# Regarding double well potential

• peripatein
So it is possible to arrive at sin(q(x-b)) without the use of q(a-b).In summary, the eigenvalues conditions may be written in the form:tan(q(a-b))=qα(1+tanh(αb)/(q2 -α2tanh(αb))$$andtan(q(a-b))=qα(1+coth(αb)/(q2 -α2coth(αb))$$f

#### peripatein

A particle is in the following double-well potential with E<0:
V(x)=0 for x<-a, x>a; -V0 for -a<x<-b, b<x<a; 0 for -b<x<b
I am asked to show that the eigenvalues conditions may be written in the form:
tan(q(a-b))=qα(1+tanh(αb)/(q22tanh(αb))$$and tan(q(a-b))=qα(1+coth(αb)/(q22coth(αb))$$
for the even and odd solutions, where -E=ħ2α2/2m and E+V02q2/2m.

I first tried to define the wave function in the various regions, focusing on the positive x-axis only and demanding odd solutions:
Ψ(x)=Ae-αx for x>a; Bsin(q(x-b)) for b<x<a; Ce-αx + Deαx for 0<x<b

Is that correct thus far?

Looks good so far.

Alright, but as I impose continuity of wave function and its derivatives I obtain the four following equations, from which I am unable to derive the desired expression:
(1) Ae-αa = Bsin(q(a-b))
(2) -αAe-αa = Bqcos(q(a-b))
(3) 0 = Ce-αb + Deαb
(4) Bq = -αCe-αb + Dαeαb

What am I doing wrong?

You could explain what you did. Where does equation 3 come from?
Where is the odd/even condition?

Well, equation 3 stems from the demand for continuity at x=b. I have decided to focus on the odd solutions first, hence the use of sine. Is that wrong?

Ah, I see the issue.
The way you defined the wave function in the well, you get a mixture of odd and even functions (the sine would have some non-zero value at x=0). If you want odd functions, you need something like sin(qx).

So which function ought I to choose for Ψ(x) in the well? Ought it to be a combination of sine and cosine at once (like harmonic oscillator)?

In other words, should I then simply choose sin(qx) for the odd and cos(qx) for the even, within the well?

So which function ought I to choose for Ψ(x) in the well? Ought it to be a combination of sine and cosine at once (like harmonic oscillator)?
You can do that, if you choose them without that offset it is easier to split them into odd and even contributions.
In other words, should I then simply choose sin(qx) for the odd and cos(qx) for the even, within the well?
That is the easierst approach I think.

Yet please note the expressions I am expected to arrive at. It seems I am indeed expected to use forms as sin(q(x-b)). Wouldn't you agree?

I don't see how you get that impression. Yes there is an q(a-b), but that can appear in other ways, too.
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