At 55:55 , in covariance matrix of 𝜌0 u wrote d instead of d0..also while doing cyclic...wont it be like d becomes d0..or else we wont be able to bring U to front of 𝜌0 ??
I'm not sure what you mean and don't have time to check carefully my video, but everything should be clear in the lecture notes. Let me know if you find any mistakes in there. Thanks!
Thank you so much for your lecture and your lecture notes. I have a question around 2:12:00, why can we have the relation tr{| nu >< nu | A} = at the bottom. Isn't this relation proved when the states we use are orthogonal? In addition, previously, you just show coherent state does not equal to 0, does that make the proof of tr{| nu >< nu | A} = invalid?
Hi, no, the relation is completely general for any outer product: tr{ |φ> and |ψ> are two completely arbitrary states (only requisite is that |φ>} of the Hilbert space: tr{ |φ>
Lecture starts from 8:47
Coherent State 1:00:00
At 55:55 , in covariance matrix of 𝜌0 u wrote d instead of d0..also while doing cyclic...wont it be like d becomes d0..or else we wont be able to bring U to front of 𝜌0 ??
I'm not sure what you mean and don't have time to check carefully my video, but everything should be clear in the lecture notes. Let me know if you find any mistakes in there. Thanks!
Thank you so much for your lecture and your lecture notes. I have a question around 2:12:00, why can we have the relation tr{| nu >< nu | A} = at the bottom. Isn't this relation proved when the states we use are orthogonal? In addition, previously, you just show coherent state does not equal to 0, does that make the proof of tr{| nu >< nu | A} = invalid?
Hi, no, the relation is completely general for any outer product: tr{ |φ> and |ψ> are two completely arbitrary states (only requisite is that |φ>} of the Hilbert space: tr{ |φ>
@@CarlosNBvs5 I got it. Thank you so much!