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Let $H=0.5(p^2+q^2)$ be the Hamiltonian in natural units. Let $f_{n}$ be the eigenfunctions of H. Show that $\langle f_{n},f_{m}\rangle=1$ if n=m, and equal to 0 otherwise. Do this by using the creation\annihaltor operators: $a_{+}$ , $a_{-}$

we have that $f_{n}=\frac{a_{+}^{n}}{(n!)^{0.5}}f_{0}$ so

$\langle f_{n},f_{m}\rangle=\frac{\langle f_{0},a_{-}^n(a_{+})^mf_{0}\rangle}{(n!m!)^{0.5}}$

$a_{-}f_{0}=0$ so $[a_{-}^n,a_{+}^m]=a_{-}^na_{+}^m$ and

$[a_{-},a_{+}^m] =ma_{+}^{m+1}$.

How does $[a_{-},a_{+}^m]$ relate to $[a_{-}^n,a_{+}^m]$? When I know this can I consider cases n>m, n

Thanks

Martin Sleziak
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1 Answers1

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Is this what you are looking for?

\begin{align*} [a_-^n,a_+^m]=a_-a_-^{n-1}a_+^m-a_+^ma_-^{n-1}a_-&=a_-[a_-^{n-1},a_+^{m}]a_-\\ &=\cdots\\ &=a_-^{n-1}[a_-,a_+^{m}]a_-^{n-1} \end{align*}

Alex R.
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  • Can you help me complete the question as I'm now not sure how to consider the cases n>m, n<m , n=m? –  Dec 21 '13 at 18:41
  • @user108605: $\langle f_n,g_m \rangle= \overline{ \langle g_m,f_n \rangle}$, so if you consider one case such as $n>m$, then $n<m$ follows.You also need to know what $a_+f_n$ and $a_-f_n$ give in terms of $f_{n+1}$ and $f_{n-1}$ respectively. – Alex R. Dec 21 '13 at 18:49
  • @Alex R what happens to your equation if you apply it to the vacuum? – lcv Dec 21 '13 at 18:58
  • My literature says that if n<m , then we have a factor times $a_{+}^{m-n}$ Can you explain how they know this? –  Dec 21 '13 at 18:59