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) $C_{not}$.

page revision: 0, last edited: 18 Nov 2012 18:37

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)etc.

(2)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

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

ReplyOptionsAs 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>

ReplyOptionsAlso |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.

ReplyOptions