does the wave function have a well-defined momentum? Explain.
The constant in the front multiplies the initial function and two opposing moving plane waves are added, so the momentum is not clearly defined.
The eigenvectors of a linear transformation are those vectors where the transformation only modifies the magnitude of the vectors and not the angle. The "eigenvalue" of vectors refers to the ratio by which they differ from the original, and the vectors it works with are referred to as the "eigenvectors" corresponding to the eigenvalue.
The momentum operator is:
Replace the value of the wave function in the above equation.
Hence, the constant in the front multiplies the initial functionand two opposing moving plane waves are added, so the momentum is not clearly defined.
In a study of heat transfer, we find that for a solid rod, there is a relationship between the second derivative of the temperature with respect to position along the rod and the first with respect to time. (A linear temperature change with position would imply as much heat flowing into a region as out. so the temperature there would not change with time).
(a) Separate variables this assume a solution that is a product of a function of x and a function of t plug it in then divide by it, obtain two ordinary differential equations.
(b) consider a fairly simple, if somewhat unrealistic case suppose the temperature is 0 at x=0 and , and x=1 positive in between, write down the simplest function of x that (1) fits these conditions and (2) obey the differential equation involving x. Does your choice determine the value, including sign of some constant ?
(c) Obtain the full for this case.
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