Finite quantum well, factor of 2*pi seems necessary but why?

[Villhelm]

Homework Statement

Solve for the allowed energy values E of a finite square quantum well of depth U₀ = 25eV, width a = 0.5nm that contains an electron of mass m (I'm presuming that m = 9.11*10^-31kg, the question doesn't indicate a specific value to use).

I'm defining the interior potential to be 0eV (so that the walls are +25eV).
I'm also defining the well as extending over x = [0,a]

Homework Equations

Time-independent Schrödinger equation in 1 dimension + some basic DE stuff.

The Attempt at a Solution

Starting from the TISE, I end up with the following:

2 * sqrt(E*U₀ - E^2) / (2E - U₀)
= tan(sqrt(2*m*E)*a/hbar)

Upto there I'd (apparently) done things correctly. However, when I estimated the value(s) of E which satisfy the equation numerically I did not get the results that were expected. I did the estimation using openoffice spreadsheet. The tan function is using radians.

My first thought was to see if my value assumed for the electron mass was wrong, so I tried multiples in the range m' = [0.5m,2m] just to see what happened to the results and noticed that the higher the value of m', the closer the values of E that solved the equation came to those I was indicated to have gotten. I tried m'=10m and found that it pushes the values of E too far, so I started iterating through some values until I ended up settling down to m'=6.28m which struck me as being 2*pi*m.

The RHS of the equation now looks like:

RHS = tan(sqrt(2*(2*pi)*m*E)*a/hbar)

and results in a reproduction of the given energy values (there are five, approximately at {1.123, 4.461, 9.905, 17.162, 24.782}eV).

I don't understand why this works, however, especially when the 2*pi is inside the square root.

 

[vela]

Using your original equations, I found five solutions close to the ones you listed.

What value are you using for \hbar? The factor of 2π could arise if you're using h instead of \hbar.

Here's a little trick you can use to make evaluating the argument of the tangent a bit easier:

\frac{\sqrt{2mE}}{\hbar}a = \frac{\sqrt{2(mc^2)E}}{\hbar c}a

where c is the speed of light. The mass of the electron is 511000 eV/c², and the combination \hbar c is equal to 197 eV nm (this is a useful combination to know). If you measure E in eV and a in nm, all the units cancel as you need them to.

 

[Villhelm]

What value are you using for \hbar? The factor of 2π could arise if you're using h instead of \hbar.

I've been using the value \hbar = 6.58 * 10^-16 eV.s
Also, for m I was using 9.11 * 10^-31kg

I did wonder if maybe I'd messed up with \hbar vs h, but with the 2π inside the square root rather than outside along with \hbar or h made me question that.

I'll use the constant values you've cited and see what I get.

Things seem to work out nicely! Thanks very much.

Incidentally I've just learned a very good lesson from this - I was making assumptions that the values I obtained from the fudge I described above were within reasonable error margins of the values I was given as correct - however I failed to notice that the values were infact _not_ consistent with calculation error because I'd been doing a simple linear interpolation of the value of E which was the root, which was reasonably accurate for the lower E values given the number of digits I was calculating with, but it actually diverged for the higher ones and I hadn't bothered to check more than a glance and just assumed they worked out (why wouldn't they, a factor of 2*pi just _couldn't_ be a coincidence, right?) ...