Physics 361 Principles of Modern Physics Lecture 13

Solving Problems with the Schrödinger Equation This lecture Particle in a box potential Next lectures Harmonic oscillator Tunneling and scattering in 1D

What happens as energy is lowered, or energy barrier increased in height? U E o In this case the solutions are: Which is the standing wave pattern o

Use of BCs to determine previous results U E o As barrier height increases to infinity, we can only assure continuity of the wave function, not its first derivative. This will only give one boundary condition. But there is only one variable to solve for -- in terms of. Our solution could just as easily be obtained if we just required continuity of the wave function. The above solutions are for unbound particles. Now consider the case where we have two such potential boundaries which have heights which approach infinity.

Particle in infinite box – bounded particles U E o The waves in the barriers on both sides will go to zero, as before. So we still only have the coefficients and. Since we have two infinite boundaries, we have two boundary conditions (continuity at each boundary). Also, since we expect a bound state, we can employ the normalization condition to fix the value of as well. However, we seem to have three conditions, and only two variables to determine! The energy is no longer allowed to be any value – allowed energy values must now be determined. The energy is quantized.

Solve the boundary problems at x=0 and L. Multiply the second BC by and add. Particle in infinite box – bounded particles U E o This gives The quantization condition

Quantized energy levels U E o The quantization condition requires that: The case of is just the trivial case, ie, no particle in the box. From the energy relation for wave number we have, Thus, the energy is quantized!!

Energy eigenfunctions of particle in infinite box potential E o Go back to the general solution and insert quantized values

Energy eigenfunctions of particle in infinite box potential E o Look at coefficient of last term From quantization condition which means so and

Since the phase term cancels out when we take the modulus squared, we can choose it to be whatever we want. Let’s choose a value that makes the wave function completely real. That is, This gives an eigenfunction Eigenfunction and Normalization E o We have an energy eigenfunction We must now determine the constant Assume form of constant Normalization condition is The modulus square is that is o o o

Now Require Normalization E o Normalization condition is From trig identities we have Since the integral over cos term gives zero, we have and the coefficient is given by o o o This gives a normalized eigenfunction