Ex5 5

$\def\abs#1{|#1|}\def\i{\mathbf {i}}\def\ket#1{|{#1}\rangle}\def\bra#1{\langle{#1}|}\def\braket#1#2{\langle{#1}|{#2}\rangle}\def\tr{\mathord{\mbox{tr}}}\mathbf{Exercise\ 5.5}$

A vector $\ket\psi$ is stabilized by an operator $U$ if $U\ket\psi = \ket\psi$. Find the set of vectors stabilized by

a) the Pauli operator $X$,

b) the Pauli operator $Y$,

c) the Pauli operator $Z$,

d) $X\otimes X$,

e) $Z\otimes X$,

f) $C_{not}$.

enough?

Laker4 22 Jan 2014 01:59

So for a) through c) this question seems pretty straight forward, basically and eigenvector problem. You put the operators into matrix form and figure out the eigenvectors. For d) and e) I did the same thing where $X \otimes X$ is

(1)

\begin{align} \left( \begin{array}{cccc} 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0\\ 0 & 1 &0 & 0\\ 1 & 0 & 0 & 0 \end{array} \right) \end{align}

etc.
However, for the $C_{not}$ operator I had some difficulty solving for the eigenvectors, I think it's because there are essentially infinite solutions. Basically any vector of the form

(2)

\begin{align} |0>\otimes \left(a|0> + b|1>\right) = a|00> + b|11> \end{align}

is going to be stable when acted on with $C_{not}$

Last edited on 22 Jan 2014 02:01 by Laker4 Show more

Hephaestis 13 Jan 2017 23:58

As Laker4 said, except for f. this is straight-forward. Just set up a matrix-vector product M|v> and solve for |v> = M|v>. (I'm using |v> in place of |psi> in the problem statement. I'm also occasionally writing the answers as transposes because I don't have Tex.) We get
a. |v> = a(|0> + |1>), for any scalar a. [I.e., the 2-D vector with both coordinates equal to each other.]
b. Only the null vector, because the product shows you that the first coordinate of |v> must equal the second coordinate and it must also equal the negative of the second.
c. |v> = a|0>, for any scalar a. [I.e., any 2-D vector with second coordinate = zero.]
d. <v| = <u|<w|, where <u| = <a,a| and <w|= <b,b|, for any coordinates a and b.
e. <v| = <u|<w|, where <w| = <a,0| and <u| = <b,b| for any coordinates a and b
f. |v> = |0>|u> for any vector |u> = a|0> + b|1>, for any scalars a and b, because C-not with control |0> does not change either the control or the target. In Laker4's answer for f., there is a typo, I think. I believe the last term should be a|00> + b|01>

More eigenvectors for (f)

Joshua Holden 14 Sep 2018 19:51

Also |0>|+> and |1>|+> are stabilized by CNOT, so anything of the form a|00> + b|01> + c|1>|+> is stabilized. That's the complete three-dimensional eigenspace.

HornetMaX 18 Mar 2020 17:44

To me they are all the same: you have to solve for $U\ket{\psi}=\ket{\psi}$ and you can do this with a matrix representation or with the bra-ket notation. On point (f) (the same apply to all the others):

1. With matrix representation (standard base and ordering):

(1)

\begin{align} C_{not} = \begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 1 & 0 \end{pmatrix}, \ket{\psi} = a\ket{00} + b\ket{01} + c\ket{10} + d\ket{11} = \begin{pmatrix} a \\ b \\ c \\ d \end{pmatrix}\\ \end{align}

Solving for $C_{not}\ket{\psi} = \ket{\psi}$ gives you the costraints $a=a, b=b, c=d, d=c$, i.e. any vector of the form:

(2)

\begin{align} \ket{\psi} = a\ket{00} + b\ket{01} + c\ket{10} + c\ket{11} = \begin{pmatrix} a \\ b \\ c \\ c \end{pmatrix}\\ \end{align}

2. With bra-ket notation:

(3)

\begin{align} C_{not}=\ket{00}\bra{00}+\ket{01}\bra{01}+\ket{11}\bra{10}+\ket{10}\bra{11} \end{align}

Now again solving for $C_{not}\ket{\psi} = \ket{\psi}$ we get:

(4)

\begin{align} a\ket{00}+b\ket{01}+c\ket{11}+d\ket{10} = a\ket{00}+b\ket{01}+c\ket{01}+d\ket{11} \end{align}

Identifying the terms gives again the same constraints $a=a, b=b, c=d, d=c$, i.e. again any vector of the form:

(5)

\begin{align} \ket{\psi} = a\ket{00} + b\ket{01} + c\ket{10} + c\ket{11} = \begin{pmatrix} a \\ b \\ c \\ c \end{pmatrix}\\ \end{align}