# Klein Gordon & spacelike/timelike virtual photon

• Lengalicious
In summary, the student is trying to find a relation between energy, mass, and momentum, but is struggling. He has tried substituting in the wave equation, but doesn't seem to be getting anywhere. He is also having trouble with the momentum equation. He has obtained the right relation for the energy, but is still having trouble with the momentum. He has also been told to use a reference frame where the first electron is at rest, but he is confused about how to do that.
Lengalicious

See attachment.

## The Attempt at a Solution

So I have shown that the plane wave sol'n satisfies the Klein-Gordon equation by subbing in and reducing the equation to:

$$E^2 = p^2c^2 + m^2c^4$$

which reduces to:

$$E = pc$$

for an m = 0 particle, however I don't really know how to show that this would correspond with the wave traveling at the speed of light.

Next I am asked to act on the wave function with the Hamiltonian and momentum operators, I assume I should act with the relativistic Hamiltonian so:

$$\hat{H} = (-\hbar^2\frac{\partial^2 }{\partial x^2}c^2 + m^2c^4)^{1/2}$$

So I get:

$$\hat{H}\psi = \sqrt{4\hbar^2c^2k^2 + m^2c^4}(e^{i(wt-kx)})$$

So:

$$E = \sqrt{4\hbar^2c^2k^2 + m^2c^4}$$

$$h\nu = \sqrt{4\hbar^2c^2k^2 + m^2c^4}$$

Reduces to:

$$-3h^2\nu^2 = m^2c^4$$

So I'm not sure what I have done wrong,
now for the momentum:

$$\hat{p}\psi = -i\hbar\frac{\partial }{\partial x}(e^{i(wt-kx)})$$

Reduces to:

$$= -\hbar k(e^{i(wt-kx)})$$

$$\frac{h}{\lambda} = -\hbar k$$

$$\frac{h}{\lambda} = -\frac{h}{\lambda}$$

Once again something has gone wrong, not sure what. . .

On the very last bit 2.(a) I basically don't know how to show that it does not satisfy the energy-mass-momentum relation. I have tried separating components and solving for E3 with,

$$(m_ec,\bar{0}) = ((|\bar{p}_2|^2 + m_e^2c^2)^{1/2},\bar{p}_2) + (E_3,\bar{p}_3)$$

to no avail, so I'm pretty lost on this one any general advice or direction would help.

#### Attachments

• Questions.jpg
49.4 KB · Views: 413
Hi, hope it's not too late to help..
So first, you can't use the relativistic Hamiltonian the way you do because you can't ignore the square-root. in this particular example, you can start with the relation:
H = pv – L = pv + mc2/γ,
then express v in terms of p and E (γmc2) to get a Hamiltonian you can work with. I get the right relation at the end.

Then, i think obtaining p = – h/λ is no big deal because |p| = h/λ, which is the significant relation.

For (2), i would start with expressions in terms of ω and k for the electrons to obtain E and p for the photon and show that the usual relation doesn't hold.

1 person
Awsome, no you're not too late so I will take your advice and retry but I didn't think I ignored the square root because I set $$e^{i(wt-kx)}=\sqrt{e^{2i(wt-kx)}}$$, thought that was mathematically consistent but I guess not anyway thanks again, will report back if I still run into any problems.

Yeah, operators and state-functions don't work this way...
Also, in the first question you're not required to find a relation in terms of E, p and m like you apparently did: once you substitute the wave solution into the Klein-Gordon equation you just obtain a relation between w and k. Then once you set m = 0 and remember that v = w/k you get the speed of your wave, which should be the desired result for that question.

1 person
Ok, so only just got round to carrying out your advice due to project work taking up my time anyway, converting the Hamiltonian using the relativistic Lagrangian H = pv – L and then applying the new Hamiltonian I get:

$$E = h\nu + \frac{mc^2}{\gamma}$$

You definitely end up with just $$E = h\nu$$
?

Also on Q2 when you say use the expressions in terms of ω & k do you mean the relativistic energy in terms of ω & k and sub that into the 4-momentums of the electrons?

Ok, so first:
in special relativity you have: v = pc2/E (and E = γmc2),
Take E as the energy eigenvalue, so H = pv – L = p2c2/E +m2c4/E.
Looks familiar, right? now see what you get when you apply the Hamiltonian to ψ.
What did you obtain in the first question (relation between k and ω)? Plug it in and see how it simplifies...

Q2: I read it quickly the first time and thought it was related to the first question, but apparently it's not. So forget what i said, but take this hint: you can use a reference frame where the first electron is at rest...

Ahh.. in Q2 you did that already but the way you put it is confusing. Separate the equations: first the energy conservation then momentum, plug the second into the first and everything becomes clear.

So here's my working for hamiltonian in Q1:

using:
$$H = \vec{p}\cdot \vec{v} - L$$

I end up with,

$$H = \frac{|\vec{p}|^2c^2}{\gamma mc^2}+\frac{mc^2}{\gamma}$$

so,

$$\hat{H} = \frac{1}{\gamma mc^2}(-\hbar^2\frac{\partial^2 }{\partial x^2}c^2+m^2c^4)$$

$$\hat{H}\psi = \frac{1}{\gamma mc^2}(\hbar^2k^2c^2+m^2c^4)e^{-i(\omega t-kx)}$$

$$E = \frac{1}{\gamma mc^2}(\hbar^2k^2c^2+m^2c^4)$$

using the relation $$\omega = 2\pi \nu$$ &
$$k = \frac{2\pi}{\lambda}$$ & that $$\frac{h\nu}{mc^2} = \gamma$$

i end up with

$$E = h\nu + \frac{mc^2}{\gamma}$$

Notes: the wave solution should be $$e^{-i(\omega t-kx)}$$ apparently the one on the sheet is missing a minus sign, typo.

Last edited:
Lengalicious said:
using the relation $$\omega = 2\pi \nu$$ &
$$k = \frac{2\pi}{\lambda}$$ & that $$\frac{h\nu}{mc^2} = \gamma$$

No! use the relation you've found in the very first question and express k in terms of ω...

1 person
Aha, yep sorry about that, it does work, thanks! As for Q2 I do what you say but I get $$m_e^2-2E_3m_e+E_3^2/c^2=p_2^2+m_e^2c^2$$ and I can't really solve for E_3 here? Or am I not supposed to be solving for that to show that it doesn't satisfy the relation :S?

Almost done: the last c^2 on the right shouldn't be there, so two of your terms cancel...
then what did you obtain for the momentum conservation equation? you should be able to use it and get an equation only in terms of the photon's energy and momentum (and the constants m and c). Does it satisfy the usual relation?

ah ok sweet, should have it from here thanks for all the help! :D

## 1. What is the Klein Gordon equation?

The Klein Gordon equation is a relativistic wave equation that describes the behavior of a quantum scalar field. It was first proposed by physicist Oskar Klein and physicist Walter Gordon in the 1920s. The equation takes into account both space and time, and is used to describe particles with zero spin, such as mesons.

## 2. What is a virtual photon?

A virtual photon is a particle that exists for a very short period of time, according to the uncertainty principle in quantum mechanics. These particles are not directly observable, but they play an important role in fundamental interactions, such as the electromagnetic force. Virtual photons can be either spacelike or timelike, depending on their energy and momentum.

## 3. What is the difference between a spacelike and timelike virtual photon?

A spacelike virtual photon has a higher energy and momentum than a timelike virtual photon. This means that a spacelike photon can travel faster than the speed of light, while a timelike photon travels slower than the speed of light. The distinction between spacelike and timelike photons is important in understanding the behavior of particles and the fundamental forces of nature.

## 4. How are Klein Gordon and virtual photon theory related?

The Klein Gordon equation is used to describe the behavior of virtual particles, such as virtual photons. It takes into account both space and time, and is a key component in understanding the behavior of particles at the quantum level. Virtual photon theory is a fundamental aspect of quantum mechanics, and it is closely related to the Klein Gordon equation.

## 5. What are the practical applications of studying Klein Gordon and virtual photon theory?

Studying Klein Gordon and virtual photon theory has many practical applications in fields such as high-energy physics and quantum computing. These theories help us understand the behavior of particles and the fundamental forces of nature, which can lead to advancements in technology and medicine. For example, understanding virtual photons is crucial in developing new imaging techniques for medical diagnostics.

Replies
13
Views
1K
Replies
10
Views
1K
Replies
11
Views
4K
Replies
1
Views
2K
Replies
8
Views
1K
Replies
5
Views
2K
Replies
2
Views
429
Replies
2
Views
1K
Replies
9
Views
1K
Replies
1
Views
840