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kasse
Nov14-08, 10:23 AM
Re [e^{iE_{0}t/\hbar} \cdot e^{-iE_{1}t/\hbar}] = Re[ e^\frac{it}{\hbar}(E_{0}-E_{1})] = cos \frac{t}{\hbar}(E_{o}-E_{1})

Correct?

Mark44
Nov14-08, 10:36 AM
Check your multiplication. You ended up with two terms, each with two factors. You should have ended up with four terms, each with two factors.

Another approach is to multiply the two exponentials first before converting them to cos X + i sin X. Remember that e^A * e^B = e^(A + B).

kasse
Nov14-08, 11:00 AM
I've updated it. Is this what you meant?

Mark44
Nov14-08, 02:38 PM
That's what I meant in "another approach." Also, your answer looks to be correct this time.