- #1

- 488

- 4

[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]