Question:
Investigate the value of Ψ(x,T2).
Solution:
Just like homework 3 bonus question, we have the exponential term reverse the sign of all the odd terms and leave the even terms alone.
The odd terms are odd functions, so ψ2n+1(−x)=−ψ2n+1(x).
The even terms are even functions, os ψ2n(x)=ψ2n(−x).
So we see by simply substituting -x, we get what we want!
The last thing to notice is that we need to compensate for the extracted out phase term (which I forget), therefore the end result is Ψ(x,T2)=−Ψ(−x,0).
Investigate the value of Ψ(x,T2).
Solution:
Just like homework 3 bonus question, we have the exponential term reverse the sign of all the odd terms and leave the even terms alone.
The odd terms are odd functions, so ψ2n+1(−x)=−ψ2n+1(x).
The even terms are even functions, os ψ2n(x)=ψ2n(−x).
So we see by simply substituting -x, we get what we want!
The last thing to notice is that we need to compensate for the extracted out phase term (which I forget), therefore the end result is Ψ(x,T2)=−Ψ(−x,0).
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