PDA

View Full Version : Probability Density in Quantum Mechanics


JamesJames
Sep8-04, 08:50 PM
Consider the wave function corresponding to a free particle in one dimension. Construct the probability density and graph it as a function of position. Is this wavefunction normalizable?

Now, I think that the function should be Psi = C1*exp(ikx-iEt). Thus, the probability density should be the integral over x of C1^2. What I am confused about is what are the limits of the integration...0 and inifinity? And also won' t the answer come out to be a constant if the upper limit is not infinity ? I am totally lost on the normalizable issue. :confused:

Thanks,
james

HallsofIvy
Sep9-04, 08:10 AM
A probability density is NOT an integral- it is the density function itself which is what you give. The probability density is not a number, it is a function.

JamesJames
Sep9-04, 01:14 PM
So, it should then be
[C1*exp(ikx-iEt)][C1*exp(-ikx+iEt)] which equals C1^2. I don' t see how you can get a function and what about the normalization question?

Any help would be great

Tide
Sep9-04, 01:29 PM
Note, your C1 may be complex.

With regard to normalization, if you KNOW your particle is absolutely contained in some region then \left| C1\right|^2 = \frac {1}{L} where L is the size (length) of the region. Ask yourself what happens when L tends toward infinity.

JamesJames
Sep10-04, 09:42 AM
I am still confused about the normalization...does it have to be of length L? Also the probability density would then produce a straight line rather than some curve which seems unusual to me.

Tide
Sep10-04, 10:09 AM
Normalization means that you want to find the amplitude of the wave function and you do that by requiring the probability of finding the particle anywhere to be 1. If you KNOW it's in a box then the size of the box will be part of the normalization.