Hamiltonian in second quantization

[hello_world30]

TL;DR Summary

Proving Hamiltonian of a simple harmonic oscillator in second quantization

Hello ! I require some guidance on this prove :

I normally derive the Hamiltonian for a SHO in Hilbert space with a term of 1/2 hbar omega included. However, I am unsure of how one derives this from Hilbert space to Fock space. I have attached my attempt at it as an image below. Any input will be of great help. Cheers.

 

[vanhees71]

You just have to express $\hat{x}$ and $\hat{p}$ in terms of the operators $\hat{a}$ and $\hat{a}^{\dagger}$ and subtract, without essentially changing the physics, the vacuum-energy contribution by "normal ordering".

In this case of a single harmonic oscillator your "Fock space" is just the single-particle Hilbert space you started with, and there is no "2nd quantization done". This you achieve by quantizing the Schrödinger field, leading to a real Fock space.

 

[hello_world30]

You just have to express $\hat{x}$ and $\hat{p}$ in terms of the operators $\hat{a}$ and $\hat{a}^{\dagger}$ and subtract, without essentially changing the physics, the vacuum-energy contribution by "normal ordering".

In this case of a single harmonic oscillator your "Fock space" is just the single-particle Hilbert space you started with, and there is no "2nd quantization done". This you achieve by quantizing the Schrödinger field, leading to a real Fock space.

Thank you for your prompt reply. Why is it that if I substitute x and p operators that are exressed in terms of a and a^(+) into the Hamiltonian , it does not have the 1/2 hbar omega term, but when I use the conventional way of deriving the Hamiltonian (starting from aa^+) then I get a Hamiltonian with 1/2 hbar omega ?

 

[vanhees71]

You must be more careful with operator ordering! You should get the last equation on your scanned calculations (BTW, it's much less work and better for the forum to use the built-in LaTeX feature. Click the LaTeX guide (link at the left directly under the text editor):

https://www.physicsforums.com/help/latexhelp/

Concerning the calculation, note that
$$(\hat{a}-\hat{a}^{\dagger})^2=\hat{a}^2 - \hat{a} \hat{a}^{\dagger} - \hat{a}^{\dagger} \hat{a}+ \hat{a}^{\dagger 2}$$
and
$$(\hat{a}+\hat{a}^{\dagger})^2=\hat{a}^2 + \hat{a} \hat{a}^{\dagger} + \hat{a}^{\dagger} \hat{a} + \hat{a}^{\dagger 2}.$$
From that you get
$$(\hat{a}+\hat{a}^{\dagger})^2-(\hat{a}-\hat{a}^{\dagger})^2=2 (\hat{a}\hat{a}^{\dagger} + \hat{a}^{\dagger} \hat{a})=2([\hat{a},\hat{a}^{\dagger}]+2 \hat{a}^{\dagger} \hat{a}]=4 \left (\hat{a}^{\dagger} \hat{a} + \frac{1}{2} \right).$$
Plugging this into your equation for the Hamiltonian, you get your final equation (1),
$$\hat{H}=\hbar \omega \left (\hat{a}^{\dagger} \hat{a} + \frac{1}{2} \right).$$
The additive piece $\hbar \omega/2 \hat{1}$ is just a constant operator commuting with everything, and just counting the energy of the ground state as zero, you get the equivalent Lagrangian
$$\hat{H}'=\hbar \omega \hat{a}^{\dagger} \hat{a},$$
which describes the same physics as the original Hamiltonian, except that your zero level for energy is shifted.

 

[PeterDonis]

@hello_world30 please do not enter your equations as images. Use the PF LaTeX feature to enter them directly into your post. (You will see a "LaTeX Guide" link at the lower left of the post window.)