# Homework Help: Quantum mechanics Hermitian operator

1. Jul 2, 2017

### binbagsss

1. The problem statement, all variables and given/known data

I have the criteria:

$<p'| L_{n} |p>=0$,for all $n \in Z$

$L$ some operator and $|p>$, $|p'>$some different physical states

I want to show that given $L^{+}=L_{-n}$ this criteria reduces to only needing to show that:

$L_n |p>=0$ for $n>0$

2. Relevant equations
look up , look down,

3. The attempt at a solution

$<p'|L_n|p>^{+}=<p|L_n^{+}|p'>=<p|L_{-n}|p'>$

So from this I can deduce that showing :

$<p'|L_n|p>^{+}$ is satisfied is equivalent to showing that $<p|L_{-n}|p'>$ is satisfied.

I.e I can reduce the criteria from showing for all $n \in Z$ to showing for $n>0$ only, BUT I am still sandwiched between two different physical states, I don't understand how this means you can reduce further to showing only that $L_n |p>=0$, $n >0$ acting on solely a ket/bra...

2. Jul 6, 2017

### Dazed&Confused

So I don't understand some of your notation. Do you mean $L_n^\dagger = L_{-n}$? Also are $|p>$ and $|p'>$ arbitrary? It seems like having an operator act on a ket and then taking the inner product always giving zero would mean the operator gives zero on the ket. All the extra parts about $n$ aren't really relevant. You can consider one $n$. If $L_n=0$ the you can show $L_{-n}$ =0 if they are related by conjugation.

3. Jul 10, 2017

### binbagsss

yes
mm I thought the idea for using two different states $|p'>$ and $|p>$ is to show that this result holds for general arbitrary states, rather than showing that $<p|L_n|p> = 0$ . I agree that if we wanted to show that this was zero then $L_n|p> = 0$ is enough for the inner product to be zero.

yup

4. Jul 10, 2017

### Dazed&Confused

I just can't imagine a situation where if $<p|L_n|p'>=0$ for all $|p>$ and $|p'>$ and they aren't equal that $L_n \neq 0$. For example say $|p'> = L_n|p>$. I mean it must be some state right? Then the condition implies $<p'|p'> =0$ which by the properties of vector implies $|p'> = L_n |p> =0$.