FYS2140 - oblig 6
Jørgen Eriksson Midtbø
University of Oslo
jorgeem@student.matnat.uio.no
Candidate #143
Group #1
February 27, 2011
Task 1
a) We have that the ground state Ψ0 for the harmonic oscillator is
mω 1/4 − mω x2
e 2~
Ψ0 =
π~
And the ladder operator
1
d
(−~ + mωx)
â+ = √
dx
2~mω
So to construct Ψ2 , we go like this:
d
mω 1/4 − mω x2
+ mωx)
e 2~
dx
π~
2~mω
1/4 mω
2mωx − mω x2
2
− mω
2~
2~ x
=
e
+
mωxe
2
π22 ~3 m2 ω2
!
2 3 3 1/4
mω 2
2 m ω
xe− 2~ x
=
π~3
â+ Ψ0 = √
1
(−~
!1/4
mω 2
d
22 m3 ω3
â+ â+ ψ0 = √
(−~ + mωx)
xgi f e− 2~ x
3
dx
π~
2~mω
!
!1/4
2 3 3
mω 2
2 m ω
d
= 2 5 2 2
−~ + mωx xe− 2~ x
dx
2 ~m ωπ
1/4 mω 2
mω 2
mω 2
1 mω
−mω
=
−~e− 2~ x − ~x
2xe− 2~ x + mωx2 e− 2~ x
~ π~
2~
mω 2
1 mω 1/4
=
(−~ + mωx2 + mωx2 )e− 2~ x
~ π~
mω 2
mω 1/4 2mω 2
=
x − 1 e− 2~ x
π~
~
1
Then, by equation (2.67) of Griffiths,
mω 2
1 mω 1/4 2mω 2
Ψ2 (x) = √
x − 1 e− 2~ x
π~
~
2
1
b) I’ll cheat a bit and make Wolfram Alpha draw for me. See figures 1, 2 and 3. (Obviously, Ψ2 is a bit
tricky since the mass is not specified, so the two terms could have different weighting.)
Figure 1: Ψ0
Figure 2: Ψ1
Figure 3: Ψ2
c) Orthogonality. If two wavefunctions are orthogonal, then by definition hΨn | Ψm i = 0 for n , m. This is
easily checked:
Z
∞
hΨ0 | Ψ1 i =
−∞
Ψ∗0 Ψ1 dx
Ψ0 is an even function, and Ψ1 is odd. Therefore, this integral is zero. The same is true for
hΨ1 | Ψ2 i. The only one needing further attention is
2
Z ∞
hΨ0 | Ψ2 i =
Ψ∗0 Ψ2 dx
−∞
Z ∞
mω 2
mω 1/4 − mω x2 1 mω 1/4 2mω 2
2~
=
e
x − 1 e− 2~ x dx
√
π~
π~
~
−∞
2
Z mω 2
1 mω 1/2 ∞ 2m 2
x − 1 e− 2~ x
= √
~
−∞
2 π~
I’ll drop the constant factor.
Z
∞
mω 2
2m 2
x − 1 e− 2~ x dx
~
−∞
Z ∞
2m 2 − mω x2
− mω
x2
2~
2~
x e
−e
=
dx
~
−∞
r
r
π~
π~
−
= 0.
mω
mω
All three wavefunctions are orthogonal to each other.
Task 2
r
a) Let ξ =
mω
mω 1/4
x, α =
. This gives
~
π~
ξ2
Ψ0 = αe− 2
√
ξ2
Ψ1 = 2αξe− 2
ξ2
1
Ψ2 = √ α(2ξ2 − 1)e− 2
2
Take Ψ0 first:
Z
hxi = hΨ0 | x̂ | Ψ0 i =
−∞
α2
~
mω
Z
∞
ξe
2
− ξ2
r
∞
Ψ∗0 xΨ0
~
x=
ξ
mω
r
~
dx =
dξ
mω
dx
dξ
−∞
The integrand is an odd function, so this integral is 0.
hx2 i = hΨ0 | x̂2 | Ψ0 i =
!3/2 Z ∞
ξ2
~
2
=α
ξ2 e− 2 dξ
mω
−∞
!3/2 √
1/2
π
mω
~
=
π~
mω
2
~
=
2mω
3
Z
∞
Ψ∗0
hpi = hΨ0 | p̂ | Ψ0 i =
−∞
Z ∞
ξ2 ~ d
ξ2
=
αe− 2
αe− 2 dx
i dx
−∞
Z ∞
ξ2
~
= α2
−ξe− 2 dξ = 0
i −∞
!
~ d
Ψ0 dx
i dx
Since the integrand is again odd.
∞
Z
Ψ∗0
~ d
i dx
!2
hp i = hΨ0 | p̂ | Ψ0 i =
Ψ0 dx
−∞
Z ∞
√
ξ2
ξ2
d2
αe− 2 (−~2 ) 2 αe− 2
~mω dξ
=
dx
−∞
r
!
Z ∞
2
d − ξ2 dξ
~
2 2
− ξ2 d
2
= −α ~
e
e
dξ
mω −∞
dx dξ
dx
!
Z ∞
ξ2 d
ξ2 dξ
e− 2
= −~2 α2
−ξe− 2
dξ
dξ
dx
−∞
r
Z
ξ2
ξ2
mω ∞ − ξ2
2 2
e 2 −e− 2 + ξ2 e− 2 dξ
= −~ α
~ −∞
r
Z
mω ∞ −ξ2 2
= −~2 α2
e (ξ − 1) dξ
~ −∞
 p 
r
pi  mω~
mω 
2 2
 =
.
= −~ α
−
~
2 
2
2
2
And for Ψ1 :
∞
Z
hxi =
−∞
~
= 2α
mω
∞
Z
Ψ∗1 x̂Ψ1
Z
dx =
√
√
ξ2
ξ2
2αξe− 2 x 2αe− 2 dx
−∞
∞
3 −ξ2
ξe
2
dξ = 0
−∞
Since the integrand is again again odd.
∞
Z
hx i = hΨ1 | x̂ | Ψ1 i =
2
Ψ∗1 x2 Ψ1 dx
r
!
√
ξ2
~
~
ξ2 2αξe− 2
dξ
mω
mω
2
−∞
Z
∞
=
√
ξ2
2αξe− 2
−∞
= 2α2
= 2α2
=
~
mω
~
mω
!3/2 Z
∞
2
ξ4 e−ξ dξ
−∞
!3/2
p
3 pi
2
=2
mω
π~
1/2
~
mω
!3/2
3 ~
2 mω
4
√
3 π
4
∞
!
~ d
Ψ1 dx
hpi = hΨ1 | p̂ | Ψ1 i =
i dx
−∞
!
Z ∞
√
√
2
2
− ξ2 ~ d
− ξ2 dξ dx
=
dξ
2αξe
2αξe
i dξ
dx dξ
−∞
Z
ξ2
~ ∞ − ξ2 − ξ2
= 2α2
ξe 2 e 2 − ξ2 e− 2 dξ
i −∞
Z ∞h
i
~
2
2
= 2α2
ξe−ξ − ξ3 e−ξ dξ = 0
i −∞
Z
Ψ∗1
Since the integrand is again again again odd. (And because stationary states are characterized
by 0 momentum).
∞
Z
2
−∞
Ψ∗1
~ d
i dx
!2
Ψ1 dx
!!
Z ∞
√
ξ2
ξ2 dξ
d d √
dx
2αξe− 2 (−~2 )
2αξe− 2
dξ
dx
dξ
dx
dξ
−∞
Z ∞
!
ξ2
2
2
2 − ξ2 dξ
2 2
− ξ2 d
− 2
−ξ e
= −2α ~
ξe
e
dξ
dξ
dx
−∞
r
Z
ξ2
ξ2
ξ2
mω ∞ − ξ2
ξe 2 −ξe− 2 − 2ξe− 2 + ξ3 e− 2 dξ
= −2α2 ~2
~ −∞
r
Z
ξ2
mω ∞ 4
2 2
= −2α ~
(ξ − 3ξ2 )e− 2 dξ
~ −∞
r
√ !
mω 3 π
= −2α2 ~2
−
~
4
r
r
√
mω 2 mω 3 π 3
=2
~
= mω~
π~
~
4
2
hp i = hΨ1 | p̂ | Ψ1 i =
2
b) Uncertainty. Need standard deviations.
For Ψ0 :
~
σ2x = hx2 i − hxi2 =
2mω
r
~
σx =
2mω
mω~
σ2p = hp2 i − hpi2 =
2
r
mω~
σp =
2
q
q
~
mω~
Giving σx σp =
=
2mω
2
lowest possible energy. Nice!
~
2!
Exactly at the limit, and the wavefunction corresponds to the
For Ψ1 :
5
3~
σ2x = hx2 i − hxi2 =
2mω
r
3~
σx =
2mω
3mω~
σ2p = hp2 i − hpi2 =
2
r
3mω~
σp =
2
q
q
3~
3mω~
Giving σx σp = 2mω
=
2
3~
2
≥ ~2 .
c) Kinetic and potential energy. For Ψ0 :
1
1 mω~ ~ω
mhp2 i =
=
2
2m 2
4
1
1
~ω
2 2
2 ~
hVi = mω hx i = mω
=
2
2
2mω
4
hKi =
So hKi + hVi =
~ω
2
= E0 , which is how it should be.
For Ψ1 :
1 3mω~ 3~ω
1
mhp2 i =
=
2
2m 2
4
1
1
3~ω
3~
hVi = mω2 hx2 i = mω2
=
2
2
2mω
4
hKi =
So hKi + hVi =
3~ω
2
= E1 , again good.
End of oblig 6.
6