**Finite Potential Barrier in an Infinite Potential Well (Part One):**

Put potential barrier of width a and height V_0 in infinite well of width b>a

V(x)=∞, x<-b/2

V(x)=0, -b/2<x<-a/2

V(x)=V_0, -a/2<x<a/2

V(x)=0, a/2<x<b/2

V(x)=∞, x>b/2

Let us examine this one-dimensional quantum mechanics problem:

The infinities tell us that φ(x)=0 for

And we also require that φ(x) and φ'(x) must match (must be continuous) at .

(Time independent) Scrödinger’s equation in one dimension

or

For the regions , V(x)=0, and the above becomes

Define . Then the solution for those regions is in the form

We now note that the potential is symmetric with respect to reflection. Thus φ(x) is either an odd or an even function: φ(-x)=φ(x) or φ(-x)=-φ(x).

We will now consider separately the cases , , and .

I.

In the region , we have

Defining , we have that the solution for that region is of the form

Thus, we have

FIrst, we use the symmetry: either φ(x) is even or φ(x) is odd

A. φ(x) even:

If φ(x) is even, then we have , , and

Thus:

Now, we match the ends of the regions:

For x=-b/2, we need continuity:

For x=-a/2, we have:

and

Which we can solve for and in terms of :

For x=a/2, we obtain the same equations as from x=-a/2, as expected from our symmetry of φ(x). Similarly, x=b/2 gives the same condition as x=-b/2.

Plugging the values for and into that condition gives:

which, combined with and , gives a transcendental equation in the energy E.

B. φ(x) odd:

If φ(x) is odd, then we have , , and

Thus:

Now, we match the ends of the regions:

For x=-b/2, we need continuity:

For x=-a/2, we have:

and

Which we can solve for and in terms of :

For x=a/2, we obtain the same equations as from x=-a/2, as expected from our symmetry of φ(x). Similarly, x=b/2 gives the same condition as x=-b/2.

Plugging the values for and into that condition gives:

, which, combined with and , also gives a transcendental equation in the energy E.

II.

In the region , we have

Defining , we have that the solution for that region is of the form

Thus, we have

FIrst, we use the symmetry: either φ(x) is even or φ(x) is odd

A. φ(x) even:

If φ(x) is even, then we have , , and

Thus:

Now, we match the ends of the regions:

For x=-b/2, we need continuity:

As in part I.

For x=-a/2, we have:

and

Which we can solve for and in terms of :

For x=a/2, we obtain the same equations as from x=-a/2, as expected from our symmetry of φ(x). Similarly, x=b/2 gives the same condition as x=-b/2.

Plugging the values for and into that condition gives:

which, combined with and , gives a transcendental equation in the energy E.

B. φ(x) odd:

If φ(x) is odd, then we have , , and

Thus:

Now, we match the ends of the regions:

For x=-b/2, we need continuity:

For x=-a/2, we have:

and

Which we can solve for and in terms of :

Once again, we find that for x=a/2, we obtain the same equations as from x=-a/2, again from our symmetry of φ(x). Similarly, x=b/2 gives the same condition as x=-b/2.

Plugging the values for and into that condition gives:

, which, combined with and , also gives a transcendental equation in the energy E.

III.

In the region , we have

And we have that the solution for that region is linear:

Thus, we have

FIrst, we use the symmetry: either φ(x) is even or φ(x) is odd

A. φ(x) even:

If φ(x) is even, then we have , , and

Thus:

Now, we match the ends of the regions:

For x=-b/2, we need continuity:

once again.

For x=-a/2, we have:

and

Which we can solve for and in terms of :

(Note that these equations are both the limit of those for I. A. as and the limit of those for II. A. as )

Again, symmetry automatically gives us the conditions for x=a/2 and x=b/2 from those for x=-a/2 and x=-b/2. Substituting the x=-a/2 conditions into that for x=-b/2, we get: , which requires , where n is a positive integer. Using , we get . Thus, an even solution only exists for being of that form.

B. φ(x) odd:

If φ(x) is odd, then we have , , and

Thus:

Now, we match the ends of the regions:

For x=-b/2, we need continuity, which again gives us

For x=-a/2, we have:

and

Which we can solve for and in terms of :

Putting these in the x=-b/2 condition, we get

Which using , we get a transcendent equation for , which gives the allowed values for for an odd solution.

Note overall that each even solution wavefunction is multiplied overall by the constant , and each odd one by . To give proper wavefunctions giving the correct probabilities, we choose or such that the function is normalized:

Later, we’ll look at limiting behaviors and numerical values for E in the different solutions.

Tags: Friday Physics, Science

February 15, 2008 at 2:01 pm |

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