Worked Problems in Physics

Nielsen and Chuang Exercise 2.26

Posted in Nielsen/Chuang by rpmuller on March 4, 2010

Exercise 2.26 Let {\left|\psi\right> = \frac{1}{\sqrt{2}}(\left|0\right>+\left|1\right>)}. Write out {\left|\psi\right>^{\otimes 2}} and {\left|\psi\right>^{\otimes 3}} explicitly, both in terms of tensor products like {\left|01\right>}, and using the Kronecker product.

In terms of tensor products, {\left|\psi\right>^{\otimes 2}=\frac{1}{2}(\left|00\right>+\left|10\right>+\left|01\right>+\left|11\right>)}. Writing this as a Kronecker product yields:

\displaystyle \left|\psi\right>^{\otimes 2}= \frac{1}{2}\left[ \begin{array}{r} 1 \\ 1 \\ 1 \\ 1 \end{array} \right].

Similarly, {\left|\psi\right>^{\otimes 3}=\frac{1}{2\sqrt{2}}(\left|000\right>+\left|100\right>+\left|010\right>+\left|110\right>+\left|001\right>+\left|101\right>+\left|011\right>+\left|111\right>)}. Writing this as a Kronecker product yields:

\displaystyle \left|\psi\right>^{\otimes 2}= \frac{1}{2\sqrt{2}}\left[ \begin{array}{r} 1 \\ 1 \\ 1 \\ 1\\1 \\ 1 \\ 1 \\ 1 \end{array} \right].

Nielsen and Chuang Exercise 2.25

Posted in Nielsen/Chuang by rpmuller on March 4, 2010

Exercise 2.25 Show that for any operator {A}, {A^\dagger A} is positive.

Because {B=A^\dagger A=B^\dagger}, {BB^\dagger=B^\dagger B} trivially, and thus {A^\dagger A} is normal, and thus has an eigendecomposition. We can therefore write:

\displaystyle  A^\dagger Au=\lambda u.

Multiplying both sides on the left by {u^\dagger}:

\displaystyle  u^\dagger A^\dagger Au=\lambda u^\dagger u=\lambda.

Defining {v=Au} yields

\displaystyle  v^\dagger v = |v|^2 = \lambda

thus any eigenvalue may be taken to be nonnegative, and thus the matrix {A^\dagger A} is positive.

Nielsen and Chuang Exercise 2.24

Posted in Nielsen/Chuang by rpmuller on March 4, 2010

Exercise 24 Show that a positive operator is necessarily Hermitian. Hint: show that an arbitrary operator {A} can be written {A=B+iC}, where {B} and {C} are Hermitian.

I don’t really get the point of this—obviously if a matrix has all nonnegative eigenvalues they are trivially real, and the matrix is Hermitian. However, I think this is something along the lines of what they want:

By defining the matrices {B=\frac{1}{2}(A+A^\dagger)} and {C=\frac{-i}{2}(A-A^\dagger)} we can write {A=B+iC} where {B} and {C} is Hermitian. We can then show that the eigenvalues of {B,C} are given by:

\displaystyle Bu_i = \lambda_iu_i

\displaystyle Cv_i = \mu_iv_i

where {\lambda_i, \mu_i} are real. Thus the eigenvalues of {A} can be written as the sum of {\lambda_i + i\mu_i}. If the matrix is positive, the imaginary part must be zero, so that the eigenvalues are all real and the matrix is Hermitian

Nielsen and Chuang Exercise 2.23

Posted in Nielsen/Chuang by rpmuller on March 4, 2010

Exercise 2.23 Show that the eigenvalues of a projector are all either 0 or 1.

We can write a projector as

\displaystyle  P=\sum_i \left|i\right>\left<i\right| \ \ \ \ \ (49)

We may also define the projector of the orthogonal space {Q=(1-P)} in a diagonal basis (exercise 2.22)

\displaystyle  Q = \sum_j \left|j\right>\left<j\right| \ \ \ \ \ (50)

and can thus expand {P} in an eigenbasis with either eigenvalue 1 (for the {P} space), or eigenvalue 0 (for the {Q} space)

\displaystyle  P = \sum_i 1\left|i\right>\left<i\right| + \sum_j 0\left|j\right>\left<j\right| \ \ \ \ \ (51)

and thus the eigenvalues are either 0 or 1.

Nielsen and Chuang Exercise 2.22

Posted in Nielsen/Chuang by rpmuller on March 4, 2010

Exercise 2.22 Prove that two eigenvectors of a Hermitian operator with different eigenvalues are necessarily orthogonal.

A Hermitian operator can act to the right or the left. Thus, if we have

\displaystyle  Av_i = \lambda_i v_i \ \ \ \ \ (44)

\displaystyle  Av_j = \lambda_j v_j, \ \ \ \ \ (45)

then

\displaystyle  \left<v_j|A|v_i\right> = \lambda_i \left<v_j|v_i\right>, \ \ \ \ \ (46)

(acting to the right), or

\displaystyle  \left<v_j|A|v_i\right> = \lambda_j \left<v_j|v_i\right>, \ \ \ \ \ (47)

(acting to the left). Subtracting the two:

\displaystyle  0=\left<v_j|A|v_i\right>-\left<v_j|A|v_i\right>=(\lambda_i-\lambda_j)\left<v_j|v_i\right> \ \ \ \ \ (48)

so that either {\lambda_i=\lambda_j} or {\left<v_j|v_i\right>=0}