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    Solutions: Homework Set 2: Due August 21

    Uploaded by

    Michel Rodrigues Andrade
    1) The document provides homework problems on normal operators in quantum mechanics. 2) Problem 1 asks students to show properties of normal operators, including that eigenbras are Hermitian conjugates of eigenkets. 3) Problem 2 asks about conditions for a superposition of eigenstates to also be an eigenstate. 4) Problem 3 considers properties of the operator given by the product of differences between another operator and its nondegenerate eigenvalues.

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    Solutions: Homework Set 2: Due August 21

    Uploaded by

    Michel Rodrigues Andrade
    144 views4 pages
    1) The document provides homework problems on normal operators in quantum mechanics. 2) Problem 1 asks students to show properties of normal operators, including that eigenbras are Hermitian conjugates of eigenkets. 3) Problem 2 asks about conditions for a superposition of eigenstates to also be an eigenstate. 4) Problem 3 considers properties of the operator given by the product of differences between another operator and its nondegenerate eigenvalues.

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    Quantum mechanics

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    HW2solutions

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    1) The document provides homework problems on normal operators in quantum mechanics. 2) Problem 1 asks students to show properties of normal operators, including that eigenbras are Hermitian conjugates of eigenkets. 3) Problem 2 asks about conditions for a superposition of eigenstates to also be an eigenstate. 4) Problem 3 considers properties of the operator given by the product of differences between another operator and its nondegenerate eigenvalues.

    Copyright:

    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
    Download as pdf or txt
    144 views4 pages

    Solutions: Homework Set 2: Due August 21

    Uploaded by

    Michel Rodrigues Andrade
    1) The document provides homework problems on normal operators in quantum mechanics. 2) Problem 1 asks students to show properties of normal operators, including that eigenbras are Hermitian conjugates of eigenkets. 3) Problem 2 asks about conditions for a superposition of eigenstates to also be an eigenstate. 4) Problem 3 considers properties of the operator given by the product of differences between another operator and its nondegenerate eigenvalues.

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    © All Rights Reserved
    Download as PDF, TXT or read online from Scribd
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    Solutions:

    Homework Set 2
    Due August 21

    1. The problem of finding the eigenkets and eigenbras of an arbitrary operator is more compli-
    cated and full of exceptions than in the case of Hermitian operators. There are, however,
    other classes of operators that share many of the nice properties of Hermitian operators.
    These include anti-Hermitian and unitary operators.
    We define an operator to be normal if it commutes with its Hermitian conjugate, [A, A† ] = 0.
    Notice that Hermitian, anti-Hermitian, and unitary operators are normal. In the following
    you may assume that you are working in a finite-dimensional Hilbert space.

    (a) Show that if A is normal, and A|ui = a|ui for some nonzero |ui, then A† |ui = a∗ |ui.
    Thus, the eigenbras of A are the Hermitian conjugates of the eigenkets, and the left
    spectrum is identical to the right spectrum. Hint: it is not necessary to introduce
    orthonormal bases or anything similar.
    Suppose [A, A† ] = 0 and A|ui = a|ui ( =⇒ hu|A† = hu|a∗ ).
    Let

    |φi ≡ (A† − a∗ )|ui, (1)

    Consider

    hφ|φi = hu|(A − a)(A† − a∗ )|ui (2)


    = hu| AA† − Aa∗ − A† a + |a|2 |ui
    
    (3)
    = hu| |a|2 − |a|2 − |a|2 + |a|2 |ui = 0
    
    (4)

    =⇒ |φi = |0i, or equivalently A† |ui = a∗ |ui


    (b) Show that the eigenspaces corresponding to distinct eigenvalues of a normal operator
    are orthogonal. This is a generalization of the easy and familiar proof for Hermitian
    operators.

    1
    Let

    A|u1 i = a1 |u1 i (5)


    A|u2 i = a2 |u2 i (6)

    Eq. (5) implies

    hu2 |A|u1 i = a1 hu2 |u1 i (7)

    From the results of part 1a and Eq. (6):

    A† |u2 i = a∗2 |u2 i (8)


    =⇒ hu2 |A = a2 hu2 | (9)
    =⇒ hu2 |A|u1 i = a2 hu2 |u1 i (10)

    Subtracting Eq. (7) from (10) gives

    (a1 − a2 )hu2 |u1 i = 0 (11)

    =⇒ either a1 = a2 , or hu2 |u1 i = 0.

    2. If the kets |αi and |βi are eigenkets of the Hermitian operator Â, under what conditions will
    the superposition |ψi = |αi + |βi be an eigenstate of Â? Explain.
    Denote the eigenvalues α and β, respectively:

    Â|αi = α|αi (12)


    Â|βi = β|βi (13)

    Â|ψi = α|αi + β|βi (14)


    ?
    = γ|ψi = γ (|αi + |βi) (15)

    This is only possible when the eigenvalues are degenerate α = β

    3. Consider an N -dimensional ket space spanned by the N eigenkets {|ai i} (where i = 1, . . . ,


    N ) of a Hermitian operator Â. Assume that the eigenvalues ai are nondegenerate.

    2
    (a) What ket(s) are produced by the action of the operator

    N 
    Y 
    θ̂ ≡ Â − ai = (Â − a1 )(Â − a2 )(Â − a3 ) . . . (Â − aN ) (16)
    i=1

    on ANY arbitrary ket in the ket space?


    The eigenkets of  span the space, and any ket can be written as a superposition of
    them:

    N
    X
    |ψi = ci |ai i (17)
    i−1

    with

    ci = hai |ψi. (18)

    θ̂ operating on an arbitrary state gives:

    N
    X N
    Y
    θ̂|ψi = cj (Â − ai )|aj i (19)
    j−1 i=1
    N
    X N
    Y
    = cj (aj − ai )|aj i (20)
    j−1 i=1

    For every term in the sum, there will be one factor with i = j which is zero, and the
    final result is the null ket.

    θ̂|ψi = |0i (21)

    Since this is true for any ket, θ̂ is the null operator


    (b) Setting  = Ŝz and acting on the ket space associated with a spin- 21 particle, demon-
    strate your answer to part (a).

    3
    Using the eigenbasis of Sx :

    ~
    Sx |+i = |+i (22)
    2
    ~
    Sx |−i = − |+i (23)
    2
    ~ ~
    =⇒ θ̂ = (Sx − )(Sx + ) (24)
    2 2

    Consider an arbitrary ket, written in the eigenbasis

    |ψi = a+ |+i + a− |−i (25)


    ~ ~
    θ̂|ψi = (Sx − )(Sx + ) (a+ |+i + a− |−i) (26)
    2 2
    ~
    = [a+ (1 − 1)(1 + 1)|+i + a− (1 + 1)(1 − 1)|−i] (27)
    2
    =0 (28)

    (c) What is the commutator of θ with itself?

    [θ̂, θ̂] = 0 (29)

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