Product state and orthorgonality

  • Thread starter KFC
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  • #1
KFC
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Main Question or Discussion Point

For two level atom trapped in a box (or cavity), initially at excited state without any photon inside, all possible states are

[tex]|0, e\rangle =|0\rangle|e\rangle \qquad and \qquad |1, g\rangle = |1\rangle|g\rangle[/tex]

e stands for excitated state, g stands for ground state.

Obviously,

[tex]\langle 0, e|0, e\rangle = 1[/tex]

and

[tex]\langle 1, g|1, g\rangle = 1[/tex]

I wonder if these two states [tex]|0, e\rangle[/tex] and [tex]|1\rangle|g\rangle[/tex] are orthorgonal? Why?

By the way, if I know the density operator at time T be [tex]\rho(t)[/tex], how to interpret [tex]\langle 0, e|\rho(t)|1, g\rangle[/tex] and [tex]\langle 1, g|\rho(t)|0, e\rangle[/tex]
 

Answers and Replies

  • #2
diazona
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Well, if the atom is known to be in the excited state, the probability that it you could measure it in the ground state is 0, right? If so, that means the two states are orthogonal. Similarly, if you know that the box is in the one-photon state, the probability of measuring zero photons in the box is 0, so those two states are orthogonal.
 
  • #3
Fredrik
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I wonder if these two states [tex]|0, e\rangle[/tex] and [tex]|1\rangle|g\rangle[/tex] are orthorgonal? Why?
The definition of [itex]\langle 0,e|1,g\rangle[/itex] is

[tex]\langle 0,e|1,g\rangle=\langle e|\otimes\langle 0|\Big(|1\rangle\otimes|g\rangle\Big)=\langle 0|1\rangle\langle e|g\rangle[/tex]

and what diazona said explains why both factors on the right are zero.
 
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