Quantum Well - Two Potentials
A particle in a one-dimensional quantum well is governed by a variant of the time-independent Schrodinger equation, expressed in terms of a non-normalized wave function Ψ N ( x ) .
The quantities V and E are the potential energy and total energy, respectively.
− d x 2 d 2 Ψ N ( x ) + V ( x ) Ψ N ( x ) = E Ψ N ( x )
The potential varies as follows:
V ( x ) = ∞ x < − 1 V ( x ) = 1 − 1 ≤ x ≤ 0 V ( x ) = 5 0 < x ≤ 1 V ( x ) = ∞ x > 1
The boundary conditions on Ψ N ( x ) are:
Ψ N ( x ) = 0 x ≤ − 1 Ψ N ( x ) = 0 x ≥ 1
The probability of detecting the particle within the region a ≤ x ≤ b is:
P ( a ≤ x ≤ b ) = ∫ − 1 1 ∣ Ψ N ( x ) ∣ 2 d x ∫ a b ∣ Ψ N ( x ) ∣ 2 d x
For the smallest possible positive non-zero value of E , what is the probability of detecting the particle in the region ( − 1 ≤ x ≤ 0 ) ?
Note: Simply take it for granted that the wave function is zero within the infinite potential region. Don't try to apply the Schrodinger equation within that region.
The answer is 0.685.
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1 solution
Thank you for your comments which re-assured me that I was on the right track. My initial attempts failed as I first I made an algebraic error and secondly, I chose the second permissible value of energy instead of the minimum.
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Glad it worked out, and thanks for the solution. It was interesting to see that the allowable energy values are the ones which make the matrix singular.
I was just re-reading your comment and another question comes to mind. You mentioned that you numerically integrated from x = − 1 . How did you do that without knowing the value of the derivative of the wave function at that point? I took a semi-analytical route for this reason, actually.
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I just declare a value of the derivative, and run several times with the initial derivative generated as a random number. I observe that the lowest allowable energy has no dependence on the initial choice of derivative.
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Thanks for the clarification. Now that I think of it since the linear system defined above has infinite solutions, this argument makes sense
Finally got it right!
Consider the given wave equation. Its general solution is as follows:
For negative x: ψ ( x ) = A cos ( x E − 1 ) + B sin ( x E − 1 )
For positive x: ψ ( x ) = C cos ( x E − 5 ) + D sin ( x E − 5 )
Applying boundary conditions gives:
A cos ( E − 1 ) − B sin ( E − 1 ) = 0 C cos ( E − 5 ) + D sin ( E − 5 ) = 0
At x = 0 the wave function must be continuous and differentiable which leads to two more equations. Thus, 4 linear equations for A , B , C and D and the value of energy for which a system has non-trivial solutions has to be evaluated. Additional two equations based on continuously differentiable conditions are:
A = C B E − 1 = D E − 5
To compute the permissible values of energy, we have four equations which can be re-written in a matrix form:
⎣ ⎢ ⎢ ⎡ 1 0 cos ( E − 1 ) 0 0 E − 1 − sin ( E − 1 ) 0 − 1 0 0 cos ( E − 5 ) 0 − E − 5 0 sin ( E − 5 ) ⎦ ⎥ ⎥ ⎤ ⎣ ⎢ ⎢ ⎡ A B C D ⎦ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎡ 0 0 0 0 ⎦ ⎥ ⎥ ⎤
⟹ W ⎣ ⎢ ⎢ ⎡ A B C D ⎦ ⎥ ⎥ ⎤ = ⎣ ⎢ ⎢ ⎡ 0 0 0 0 ⎦ ⎥ ⎥ ⎤
To obtain the permissible energy values, W must have non-trivial solutions. That can only happen when the determinant of W is zero. The first two values of energy at which this happens are approximately:
E 1 ≈ 5 . 0 0 0 1 E 2 ≈ 1 3 . 1 5
Having found the lowest possible value of energy, that value is used in the general solution and the constants A , B , C and D are chosen such that the above matrix equation is satisfied. Note that the system has infinite solutions.
The probability can be computed analytically however, I took a numerical route. The result is:
P ( − 1 ≤ x ≤ 0 ) ≈ 0 . 6 8 3 3