# Momentum state normalisation [on hold]

Is the momentum state $$|p>=a^{\dagger}|0>$$ normalized? If yes then how do we show that? I tried using the ladder operator commutation relations but it does not give me 1.

## put on hold as off-topic by Aaron Stevens, Kyle Kanos, ZeroTheHero, glS, Jon CusterFeb 15 at 16:14

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "Homework-like questions should ask about a specific physics concept and show some effort to work through the problem. We want our questions to be useful to the broader community, and to future users. See our meta site for more guidance on how to edit your question to make it better" – Aaron Stevens, ZeroTheHero, glS
If this question can be reworded to fit the rules in the help center, please edit the question.

• What is the context here? If you show your work it will be easier to answer – Hugo V Feb 7 at 11:04
• Or first, tell us what this $\langle r | p \rangle$ is in your problem? – rnels12 Feb 7 at 11:34
• $|p>$here is just a state vector – user142756 Feb 7 at 12:36
• It is defined in terms of ladder operator acting on a ground state |0>. – user142756 Feb 7 at 12:36
• Is your $a^\dagger$ acting on a vacuum to create a particle with momentum $p$? Anyway, momentum eigenstates are "Dirac's delta" normalized: $\langle p| p'\rangle = \delta(p-p')$ – ErickShock Feb 7 at 12:59

In Dirac's algebraic approach to the quantum mechanical harmonic oscillator problem with ladder operators, the fact that the ground state wave function has to correspond to finding the oscillating particle somewhere with overall 100% probability (i.e. the wave function has to be normalized) is used to determine the prefactors entering $$a^\dagger$$. The conjugate operator $$a$$ on the other hand is defined to destroy the ground state when acting on it, i.e. it yields zero in this case.
To show normalization, calculate $$\langle p|p\rangle = \langle 0 | a a^\dagger | 0 \rangle.$$ Using the commutator for raising/lowering operators (equivalent to bosonic creation/annihilation operators) (or alternatively use anticommutator in the case of fermionic operators): $$\langle 0 | a a^\dagger | 0 \rangle = \langle 0 | 1 | 0 \rangle \pm \langle 0 | a^\dagger a | 0 \rangle,$$ where the plus sign holds for bosons (as e.g. excitations of the harmonic oscillator) and the minus sign for fermions.
There's nothing to annihilate in the vacuum state $$|0\rangle$$, hence $$\langle 0 | a^\dagger a | 0 \rangle=0$$ (or in the language of the harmonic oscillator problem: the ground state cannot be lowered). The vacuum state (or harmonic oscillator ground state) is normalized and thus: $$\langle p|p\rangle = \langle 0 | 0 \rangle = 1,$$ both for bosons and fermions.