Q6:4 We showed in the Lecture notes that the normalised energy eigenfunction for the lowest energy eigenvalue of the harmonic oscillator (E = ħw/2) is 1/2 exp a²r² to (x) = 2 where a = . As we encountered in the group work, in quantum mechanics the uncertainty in the position of the particle may be described by Ar = / (x²) – (x)², for a particle represented by a given wavefunction. Calculate Ax for the wavefunction yo (x). In addition, calculate the uncertainty in the momentum of the particle for the same wavefunction, Ap, namely Ap = / (p²) – (p)² and hence show that ArAp = (This is an example of a Heisenberg uncertainty relation) Note that the operator p= -iħ and p² = –ħ² . [You may assume that ſ y² exp (–y³)dy = Vñ/2.]

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y = αx

Q6:4 We showed in the Lecture notes that the normalised energy eigenfunction for the lowest
energy eigenvalue of the harmonic oscillator (Eo = ħw/2) is
1/2
exp
a?r?
th (1) = ()" p (-)
a
-1/2
2
mw
where a = .
As we encountered in the group work, in quantum mechanics the uncertainty in
the position of the particle may be described by
Ar = V (r²) – (x)²,
for a particle represented by a given wavefunction.
Calculate Ar for the wavefunction to (x). In addition, calculate the uncertainty in the momentum
of the particle for the same wavefunction, Ap, namely
Ap = (p²) – (p)²
and hence show that
AxAp =
2
(This is an example of a Heisenberg uncertainty relation)
Note that the operator p= -iħ and p² = –h° .
[You may assume that ſ y² exp (-y²)dy = /2.]
Transcribed Image Text:Q6:4 We showed in the Lecture notes that the normalised energy eigenfunction for the lowest energy eigenvalue of the harmonic oscillator (Eo = ħw/2) is 1/2 exp a?r? th (1) = ()" p (-) a -1/2 2 mw where a = . As we encountered in the group work, in quantum mechanics the uncertainty in the position of the particle may be described by Ar = V (r²) – (x)², for a particle represented by a given wavefunction. Calculate Ar for the wavefunction to (x). In addition, calculate the uncertainty in the momentum of the particle for the same wavefunction, Ap, namely Ap = (p²) – (p)² and hence show that AxAp = 2 (This is an example of a Heisenberg uncertainty relation) Note that the operator p= -iħ and p² = –h° . [You may assume that ſ y² exp (-y²)dy = /2.]
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