Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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 have solved the following problem from Griffiths "Introduction to Quantum Mechanics".

Consider the wavefunction: $\Psi (x,t) = A e^{-\lambda |x|} e^{-i\omega t} $

Normalize $\Psi$.

Now, we want $ \int_{-\infty}^\infty |\Psi (x,t)|^2 dx = 1$

It is fairly straightforward, where the modulus is $|\Psi (x,t)|= r = A e^{-\lambda |x|}$. Therefore I square $r$ and integrate. I deal with the absolute value sign by multiplying by $2$ and integrating from 0 to $\infty$, while dropping the absolute value sign, to get:

$ 2\int_0^\infty (A e^{-\lambda x})^2 dx$

This should give me a factor of $A^2$ which I can take outside the integral sign. However, instead of a simple $A^2$, the solution gives an $|A|^2$. I don't understand where the absolute value sign came from. After all, taking the above expression $r$ as being equal to $|\Psi(x,t)$|, the modulus has already been dealt with.

share|cite|improve this question
You are assuming A is real. – Ron Maimon Jul 14 '12 at 8:57
up vote 0 down vote accepted

Well, $\left|\Psi(x,t)\right|^2=\Psi^*(x,t)\Psi(x,t)=|A|^2e^{-2\lambda \left|x\right|}$, isn't it?

share|cite|improve this answer
I still don't see where the | | came from. It is true that $A^2 = |A|^2$ but I'm not that satisfied with placing an absolute value sign where there needn't be one. I get $\left|\Psi(x,t)\right|^2=\Psi^*(x,t)\Psi(x,t)=Ae^{-\lambda \left|x\right|}Ae^{-\lambda \left|x\right|}=A^2e^{-2\lambda \left|x\right|}$. Note that I'm assuming A is a real constant, which seems reasonable. – Joebevo Jul 14 '12 at 8:41
I think there's my problem. If A can be complex then your expression seems to make sense. – Joebevo Jul 14 '12 at 8:47
It is not true that $A^2=|A|^2$ and it is not true that $A$ is real. A general amplitude isn't real; and your particular $A$ clearly isn't real, either. The probabilities are always expressed as squared absolute values of the amplitudes. A formula without the absolute value would clearly be wrong as probabilities are real. – Luboš Motl Jul 14 '12 at 8:48

Introduction to Quantum Mechanics -Second edition - David J. Griffiths

Problem 1.5 Consider the wave function $$\Psi(x,t)=Ae^{-\lambda|x|}e^{-i\omega t}$$ where $A$, $\lambda$, and $\omega$ are positive real constants.

(a) Normalize $\Psi$.

$$\begin{align*} 1&=\int\limits_{-\infty}^\infty|\Psi(x,t)|^2\,\mathrm dx\\ &=\int\limits_{-\infty}^\infty|Ae^{-\lambda|x|}e^{-i\omega t}|^2\,\mathrm dx\\ &=\int\limits_{-\infty}^\infty |A|^2(e^{-\lambda|x|-\lambda|x|}e^{+i\omega t-i\omega t})\mathrm dx\\ &=|A|^2\int\limits_{-\infty}^\infty e^{-2\lambda|x|}\,\mathrm dx\\ &=2|A|^2\int\limits_{0}^\infty e^{-2\lambda|x|}\,\mathrm dx\qquad\text{(even integrand)}\\ &=2|A|^2\left.\left(\frac{e^{-2\lambda|x|}}{-2\lambda}\right)\right|_0^\infty\\ &=2|A|^2\left(0-\frac1{-2\lambda}\right)\\ &=\frac{|A|^2}{\lambda}\\ \\ \frac1{|A|^2}&=\frac1\lambda\\ \\ |A|^2&=\lambda\\ A^*A&=\lambda\\ A^2&=\lambda\qquad\text{($A$ is real )}\\ \\A&=\sqrt\lambda\\ \\ \therefore \boxed{\Psi(x,t)=\sqrt\lambda e^{-\lambda|x|}e^{-i\omega t}} \end{align*}$$

share|cite|improve this answer
It is not generally true that the amplitude ($A$) is real. – dmckee Dec 30 '12 at 22:05

$$\int_{-\infty}^{\infty} \psi^{\dagger}\psi dx = \int_{-\infty}^{\infty} (A e^{-\lambda |x|} e^{-i\omega t})^{\dagger}(A e^{-\lambda |x|} e^{-i\omega t})dx$$ $$= A^{\dagger}A\int_{-\infty}^{\infty} (e^{-\lambda |x|} e^{-i\omega t})^{\dagger}(e^{-\lambda |x|} e^{-i\omega t})dx$$

Where $\dagger$ represents the Hermitian conjugate, or the complex conjugate in the case of $A$, so $$A^{\dagger}A = |A|^2$$

and that is where the $|A|^2$ comes from, regardless of whether $A$ is real or not.

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.