Controlled-Z gate as a product of exponentials

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
1 reply · 2K views
EightBells
Messages
9
Reaction score
1
Homework Statement
**see attached photo** sorry, it was impossible to format in a readable manner
Relevant Equations
sigma_0 is the identity matrix, and superscripts such as (1) and (2) indicates qubit 1 or 2 respectively
I have numerous points of confusion: what does it mean that the matrices are within the exponential? How do I go about doing the matrix multiplication to prove the given form of CZ matches the common form, the 4x4 matrix?

Update: using the fact that exp(At)=∑ ((t^n)/n!)*A^n, where A is a matrix. In this case I used three different A's for the three exponentials that form CZ, and found:

CZ= (exp(t))A'A ''A''', where A'=diag(exp(t),exp(-t),exp(t),exp(-t)), A''=diag(exp(t),exp(t),exp(-t),exp(-t)), and A'''=diag(exp(-t),exp(t),exp(t),exp(-t))
therefore, CZ=diag(1,1,1,-1) as expected.
 

Attachments

  • HW4.3.PNG
    HW4.3.PNG
    20.9 KB · Views: 352
Last edited:
Physics news on Phys.org
EightBells said:
Homework Statement:: **see attached photo** sorry, it was impossible to format in a readable manner
See Learn LaTeX for Math Equations.

EightBells said:
Update: using the fact that exp(At)=∑ ((t^n)/n!)*A^n, where A is a matrix.
This is indeed the correct approach.