# Orthogonality by Induction

1. Sep 22, 2014

### fuchini

Hello,

I'm currently studying second quantization. I need to prove $<n^\prime| n>=\delta_{n^\prime n}$ by mathematical induction in the number of particles representation. However I don't know how to do this proof having two natural numbers $n$ and $n^\prime$. Must I prove it holds for $<0|0>$, $<0|1>$ and $<1|1>$. Then assuming it holds true for $<n^\prime|n>$, prove it for $<n^\prime|n+1>$ and $<n^\prime +1|n+1>$. Excuse me if this is an obvious question, please help me.

Thanks a lot.

2. Sep 23, 2014

### vanhees71

This seems to belong more to the homework section of this forum, but let me give some hints. In 2nd quantization you use creation and annihilation operators to build the many-particle Hilbert space in terms of the corresponding Fock space. For a single mode this space is spanned by your vectors $|n \rangle$. By definition these states are normalized to 1 (suppose we work in a finite quantization volume with periodic boundary conditions leading to discrete momenta), i.e.,
$$\langle n | n \rangle=1.$$
These states are built from the vacuum (ground) state, $|\Omega$ by successive application of creation operators,
$$|n \rangle=\frac{1}{\sqrt{n!}} (a^{\dagger})^{n} |\Omega \rangle.$$
Now you can use the commutation or anti-commutation relations for the creation and annihilation operators to prove the orthonormality of the number states. To that end you have to express $\langle n'|$ with help of the above given forumula and then write down the corresponding expression
$$\langle n'|n \rangle=\langle \Omega|???|\Omega \rangle.$$
The "???" indicate some annihilation and creation operators, which you successively bring into an order such as to use $a |\Omega \rangle=0$, using the (anti-)commutation relations for the creation and annihilation operators. This will prove the claim and can be done by induction.

3. Sep 23, 2014

### fuchini

I've done that but I'm still stuck. I forgot to mention this was for bosons so the commutation relation must hold $[a,a^\dagger]=1$. From it I got:
$$a(a^\dagger)^n=n(a^\dagger)^{n-1}+(a^\dagger)^n a$$
I suppose the relation holds for $\langle m | n \rangle$ and I must prove it for $\langle m | n+1 \rangle$ and $\langle m+1 | n \rangle$:
$$\langle m | n+1 \rangle=\frac{1}{\sqrt{m!(n+1)!}}\langle 0|a^m(a^\dagger)^{n+1}| 0 \rangle=\frac{1}{\sqrt{m!(n+1)!}}\langle 0|a^{m-1}a(a^\dagger)^{n+1}| 0 \rangle$$
Using the above property and $a|0\rangle=0$:
$$\langle m | n+1 \rangle=\sqrt{\frac{n+1}{m!n!}}\langle 0|a^{m-1}(a^\dagger)^n | 0 \rangle$$
This is as far as I go. How can I use the $\langle m | n \rangle$ relation here?