Worked Problems in Physics

Nielsen and Chuang Exercise 2.47

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.47 Suppose that {A} and {B} are Hermitian. Show that {i[A,B]} is Hermitian.

\displaystyle (i[A,B])^\dagger = [iA,iB]^\dagger = [(iB)^\dagger,(iA)^\dagger] = [-iB^\dagger,-iA^\dagger]=-i[B,A]=i[A,B].

Nielsen and Chuang Exercise 2.46

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.46 Show that {[A,B]=-[B,A]}.

\displaystyle -[B,A]=-(BA-AB)=(AB-BA)=[A,B].

Nielsen and Chuang Exercise 2.45

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.45 Show that {[A,B]^\dagger=[B^\dagger,A^\dagger]}.

\displaystyle [A,B]^\dagger=(AB-BA)^\dagger=(AB)^\dagger-(BA)^\dagger=B^\dagger A^\dagger-A^\dagger B^\dagger = [B^\dagger,A^\dagger]

Nielsen and Chuang Exercise 2.44

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.44 Suppose that {[A,B]=0} and {\{A,B\}=0} and {A} is invertible. Show that {B} must be 0.

If {[A,B]=0}, then {AB=BA} and if {\{A,B\}=0}, then {AB=-BA}. The only way for {AB=\pm BA} is for either {A} or {B} to be zero, and if {A} is invertible it isn’t zero, and thus {B} is zero.

Nielsen and Chuang Exercise 2.43

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.43 Show that for {j,k=1,2,3,}

\displaystyle \sigma_j\sigma_k = \delta_{jk}I + i\sum_{l=1,3}\epsilon_{jkl}\sigma_l.

If {j=k}, {\sigma_j^2=I}, and the {\sum_l} terms are zero. Otherwise, the terms are give by the matrix products from exercise 2.40.

Nielsen and Chuang Exercise 2.42

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.42 Verify that {AB=\frac{1}{2}([A,B]+\{A,B\})}.

\displaystyle \frac{1}{2}([A,B]+\{A,B\}) = \frac{1}{2}(AB-BA+AB+BA) = AB

Nielsen and Chuang Exercise 2.41

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.41 Anti-commutation relations for the Pauli matrices. Verify the anti-commutation relations {\{\sigma_i,\sigma_j\}=0}, where {i\neq j} are both chosen from the set 1,2,3. Also verify that ({i=0,1,2,3}) {\sigma_i^2=I}.

The anti-commutation relations follow from the Pauli products worked out in exercise 2.40. The Pauli squares were worked out in exercise 2.19.

Nielsen and Chuang Exercise 2.40

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.40 Commutation relations for the Pauli matrices. Verify the commutation relations {[X,Y]=2iZ}, {[Y,Z]=2iX}, {[Z,X]=2iY}. There is an elegant way of writing this using {\epsilon_{jkl}}, the antisymmetric tensor on three indices, for which {\epsilon_{jkl}=0}, except for {\epsilon_{123}=\epsilon_{231}=\epsilon_{312}=1} and {\epsilon_{321}=\epsilon_{213}=\epsilon_{132}=-1}:

\displaystyle [\sigma_j,\sigma_k]=2i\sum_{l=1}^3\epsilon_{jkl}\sigma_l

\displaystyle  XY = \left[\begin{array}{rr}0 & 1 \\ 1& 0\end{array}\right]\left[\begin{array}{rr}0 & -i \\ i&0\end{array}\right] = \left[\begin{array}{rr}i & 0 \\ 0&-i\end{array}\right] =iZ

\displaystyle  YX = \left[\begin{array}{rr}0 & -i \\ i&0\end{array}\right]\left[\begin{array}{rr}0 & 1 \\1&0\end{array}\right] = \left[\begin{array}{rr}-i & 0 \\ 0&i\end{array}\right] =-iZ

\displaystyle YZ = \left[\begin{array}{rr}0 & -i \\ i&0\end{array}\right]\left[\begin{array}{rr}1 & 0 \\ 0&-1\end{array}\right] = \left[\begin{array}{rr}0 & i \\ i&0\end{array}\right]=iX

\displaystyle ZY = \left[\begin{array}{rr}1 & 0 \\ 0&-1\end{array}\right] \left[\begin{array}{rr}0 & -i \\ i&0\end{array}\right] = \left[\begin{array}{rr}0 & -i \\ -i&0\end{array}\right]=-iX

\displaystyle ZX = \left[\begin{array}{rr}1 & 0 \\ 0&-1\end{array}\right]\left[\begin{array}{rr}0 & 1 \\ 1&0\end{array}\right] = \left[\begin{array}{rr}0 & 1 \\ -1&0\end{array}\right]=iY

\displaystyle XZ = \left[\begin{array}{rr}0 & 1 \\ 1&0\end{array}\right]\left[\begin{array}{rr}1 & 0 \\ 0&-1\end{array}\right] = \left[\begin{array}{rr}0 & -1 \\ 1&0\end{array}\right]=-iY

From these it follows that:

\displaystyle [\sigma_j,\sigma_k]=2i\sum_{l=1}^3\epsilon_{jkl}\sigma_l

Nielsen and Chuang Exercise 2.39

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.39 The Hilbert-Schmidt inner product on operators: the set {L_V} of linear operators on a Hilbert space {V} is obviously a vector space—the sum of two linear operators is a linear operator, {zA} is a linear operator is {A} is a linear operator and {z} is a complex number, and there is a zero element 0. An important additional result is that the vector space {L_V} can be given a natural inner product structure, turning it into a Hilbert space.

  1. Show that the function (,) on {L_V\times L_V} defined by {(A,B)\equiv \mathrm{tr}(A^\dagger B)} is an inner product function. This inner product is known as the Hilbert-Schmidt or trace inner product.
  2. If {V} has {d} dimensions, show that {L_V} has dimension {d^2}.
  3. Find an orthonormal basis of Hermitian matrices for the Hilbert space {L_V}.

  1. Show that the function (,) on {L_V\times L_V} defined by {(A,B)\equiv \mathrm{tr}(A^\dagger B)} is an inner product function. This inner product is known as the Hilbert-Schmidt or trace inner product.

    An inner product must satisfy:

    1. {(.,.)} is linear in the second argument:

      \displaystyle  \left(\left|v\right>,\sum_i\lambda_i\left|w_i\right>\right) = \sum_i\lambda_i\left(\left|v\right>,\left|w_i\right>\right)

      We can show this via:

      \displaystyle (A,zB)=\mathrm{tr}(A^\dagger zB)=\sum_{ik}A_{ik}^\dagger zB_{ki}=z\sum_{ik}A_{ik}^\dagger B_{ki}=z(A,B)

    2. {(\left|v\right>,\left|w\right>) = (\left|w\right>,\left|v\right>)^*}

      \displaystyle (A,B)=\sum_{ik}A_{ik}^\dagger B_{ki}=\sum_{ik}A_{ki}^*B_{ki}=\sum_{ik}B_{ik}^{*\dagger}A_{ki}^*=(B,A)^*

    3. {(\left|v\right>,\left|v\right>) \geq 0}, with equality iff {\left|v\right>=0}.

      \displaystyle (A,A)=\sum_{ik}A^\dagger_{ik}A_{ki}=\sum_{ik}A^*_{ki}A_{ki}=\sum_{ik}|A_{ik}|^2.

      This is a sum of strictly positive values, so that the sum is {\geq 0}, and is equal to zero iff all entries are zero.

  2. If {V} has {d} dimensions, show that {L_V} has dimension {d^2}: For {L_V} to fully span {V}, it must be able to map any vector {v_i} into itself and into any other vector {v_j}. Thus there must be {d^2} different transformation.
  3. Find an orthonormal basis of Hermitian matrices for the Hilbert space {L_V}. Here we can do something along the lines of the Gramm-Schmidt orthogonalization. Taking the {L_V} linear operators, we can Hermitize each one via {\tilde{A_i}=(A_i+A_i^\dagger)/2}, and then orthogonalize each matrix {\tilde{A_i}} to all of the {\tilde{A_j}} for {j=1,\dots,i-1} by subtracting the inner product representation given by {(A,B)}, above.

I think my answers to 2 and 3 are still incomplete. Please comment if you have suggestions.

Nielsen and Chuang Exercise 2.38

Posted in Nielsen/Chuang by rpmuller on March 12, 2010

Exercise 2.38 Linearity of the trace: if {A} and {B} are two linear operators, show that {\mathrm{tr}(A+B)=\mathrm{tr}(A)+\mathrm{tr}(B)}, and, if {z} is an arbitrary complex number, show that {\mathrm{tr}(zA)=z\mathrm{tr}(A)}.

\displaystyle \mathrm{tr}(A+B)=\sum_i(A_{ii}+B_{ii}) = \sum_iA_{ii}+\sum_iB_{ii}=\mathrm{tr}(A)+\mathrm{tr}(B)

\displaystyle \mathrm{tr}(zA)=\sum_izA_{ii}=z\sum_{i}A_{ii}=z\mathrm{tr}(A).