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Quantum mechanics homework help 6
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PY452 Homework 6Guillem Cucurull
28 October 2021
1 Problem 8.2
Find the best bound on Egs for the one-dimensional harmonic oscillator using
a trial wave function of the form
A
ψ(x) =
x2 + b2
where A is determined by normalization and b is an adjustable parameter.
We need to first find A by normalizing ψ(x):
Z +∞ Z +∞ Z +∞
A 1
⟨ψ(x)|ψ(x)⟩ = ψ(x)ψ(x)∗ dx = ( 2 2
)2
dx = 2|A|2
dx
−∞ −∞ x + b 0 (x + b2 )2
2
We substitute the following: x = btanα, so dx = dαbsec2 α; and 1 + tan2 x =
2
cot x
+∞ Z +∞
bsec2 α 2A2 +∞
Z Z
2 1 2 1
2A 2 + b2 )2
dx = 2A 2 tan2 α + b2 )2
dx = 3 2 α)(1 + tan2 α)2
dα
0 (x 0 (b b 0 (cos
2A2 +∞ 2A2 π A2 π
Z
= 3 cos2 α dα = 3 = =1
b 0 b 4 2b3
q
3
Therefore A = 2bπ
Now we find ⟨H⟩ = ⟨T ⟩ + ⟨V ⟩
∞
h̄2 2 d2
Z
1 1
⟨T ⟩ = − A dx
2m −∞ x2 + b dx2 x2 + b2
2
d2 1
Let´s find dx2 x2 +b2
separately:
d2 1 d
= (−1)(x2 +b2 )−2 (2x) = (−2x)(−2)(x2 +b2 )−3 (2x)−2(x2 +b2 )−2
dx2 x2 + b2 dx
8x2 2 6x2 − 2b2
= − =
(x2 + b2 )3 (x2 + b2 )2 (x2 + b2 )3
1
, We plug back into ⟨T ⟩
∞
2h̄2 2 3x2 − b2 2h̄2 2b3 h̄2
Z
π
⟨T ⟩ = − A dx = − ( )(− ) =
m 0 (x2 + b2 )4 m π 16b5 4mb2
∞ ∞
x2 x2 3
Z Z
1 2 2 π 2 2b π 1
⟨V ⟩ = mw2 A2 dx = mw2 A2 dx = mw A = mw = mw2 b2
2 −∞ (x2 + b2 )2 0 (x2 + b2 )2 4b π 4b 2
Therefore
h̄2 1
⟨H⟩ = ⟨T ⟩ + ⟨V ⟩ = + mw2 b2
4mb2 2
∂
We now minimize ⟨H⟩ to get ⟨H⟩min by ∂b
∂ −h̄2
⟨H⟩ = + mw2 b = 0
∂b 2mb3
2 1
so we have b4 (2m2 w2 ) = h̄2 . Thus b = ( 2mh̄2 w2 ) 4
So
√ √ √
h̄2 1 2 2 h̄2 1 2 h̄2 1 2 2 2
⟨H⟩min = + mw b = + mw [( ) 4 ]2 = h̄w( + ) = h̄w
4mb2 2 h̄2 1
2 2 2m 2 w2 4 4 2
4m[( 2m2 w2 ) ]
4
√
2 1
We see that 2 h̄w> 2 h̄w, which is the expected value.
2 Problem 8.3
Find the best bound on Egs for the delta function potential V (x) = −αδ(x),
using a triangular trial function (Equation 8.10, only centered at the origin).
This time a is an adjustable parameter.
Ax 0 ≤ x ≤ a2
a
ψ(x) = { A(a − x) 2 ≤x≤a
0 otherwise
The trial function is centered at the origin with the same form so:
A( a2 + x) − a2 ≤ x ≤ 0
ψ(x) = { A( a2 − x) 0 ≤ x ≤ a2
0 otherwise
We need to first find A by normalizing ψ(x):
+∞ 0 +a
a3 a3 A2 a3
Z Z Z
a 2 a
⟨ψ(x)|ψ(x)⟩ = ψ(x)ψ(x)∗ dx = A2 ( +x)2 dx+ A2 ( −x)2 dx = A2 ( + ) = =1
−∞ −a
2
2 0 2 24 24 12
2
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