# Wavefunction normalisation

1. Homework Statement
Determine the constant λ in the wave equation

$\Psi(x) = C(2a^2 x^2 + \lambda)e^{-(a^2 x^2/2)}$

where $a=\sqrt{mω/\hbar}$

2. Homework Equations

Some standard integrals I guess

3. The Attempt at a Solution

So I believe the wave equation just needs to be normalised. Using the usual conditions for normalisation,

$(C2a^2 + C\lambda)^2 \int^{∞}_{-∞} | x^2 e^{-(a^2 x^2/2)} + e^{-(a^2 x^2/2)} |^2 dx =1$

From there,

$(C2a^2 + C\lambda)^2 \int^{∞}_{-∞} |2x^2 e^{-(a^2 x^2/2)}|^2 dx =1$

Then squaring the function inside the integral and moving the '4' outside the integral as it is a constant,

$4(C2a^2 + C\lambda)^2 \int^{∞}_{-∞} x^4 e^{-(a^2 x^2)} dx =1$

Now that should be a standard integral but I don't know any involving an x term to the fourth power. Or perhaps I've done something else wrong?

## Answers and Replies

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DrClaude
Mentor
$(C2a^2 + C\lambda)^2 \int^{∞}_{-∞} | x^2 e^{-(a^2 x^2/2)} + e^{-(a^2 x^2/2)} |^2 dx =1$
Since when is
$$[(ab+c) d]^2 = (a+c)^2(bd+d)^2$$

The next step is also completely wrong. Start again from $|\psi|^2 = \psi^* \psi$. You should also allow $C$ and $\lambda$ to be complex.

If the function doesn't contain any complex exponentials, then $\psi^{*}$ is the same as $\psi$, isn't it?

DrClaude
Mentor
As all physical observables depend ultimately on $| \psi |^2$, the wave function of a physical system is only defined up to a complex phase. In other words, $\psi$ and $\psi e^{i \delta}$, with $\delta$ real, decribe the same thing. Therefore, you can choose the normalization constant $C$ in
$$\psi(x) = C f(x)$$
to be real, because if it is complex, you can always do a rotation in the complex plane such that $C' = C e^{i \delta}$ is real.

But you also have the $\lambda$ in there and, unless told otherwise, you can't assume that it is real. You should appraoch the problem without restricting $C$ or $\lambda$ to be real, and see what you get.

use Maple to do the follow step:

assume(a>0)

int(C*(2*a^2*x^2+B)*exp(-(a^2*x^2)/2),x=-infinity..infinity)

where B is your λ. then we can get the result is
(B+2)*C*(2*Pi)^0.5/a