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Department of Physics University of Cambridge

Part II Experimental and Theoretical Physics

Theoretical Physics 2

N.R. Cooper and M.C. Payne Lent Term 2010

Chapter 1

Introduction/Revision

Problem 1.1 a) We are given that

ˆ ˆ Q† Q† = 0, ˆˆ ˆ ˆ ˆ QQ† + Q† Q = I, ˆ ˆˆ H = αQQ† , where α is a real constant. a) ˆ ˆ ˆ H2 = HH = ˆˆ αQQ† ˆˆ αQQ†

(A1.1)

ˆˆˆˆ = α2 QQ† QQ† ˆ ˆ ˆˆ ˆ = α2 Q I − QQ† Q† ˆˆ ˆˆ = α2 QQ† − α2 QQ ˆ = αH, 1 ˆ ˆ Q† Q†

2

where we used the condition (A1.1) in the second last line.

ˆ b) Let E be the eigenvalue of H corresponding to the eigenvector |E . Therefore ˆ ˆ H|E = E|E −→ H 2 |E = E 2 |E , ˆ ˆ but from part a), H 2 = αH, hence E 2 |E = αE|E , which implies E 2 − αE = 0, hence E = 0or α.

ˆ ˆ ˆ ˆ Problem 1.2 a) Now, A|a = a|a , A2 |a = A (a|a ) = a A|a = a2 |a , which implies ˆ ˆ An |a = an |a . Therefore, deﬁning f (A)) by its Taylor expansion ˆ f A =
n

f (n) (0) ˆn A , n!

then ˆ f A |a =
n

f (n) (0) ˆn A |a n!

=
n

f (n) (0)an n!

|a

= f (a)|a . ˆ Hence, if |a is an eigenvector of A corresponding to the eigenvalue a then it is also an ˆ eigenvectorof f (A) with eigenvalue f (a).

ˆˆ b) To prove that AB

ˆ ˆ = B † A† . Consider |α and |β to be arbitrary kets. We have ˆˆ α| AB

=

ˆˆ β|AB|α

from the deﬁnition of the adjoint, and hence ˆˆ α| AB

= =

ˆ β|A Φ |Ψ
∗ ∗

ˆ B|α

ˆ ˆ where |Ψ = B|α and Φ| = β|A. However, Φ | Ψ ˆ ˆ α|B † and |Φ = A† |β , we have Ψ |Φ = = ˆ α|B † ˆˆ α| AB ˆ A† |β
†= Ψ | Φ , and since Ψ| = (|Ψ )† =

ˆ ˆ = α| B † A† |β

ˆ ˆ |β = α|B † A† |β

Theoretical Physics 2

3

ˆˆ and since |α and |β are arbitrary, this implies AB ˆˆ ˆ ˆˆ ˆ For AB C , let B C = D, then ˆˆ ˆ AB C
† †

ˆ ˆ = B † A† .

= =

ˆ ˆ = D† A† ˆ ˆ ˆ ˆ A† = C † B † A†

where we used the result that we just proved above. Hence in general we canprove similarly that ˆ ˆ ˆ A1 A2 . . . An ˆ c) Noting that A†
† †

ˆn ˆ ˆ ˆ = A† A† . . . A† A† . n−1 2 1

ˆ = A, then ˆˆ AA†

ˆ = A†

ˆ ˆˆ A† = AA† ,

ˆˆ ˆ which implies that AA† is Hermitian even if A is not.

ˆ ˆ d) Now, A2 = Ψ|A2 |Ψ , (assuming Ψ | Ψ = 1), implying ˆ A2 = ˆ ˆ (where we used A† = A). ˆ ˆ ˆ Now, if A|a = a|a , then A2 |a = a2 |a , and since A is Hermitian, itseigenvalues are real, ˆ which implies that a2 ≥ 0 and the eigenvalue of A2 is non-negative. ˆ Ψ|A ˆ A|Ψ ˆ = A|Ψ
2

≥ 0,

ˆ ˆ ˆ ˆ ˆˆ e) Now A|ψ = a|ψ , and given that A, B = B + 2B A2 , this implies ˆˆ ˆˆ ˆ ˆˆ ˆ ˆ ˆ ˆ AB = B A + B + 2B A2 = B A + I + 2A2 , hence ˆˆ AB|ψ ˆ ˆ ˆ ˆ = B A + I + 2A2 |ψ

4

ˆ = B a + 1 + 2a2 |ψ = 1 + a + 2a2 ˆ B|ψ ,

ˆ ˆ whichimplies that B|ψ is also an eigenvalue of A with eigenvalue (1 + a + 2a2 ).

implies

ˆ Problem 1.3 Let H0 be the matrix representation of H0 in the {|1 , |2 } basis, which ˆ 1|H0 |1 H0 =  ˆ 2|H0 |1 ˆ 1|V |1 V = ˆ 2|V |1
 

ˆ Let V be the matrix representation of V in the {|1 , |2 } basis, and hence ˆ 0 V12 1|V |2 . = ∗ ˆ |2 V12 0 2|V
  

ˆ E 0 1|H0 |2 . = 0 ˆ 0 E0 2|H0 |2

ˆ ˆ ˆ If H = H0 + V , and H is the matrix representation of this operator in the {|1 , |2 } basis, then we have

To ﬁnd the eigenvalues and eigenvectors of this matrix we need to solve the secular determinant (i.e. |H − EI| = 0) E0 − E
∗ V12

H=

E0
∗ V12

V12 E0

.

V12 E0 − E

=0

which implies (E0 − E)2 − |V12 |2 = 0, which has the solutions E± = E0± |V12 | . Therefore the eigenvalues of H0 are twofold degenerate and this degeneracy is broken by the “perturbation”, V . To solve for the eigenvectors, |E+ and |E− , we proceed as follows:
 

For E = E+ we have

E0 − E +
∗ V12

V12 E0 − E +

 

x y

 =

0

Theoretical Physics 2

5

E+ E0 E−
=⇒
 

}
V12

E = E + − E− = 2|V | 12
x y

− |V12...