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University of California at San Diego – Department of Physics – TA: Shauna Kravec
Quantum Mechanics C (Physics 130C) Winter 2015
Worksheet 10 –
Solutions
Announcements
• The 130C web site is:
http://physics.ucsd.edu/∼mcgreevy/w15/ .
Please check it regularly! It contains relevant course information!
• This is going to be the last worksheet of the quarter so let’s finish with fun things.
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Problems
1. Supersymmetry?
Consider a spin- 12 particle on a line and in a magnetic field. The Hamiltonian is:
ˆ = ( 1 Pˆx2 + V (x))1 + B(x)σ z
H
2
(1)
The wavefunction of this system would have two components called a spinor.
Suppose your wrote your V (x) and B(x) as the following:
1
V (x) = (∂x W )2
2
B(x) = ∂x2 W
Pˆx = −i∂x
Where W (x) is an arbitrary function known as the superpotential. Now define the
following operators:
Q = (Pˆx − i∂x W )σ +
Q† = (Pˆx + i∂x W )σ −
Where recall σ ± = 21 (σ x ± iσ y ) are raising and lowering operators. Q and Q† are
known as supercharges.
(a) Show that Q2 = 0 = (Q† )2
(σ ± )2 = 14 (σ x ± iσ y )2 = 1 − 1 + (σ z − σ z ) = 0
1
ˆ where recall {A, B} = AB + BA
(b) Show that {Q, Q† } = 2H
This shows the algebra of the supercharges gives you the generator of (time)
translations!
Expand directly:
QQ† = (Pˆx2 + (∂x W )2 + iPˆx (∂x W ) − i(∂x W )Pˆx )σ + σ −
Q† Q = (Pˆx2 + (∂x W )2 − iPˆx (∂x W ) + i(∂x W )Pˆx )σ − σ +
Now we use σ + σ − + σ − σ + = 1 , and iPˆx (∂x W ) = ∂x2 W Sum the above:
ˆ
(Pˆx2 + (∂x W )2 )1 + (iPˆx (∂x W ) − i(∂x W )Pˆx )σ z = (Pˆx2 + 2V (x))1 + 2B(x)σ z = 2H
Note the integration by parts on −i(∂x W )Pˆx
ˆ = 0 = [Q† , H]
ˆ
(c) Show that [Q, H]
ˆ = 1 ([Q, QQ† ] + [Q, Q† Q]) = 1 (Q[Q, Q† ] + [Q, Q† ]Q)
[Q, H]
2
2
Now we expand using Q2 = 0 :
ˆ
Similarly for [Q† , H]
= 12 (−QQ† Q + QQ† Q) = 0
Now what does supersymmetry do? Consider an eigenstate |ψi with energy E.
ˆ
(d) Compute hψ|H|ψi.
What does this tell us about the ground state of the system?
1
†
ˆ
hHi = (hψ|QQ + Q† Q|ψi = 1 (|Q† |ψi|2 + |Q|ψi|2 )
2
2
This implies that E = 0 → Q|ψ0 i = 0 = Q† |ψ0 i
(e) Now also suppose that W (x) → ∞ as x → ±∞. Using
above constraint on
the ψ+ (x)
the ground state construct the wavefunction ψ0 (x) =
in terms of W (x).
ψ− (x)
Hint: It may be helpful to write:
0
0
0 Pˆx − iW 0 (x)
†
Q=
Q = ˆ
0
0
Px + iW 0 (x) 0
This amounts to solving a pair of differential equations:
−i∂x ψ± (x) ± iW 0 (x)ψ± (x) = 0
where we’ve used that supercharges annihilate the ground state.
W (x) Ae
Therefore ψ0 (x) =
where the above assumption of W (±∞) = ∞ and
Be−W (x)
normalization imply A = 0
2