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daggaz
2009-10-02, 10:00 AM
Ok was reading thru my quantum mechanics again, and skimmed over a proof real quick, thinking yeah yeah blah blah ok thats fine...

..Wait a second... :smallconfused:

(forgive me but I dont know how to type math on the internet, let alone on this forum)

C1*e^(ikx) + C2*e^(-ikx)=(C1+C2)*e^(iknx), where kn= +/- k

Whaaa? How does that work? Im not sure how to sub in both the positive and negative value of k here, let alone what the mathematical justification for this is. Help?:smallfrown:

RS14
2009-10-02, 10:27 AM
C1*e^(ikx) + C2*e^(-ikx)=(C1+C2)*e^(iknx), where kn= +/- k

Whaaa? How does that work? Im not sure how to sub in both the positive and negative value of k here, let alone what the mathematical justification for this is. Help?:smallfrown:

The notation means that kn is an element of the set {k,-k}. The statement must be satisfied for every possible kn.

The equation you typed is true if you consider only the real parts. It doesn't hold for imaginary parts. (Provided C1 and C2 are real).

Note that Re[e^(ikx)]=Re[e^(-ikx)].

Then C1*Re[e^(ikx)] + C2*Re[e^(-ikx)]=C1*Re[e^(ikx)] + C2*Re[e^(ikx)]=(C1+C2)Re[e^(ikx)]=(C1+C2)Re[e^(-ikx)]=(C1+C2)*Re[e^(iknx)], where kn= +/- k

daggaz
2009-10-02, 10:45 AM
ahhh... ahah. mmmm.... mmmhmmmm... yes. ok. thank you. =)

Mando Knight
2009-10-02, 11:14 AM
The equation you typed is true if you consider only the real parts. It doesn't hold for imaginary parts. (Provided C1 and C2 are real).

Well, even in quantum mechanics, you're typically only interested in the real number results anyway...

As for the proof of the real part of the equation being equal, e^(ikx) is defined as cos(kx)+i*sin(kx), so Re(e^(ikx))=cos(kx), and Re(e^(-ikx))=cos(-kx). For all values of k and x, cos(kx)=cos(-kx), so Re(e^(ikx))=Re(e^(-ikx)), so they can be added linearly, QED.