Question

How do I normalize this wavefunction? simple integral that I can't do!

Answer 1

ok so I am trying to with the x-representation of the ground state of the quantum harmonic oscillator. I use the ladder termination condition so say that\[a\psi_0(x)=0\]where a is the lowering operator. So\[\sqrt{\frac{m\omega}{2\hbar}}\left(\hat{x}+i\frac{\hat{p}}{m\omega}\right)\psi_0(x)=0\]\[\sqrt{\frac{m\omega}{2\hbar}}\left(x+\frac{\hbar}{m\omega}\frac{d}{dx}\right)\psi_0(x)=0\]Which gives me the differential equation\[\frac{d}{dx}\psi_0(x)=-\frac{m\omega}{\hbar}x\psi_0(x)\]\[\int\frac{d\psi_0}{\psi_0}=-\frac{m\omega}{\hbar}\int xdx\]\[\ln\psi_0=-\frac{m\omega}{2\hbar}x^2\]\[\psi_0(x)=e^{-\frac{m\omega}{2\hbar}x^2}\]I would now like to normalize \(\psi_0\). I said \[\left|\left\langle \psi_0 | \psi_0\right\rangle \right|^2=1\]\[\int^{\infty}_{-\infty}\psi_0^\star\psi_0dx=1\]\[\int^{\infty}_{-\infty}e^{-\frac{m\omega}{2\hbar}x^2}e^{-\frac{m\omega}{2\hbar}x^2}dx=1\]\[\int^{\infty}_{-\infty}e^{-\frac{m\omega}{\hbar}x^2}dx=1\]but I cannot evaluate this integral! I have tried making a u substitution, but the infinite limits are getting me lost!

Answer 2

is this a gaussian function or something?

Answer 3

I meant to have an \(|A|^2\) outside those integrals for normalization.

Answer 4

nevermind. I figured it out.

Answer 5

Yes. Gaussian, normal distribution curve equivalent.

Answer 6

thanks