# Homework Help: Wavefunctions and boundaries of a 1D-potential well

1. Apr 13, 2010

### indie452

the attachment here shows a potential well and i am practicing writing down wave functions in each region and boundary conditions (for this problem I suppose the boundaries would be x= -a, 0, b.

i think that for this problem i need 4 functions and i thought that they may be:

when
x<-a (1) = 0
-a<x<0 (2) = Asin(k1x) + Bcos(k1x)
0<x<b (3) = Csin(k2x) + Dcos(k2x)
b>x (4) = 0

where
k1 = ($$\sqrt{}2mE$$)/$$\hbar$$
k2 = ($$\sqrt{}2m(E-Vo))/\hbar$$

and i think at the boundary x = -a and b the wave function is continuous and at x=0 the wave function and the first derivative is continuous.

Is this right?
thanks

#### Attached Files:

• ###### qu7.jpg
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2. Apr 13, 2010

### indie452

|......................................|
|......................................| Vo < E
|......................................|
|....................__________|
|....................|
|....................|
|....................|
|....................|
-a...................0.................b

V(x)=
infinity --- x<-a --- region 1
0 --- -a<x<0 --- region 2
Vo --- 0<x<b --- region 3
infinity --- b<x --- region 4

if you cant see the pic

3. Apr 13, 2010

### thebigstar25

first thing:

maybe you want to think again about what you stated here when you said "b>x (4) = 0", i believe you have done simple mistake since you wrote the first three equations correctly, it actually should be "x>b (4) = 0"

second thing:

dont you think that you should make your wavefunction continuous by appliying the fact that the wavefunction and its derivative (1st) in region(1) = the wavefunction and its derivative (1st) in region(2) and that when x = -a , and the same goes with region(2) and (3) when x = 0 , also region(3) and (4) when x = b ..

hmm, well .. I think so .. what do you think?

4. Apr 13, 2010

### indie452

for the first thing your right i did write it down wrong t=it was meant to be b<x.

but with the the reason i wrote that the first derivative of the wavefunction is not continuous at the boundary x=-a, and b was cause i thought that when on the otherside of a boundary the potential->infinity, so KE->-infinty at boundaries so first derivative->infinity also. this is cause it seems to be infinite potential and so not physically realistic.

5. Apr 13, 2010

### indie452

for the first thing your right i did write it down wrong t=it was meant to be b<x.

but with the the reason i wrote that the first derivative of the wavefunction is not continuous at the boundary x=-a, and b was cause i thought that when on the otherside of a boundary the potential->infinity, so KE->-infinty at boundaries so first derivative->infinity also. this is cause it seems to be infinite potential and so not physically realistic.

6. Apr 13, 2010

### thebigstar25

thats mainly the reason behind choosing the wavefunction = 0 in both regions 1 and 4..

lets first consider the situation when we say that the wavefunction and its derivative should be continuos in region (1) and (2) when x = -a ..

you have in region (2) that the wavefunction = A sin(k1x) + B cos(k1x) .. so when x = -a

in region(2) the wavefunction = - A sin(k1a) + B cos(k1a) & you know that the wavefunction in region(1) at x = -a is zero >>> then 0 = -A sin(k1a) + B cos(k1a)
im not allowed to proceed more than this, but i think you got the point (I hope!) .. continue with the derivatives .. and dont forget the other regions ..

good luck..

7. Apr 13, 2010

### indie452

for the first thing your right i did write it down wrong t=it was meant to be b<x.

but with the the reason i wrote that the first derivative of the wavefunction is not continuous at the boundary x=-a, and b was cause i thought that when on the otherside of a boundary the potential->infinity, so KE->-infinty at boundaries so first derivative->infinity also. this is cause it seems to be infinite potential and so not physically realistic.