Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I'm following Griffiths' Introduction to Quantum Mechanics, where he's discussing the general solution to the delta-function potential problem. The solution he refers to is $$\psi(x)=Ae^{ikx}+Be^{-ikx}$$

Now, he says that neither of the the terms blows up. The case he is dealing with is $x<0$. I cannot see this, because it seems to me that the second term blows up if $x$ is negative. I suspect he is using some argument that involves the complex nature of the terms. I would like to know if indeed this is the case. If so, what is the argument to show that the terms do not blow up? Thanks.

share|cite|improve this question
One could expand these terms as sines and cosines, and then they would be finite regardless of the value of x. Is this the correct method? – Joebevo Feb 27 '13 at 3:46
up vote 3 down vote accepted



$$\psi(x)=A\cos({kx}) + iA\sin(kx)+B\cos({-kx}) + iB\sin(-kx)$$

To arrive at the above equation, I utilised the fact that $e^{\theta i} = \cos\theta + i\sin\theta$.

Now I can further simplify the above expression to give:

$$\psi(x)=A\cos({kx}) + iA\sin(kx)+B\cos({kx}) - iB\sin(kx)$$
Therefore, $$\psi(x)=(A+B)\cos({kx}) + i(A-B)\sin(kx)$$

Therefore $\psi$ is a complex function with an oscillating real part, and an oscillating imaginary part. Also, evidently, the function does not "blow up".

So in conclusion you are correct, it is due to the "i"s in the function that prevent it from blowing up, which would occur without the i.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.