kasse Messages 383 Reaction score 1 Thread starter Nov 14, 2008 #1 [tex]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})[/tex] Correct? Last edited: Nov 14, 2008
[tex]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})[/tex] Correct?
Mark44 Mentor Insights Author Messages 38,152 Reaction score 10,760 Nov 14, 2008 #2 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).
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).
Mark44 Mentor Insights Author Messages 38,152 Reaction score 10,760 Nov 14, 2008 #4 That's what I meant in "another approach." Also, your answer looks to be correct this time.