily check if you've actually found them. Recall the definition of an eigenfunction of the Hamiltonian with eigenvalue E: H phi(x) = E phi(x). Thus if you have a function phi(x) which you think is an eigenfunction of H, this definition tells you how to check your suspicion. Simply hit the eigenfunction with the Hamiltonian operator and if what you get out is a constant times your original function, then you've got an eigenfunction. The constant is just the eigenvalue. If you don't get out something proportional to your original function, then your original function is not an eigenfunction.
 

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