Example 3.9 Derive the transformation from the position-space wave function to the momentum-space wave function. (We already know the answer, of course, but I want to show you how this works out in Dirac notation.) Solution: We want to find (p, t) = (p|S (1)) given (x, 1) = (x\S (t)). We can relate the two by inserting a resolution of the identity: O (p, t) = (p|S (t)) (3.108) x \x) = | (plx) (x\S (1)) dx = | (plx) ♥ (x, 1) dx. Now, (x|p) is the momentum eigenstate (with eigenvalue p) in the position basis-what we called fp (x), in Equation 3.32. So (p|x) = (x\p)* = [fp (x)]" : i px/h_ Plugging this into Equation 3.108 gives O (p, t) = -i px/h\v (x, t) dx, 2n h which is precisely Equation 3.54.
Example 3.9 Derive the transformation from the position-space wave function to the momentum-space wave function. (We already know the answer, of course, but I want to show you how this works out in Dirac notation.) Solution: We want to find (p, t) = (p|S (1)) given (x, 1) = (x\S (t)). We can relate the two by inserting a resolution of the identity: O (p, t) = (p|S (t)) (3.108) x \x) = | (plx) (x\S (1)) dx = | (plx) ♥ (x, 1) dx. Now, (x|p) is the momentum eigenstate (with eigenvalue p) in the position basis-what we called fp (x), in Equation 3.32. So (p|x) = (x\p)* = [fp (x)]" : i px/h_ Plugging this into Equation 3.108 gives O (p, t) = -i px/h\v (x, t) dx, 2n h which is precisely Equation 3.54.
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Problem 3.30 Derive the transformation from the position-space wave function to the “energy-space” wave function using the technique of Example 3.9. Assume that the energy spectrum is discrete, and the potential is time-independent.
![Example 3.9
Derive the transformation from the position-space wave function to the momentum-space wave
function. (We already know the answer, of course, but I want to show you how this works out in
Dirac notation.)
Solution: We want to find (p, t) = (p|S (1)) given (x, 1) = (x\S (t)). We can relate the two
by inserting a resolution of the identity:
O (p, t) = (p|S (t))
(3.108)
x \x)
= | (plx) (x\S (1)) dx
= | (plx) ♥ (x, 1) dx.
Now, (x|p) is the momentum eigenstate (with eigenvalue p) in the position basis-what we called
fp (x), in Equation 3.32. So
(p|x) = (x\p)* = [fp (x)]" :
i px/h_
Plugging this into Equation 3.108 gives
O (p, t) =
-i px/h\v (x, t) dx,
2n h
which is precisely Equation 3.54.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5f916f44-2b56-40b3-9b96-44e026d40d84%2F52f1fda3-95c0-44ae-8e09-8ecbbb10ed31%2F2vs019a_processed.png&w=3840&q=75)
Transcribed Image Text:Example 3.9
Derive the transformation from the position-space wave function to the momentum-space wave
function. (We already know the answer, of course, but I want to show you how this works out in
Dirac notation.)
Solution: We want to find (p, t) = (p|S (1)) given (x, 1) = (x\S (t)). We can relate the two
by inserting a resolution of the identity:
O (p, t) = (p|S (t))
(3.108)
x \x)
= | (plx) (x\S (1)) dx
= | (plx) ♥ (x, 1) dx.
Now, (x|p) is the momentum eigenstate (with eigenvalue p) in the position basis-what we called
fp (x), in Equation 3.32. So
(p|x) = (x\p)* = [fp (x)]" :
i px/h_
Plugging this into Equation 3.108 gives
O (p, t) =
-i px/h\v (x, t) dx,
2n h
which is precisely Equation 3.54.
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