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mtak0114
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1. Show that an arbitrary density operator for a mixed state qubit may be written as
2. [tex]\rho = \frac{I+r^i\sigma_i}{2}[/tex], where [tex]||r||<1[/tex]
(Nielsen and Chuang pg 105)
3. So my attempt was as follows
Given that a [tex]\rho[/tex] is hermitian it may be written as a linear combination of the pauli matrices and the identity.
[tex]\rho = aI+r^i\sigma_i[/tex] where r^i is arbitrary
the trace condition [tex]tr(\rho)=1[/tex] implies [tex]a=1/2[/tex] and
the positivity condition [tex]\langle\varphi|\rho|\varphi\rangle \geq 0[/tex]
implies that [tex]2cos\theta |r||\langle\varphi|\sigma^i|\varphi\rangle| \geq -1[/tex] which implies that
[tex]|r|\leq 1/2[/tex], fnally redefing r above gives the result QED
is this correct or am i assuming too much?
cheers
Mark
2. [tex]\rho = \frac{I+r^i\sigma_i}{2}[/tex], where [tex]||r||<1[/tex]
(Nielsen and Chuang pg 105)
3. So my attempt was as follows
Given that a [tex]\rho[/tex] is hermitian it may be written as a linear combination of the pauli matrices and the identity.
[tex]\rho = aI+r^i\sigma_i[/tex] where r^i is arbitrary
the trace condition [tex]tr(\rho)=1[/tex] implies [tex]a=1/2[/tex] and
the positivity condition [tex]\langle\varphi|\rho|\varphi\rangle \geq 0[/tex]
implies that [tex]2cos\theta |r||\langle\varphi|\sigma^i|\varphi\rangle| \geq -1[/tex] which implies that
[tex]|r|\leq 1/2[/tex], fnally redefing r above gives the result QED
is this correct or am i assuming too much?
cheers
Mark
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