H|0> = 0? Does that mean that zero-point is ZERO?

1. Aug 24, 2011

LostConjugate

In QFT after the Hamiltonian is normal ordered I guess when acting on the lowest energy state it annihilates it. So doesn't that mean the energy is zero now instead of 1/2?

2. Aug 24, 2011

G01

Essentially, what you are doing is measuring the energy in the field relative to the energy of the vacuum. It is exactly like changing the 0 point of your potential energy in introductory physics. The vacuum energy density is constant, so one can just measure relative to it to avoid the diverging integral that appears when you try to calculate the total energy in the field. What has really happened, behind the scenes, is a redefinition of the Hamiltonian:

$$H \rightarrow H - <0|H|0>$$

3. Aug 24, 2011

LostConjugate

Well doesn't the infinity arise because your integrating over infinity. What happens when the next state is reached? Now your integrating a non-zero value over infinity again.

I thought the way of dealing with this was dividing the energy by the volume of the space in order to get the density. Essentially dividing infinity by infinity.

4. Aug 24, 2011

G01

Try working out the zero point energy for the scalar field. You'll get:

$$<0|H|0>=V\int \frac{d^4k}{(2\pi)^4} \frac{1}{2}\hbar\omega(k)$$

Even after you divide by the volume V, the energy of the vacuum still diverges because you are integrating $\omega(k)=\sqrt{\vec{k}^2+m^2}$ over an infinite number of modes.

So you may be able to cancel out the infinity arising from the spacial integral, but not the one arising from the k-space integral.

5. Aug 24, 2011

LostConjugate

Ok so making the ground state 0 solves the problem. But what about the second state?

6. Aug 24, 2011

G01

You mean the 1 particle state?

The vacuum term is the same in the states with non zero particle number. Once you subtract it off the Hamiltonian once, you will get finite results for your non-zero particle states. For instance work out: $<q|H'|q>=<q|H - H_o|q>$ where $H_o$ is the groundstate energy defined above.

You will see that the ground state contribution cancels exactly with a term from $<q|H|q>$.

Last edited: Aug 24, 2011
7. Aug 25, 2011

LostConjugate

Sorry for my poor math knowledge.

Is this correct?

$$H-H_o = \int \frac{d^4 k}{(2\pi)^4} k_p a_p^\dagger a_p$$

Which is still infinite? Is |q> the same as the scalar field integral evaluated at p=1?

8. Aug 26, 2011

G01

Your Hamiltonian will not change from the previous case. Simplified, it should look like:

$$H=\int d^3p \omega_p a^+_p a_p+H_o$$

(If you need a text reference check out Srednicki Chapter 3. )

Thus, $$H-H_o=\int d^3p \omega_p a^+_pa_p$$

Remember that $|q>=a^+(q)|0>$

So, you need to work out:

$$(H-H_o)|q>=(H-H_o)a^+_q|0>=\int d^3p \omega_p a^+_p a_p a^+_q|0>$$

Put the operators in the correct order and you should end up with a finite result.

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook