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# Two well problem

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If b = 0 you have one well of size 2a. If V_0 is very large then the wavefunctions will be very small outside the well; they'll decay very fast to zero. The ground state has no zeros inside the well and a maximum in the middle. You can sketch it by drawing cos(x) between -pi/2 and pi/2. you put the point -a slightly closer to zero than -pi/2 so that the function isn't exactly zero there and you put the point +a slightly closer to zero than pi/2. Then, outside the well, you just let the wavefunction go to zero very fast (it is already close to zero at a and -a). Note that the graph you draw has to be symmetrical.

The first excited state has one zero in the well. You can sketch it by drawing the function sin(2x) between -pi/2 and pi/2. You let the point -a correspond to a point slightly closer to zero than -pi/2 and the point +a to a point slightly closer to zero than pi/2. Outside the well you again let the graph approach zero. Note that the graph you draw has to be anti-symmetrical.

If b is increased from zero, the well is split in two parts. in between the two wells the wavefunction will decay exponentially. The split happens at x=0, the place where the ground state was maximal and the first exited state had a zero. At x = 0 the wavefunction now has to be small. The ground state get's the following shape.

Let's draw it from left to right. Outside both wells on the left it is very small. If you get close to -a it increases by a ...

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