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9.4 Normalisation and Hermitean conjugates

If you look at the expression $ \int_{-\infty}^{\infty}$f (y)*$ \hat{a}^{\dagger}$g(y) dy and use the explicit form $ \hat{a}^{\dagger}$ = $ {\frac{1}{\sqrt 2}}$$ \left(\vphantom{ y - \frac{d}{dy} }\right.$y - $ {\frac{d}{dy}}$ $ \left.\vphantom{ y - \frac{d}{dy} }\right)$, you may guess that we can use partial integration to get the operator acting on f,

$\displaystyle \int_{-\infty}^{\infty}$f (y)*$\displaystyle \hat{a}^{\dagger}$g(y) dy
  = $\displaystyle \int_{-\infty}^{\infty}$f (y)*$\displaystyle {\frac{1}{\sqrt 2}}$$\displaystyle \left(\vphantom{ y - \frac{d}{dy} }\right.$y - $\displaystyle {\frac{d}{dy}}$ $\displaystyle \left.\vphantom{ y - \frac{d}{dy} }\right)$g(y) dy  
  = $\displaystyle \int_{-\infty}^{\infty}$$\displaystyle {\frac{1}{\sqrt 2}}$$\displaystyle \left(\vphantom{ y + \frac{d}{dy} }\right.$y + $\displaystyle {\frac{d}{dy}}$ $\displaystyle \left.\vphantom{ y + \frac{d}{dy} }\right)$f (y)*g(y) dy  
  = $\displaystyle \int_{-\infty}^{\infty}$[$\displaystyle \hat{a}$f (y)]*g(y) dy        . (9.21)

This is the first example of an operator that is clearly not Hermitean, but we see that $ \hat{a}$ and $ \hat{a}^{\dagger}$ are related by ``Hermitean conjugation''. We can actually use this to normalise the wave function! Let us look at
On = $\displaystyle \int_{-\infty}^{\infty}$$\displaystyle \left[\vphantom{\left({\hat a}^{\dagger}\right)^{n} e^{-y^2/2}}\right.$$\displaystyle \left(\vphantom{{\hat a}^{\dagger}}\right.$$\displaystyle \hat{a}^{\dagger}$$\displaystyle \left.\vphantom{{\hat a}^{\dagger}}\right)^{n}_{}$e-y2/2$\displaystyle \left.\vphantom{\left({\hat a}^{\dagger}\right)^{n} e^{-y^2/2}}\right]^{*}_{}$$\displaystyle \left(\vphantom{{\hat a}^{\dagger}}\right.$$\displaystyle \hat{a}^{\dagger}$$\displaystyle \left.\vphantom{{\hat a}^{\dagger}}\right)^{n}_{}$e-y2/2 dy  
  = $\displaystyle \int_{-\infty}^{\infty}$$\displaystyle \left[\vphantom{\hat a \left({\hat a}^{\dagger}\right)^{n} e^{-y^2/2}}\right.$$\displaystyle \hat{a}$$\displaystyle \left(\vphantom{{\hat a}^{\dagger}}\right.$$\displaystyle \hat{a}^{\dagger}$$\displaystyle \left.\vphantom{{\hat a}^{\dagger}}\right)^{n}_{}$e-y2/2$\displaystyle \left.\vphantom{\hat a \left({\hat a}^{\dagger}\right)^{n} e^{-y^2/2}}\right]^{*}_{}$$\displaystyle \left(\vphantom{{\hat a}^{\dagger}}\right.$$\displaystyle \hat{a}^{\dagger}$$\displaystyle \left.\vphantom{{\hat a}^{\dagger}}\right)^{n-1}_{}$e-y2/2 dy (9.22)

If we now use $ \hat{a}$$ \hat{a}^{\dagger}$ = $ \hat{a}^{\dagger}$$ \hat{a}$ + $ \hat{1}$ repeatedly until the operator $ \hat{a}$ acts on u0(y), we find

On = nOn - 1. (9.23)

Since O0 = $ \sqrt{\pi}$, we find that

un(y) = $\displaystyle {\frac{1}{\sqrt{n!\sqrt{\pi}}}}$$\displaystyle \left(\vphantom{{\hat a}^{\dagger}}\right.$$\displaystyle \hat{a}^{\dagger}$$\displaystyle \left.\vphantom{{\hat a}^{\dagger}}\right)^{n}_{}$e-y2/2 (9.24)

Question: Show that this agrees with the normalisation proposed in the previous study of the harmonic oscillator!

Question: Show that the states un for different n are orthogonal, using the techniques sketched above.


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Next: 10. Time dependent wave Up: 9. Ladder operators Previous: 9.3 Eigenfunctions of through

© 1998 Niels Walet, UMIST
Email Niels.Walet@umist.ac.uk