# Klein Gordon 4-current from Dirac Equation

• shinobi20
In summary: Thank you for pointing that out! I got that mistake when copying and pasting from the wiki page. The extra factor of 2 is due to the fact that I'm multiplying two differentials.
shinobi20
Homework Statement
Let ##\psi_1##, ##\psi_2## be solutions of the Dirac equation such that ,

##\Big( \gamma^\mu p_\mu - m \Big) \psi_1= 0 \hspace{1cm} \Big( \gamma^\mu p_\mu - m \Big) \psi_2= 0 \hspace{1cm} (1)##

Define,

##\bar{\psi} \overleftarrow{p_\mu} = (p_\mu \bar{\psi})##, ##\hspace{1cm}## ##\bar{\psi} = \psi^\dagger \gamma_0 \hspace{1cm} (2)##

##\hat{\sigma}_{\mu\nu} = \frac{i}{2} \Big[ \gamma_\mu, \gamma_\nu \Big] \hspace{1cm} (3)##

From the equation

##\bar{\psi_2} \Big( -\overleftarrow{p_\mu} \gamma^\mu - m \Big) {\not}a \psi_1 + \bar{\psi_2} {\not}a \Big(\gamma^\mu p_\mu - m \Big) \psi_1 = 0 \hspace{1cm} (4)##

Show that

##\bar{\psi_2} \gamma_\mu \psi_1 = \frac{1}{2m} \Big( \bar{\psi_2} p_\mu \psi_1 - (p_\mu \bar{\psi_2}) \psi_1 \Big) - \frac{i}{2m} p^\nu \Big(\bar{\psi_2} \hat{\sigma}_{\mu\nu} \psi_1 \Big) \hspace{1cm} (5)##

Hint: write ##{\not}a{\not}p## in terms of ##\hat{\sigma}_{\mu\nu}##
Relevant Equations
##\Big( \gamma^\mu p_\mu - m \Big) \psi_1= 0 \hspace{1cm} \Big( \gamma^\mu p_\mu - m \Big) \psi_2= 0 \hspace{1cm} (1)##

##\bar{\psi} \overleftarrow{p_\mu} = (p_\mu \bar{\psi})##, ##\hspace{1cm}## ##\bar{\psi} = \psi^\dagger \gamma_0 \hspace{1cm} (2)##

##\hat{\sigma}_{\mu\nu} = \frac{i}{2} \Big[ \gamma_\mu, \gamma_\nu \Big] \hspace{1cm} (3)##

##\bar{\psi_2} \Big( -\overleftarrow{p_\mu} \gamma^\mu - m \Big) {\not}a \psi_1 + \bar{\psi_2} {\not}a \Big(\gamma^\mu p_\mu - m \Big) \psi_1 = 0 \hspace{1cm} (4)##

##\bar{\psi_2} \gamma_\mu \psi_1 = \frac{1}{2m} \Big( \bar{\psi_2} p_\mu \psi_1 - (p_\mu \bar{\psi_2}) \psi_1 \Big) - \frac{i}{2m} p^\nu \Big(\bar{\psi_2} \hat{\sigma}_{\mu\nu} \psi_1 \Big) \hspace{1cm} (5)##
The left side of the equality of ##(5)## is obvious from ##(4)##, however the rest of the terms are still unknown to me. I have tried adding and subtracting terms similar to the rest of the terms so as to produce a commutator and use ##(3)##, but I can't seem to figure out how to get ##(5)## correctly. Any help or guidance on how to proceed?

How is ##a## defined in equation (4). Looks to me that in order to get this form in equation (5), you need to remove ##a## from the equation. Since ##\not{a}## does not commute with gamma matrices, that's where the commutator arises from. But I still don't see what that variable ##a## is, so I can't help you further than that.

Last edited:
Antarres said:
How is ##a## defined in equation (4). Looks to me that in order to get this form in equation (5), you need to remove ##a## from the equation. Since ##\slash{a}## does not commute with gamma matrices, that's where the commutator arises from. But I still don't see what that variable ##a## is, so I can't help you further than that.

I was also thinking that ##a## is just a placeholder in order to produce some identity and in the end, ##a## can just be removed because every term will contain some vector component of it.

From the wiki page on gamma matrices,
##{\not}a{\not}b = \vec{a} \cdot \vec{b} - i a^\mu \hat{\sigma}_{\mu \nu} b^\nu##

where ##b = p## in this case.

This immediately takes care of the term with ##\hat{\sigma}_{\mu \nu}## in equation ##5##.

Yeah, I agree that it can be removed, but not along with the gamma matrix that comes with it(by the way, I have no idea how to write Feynman slash in latex). In order to produce the correct identity, slashed ##a## should be to the left of all gamma matrices in each term. Then you can use the identity that ##(\not{a})^2 = a^2##, from where you easily remove ##\not{a}##, and retrieve the final equation.

Last edited:
Antarres said:
Yeah, I agree that it can be removed, but not along with the gamma matrix that comes with it(by the way, I have no idea how to write Feynman slash in latex). In order to produce the correct identity, slashed ##a## should be to the left of all gamma matrices in each term. Then you can use the identity that \slash(##a##)^2 = ##a^2##, from where you easily remove slashed ##a##, and retrieve the final equation.
You can use {\not}a to produce slashed a.

I think I should not use ##a^2## since you will notice in the left side of equation ##5##, the ##\gamma_\mu## came from ##{\not}a = \gamma_\mu a^\mu## since each term contains ##{\not}a##.

Thank you, that works.
Okay, I will give you the example in the second term. So the reason why we can't just remove ##\not{a}## is because in the first term it's to the right of gamma matrices, and in the second term it's to the left. We can assume(I guess that's reasonable), that ##a## isn't a differential operator, since there shouldn't be double differentials in the current.
So for the second term, we get
$$\bar{\psi}_2\not{a}(\gamma^\mu p_\mu - m)\psi_1 = \bar{\psi}_2(\gamma^\mu p_\mu -m)\not{a}\psi_1 + 2i\bar{\psi}_2\sigma^{\mu\nu}p_\mu a_\nu\psi_1$$
Now you can remove the ##a## vector component from the equation since ##\not{a}## is always on the right of gamma matrices. Then proceed to rearrange terms in such a way to produce the last equality. However, be careful about the last term, since ##p_\mu## is a differential operator, so you can't pull it out of a term without using product rule.

P.S. I have here assumed that vector components of ##p## and ##a## commute, but I guess that is reasonable.

Antarres said:
Thank you, that works.
Okay, I will give you the example in the second term. So the reason why we can't just remove ##\not{a}## is because in the first term it's to the right of gamma matrices, and in the second term it's to the left. We can assume(I guess that's reasonable), that ##a## isn't a differential operator, since there shouldn't be double differentials in the current.
So for the second term, we get
$$\bar{\psi}_2\not{a}(\gamma^\mu p_\mu - m)\psi_1 = \bar{\psi}_2(\gamma^\mu p_\mu -m)\not{a}\psi_1 + 2i\bar{\psi}_2\sigma^{\mu\nu}p_\mu a_\nu\psi_1$$
Now you can remove the ##a## vector component from the equation since ##\not{a}## is always on the right of gamma matrices. Then proceed to rearrange terms in such a way to produce the last equality. However, be careful about the last term, since ##p_\mu## is a differential operator, so you can't pull it out of a term without using product rule.

P.S. I have here assumed that vector components of ##p## and ##a## commute, but I guess that is reasonable.
I noticed you have an extra factor of 2 in the ##\sigma^{\mu \nu}## term, but it should not be there if you look at the final equality.

Hmm, well that stems from the relation ##[\gamma^\mu,\gamma^\nu] = -2i\sigma^{\mu\nu}##. So it should be correct from the definition of ##\sigma^{\mu\nu}##.

Edit: Final equality is not obtained by simple substitution of this term I derived. You have to rearrange the derivatives(momenta) and the extra gamma matrices that appear in the terms. So that factor of two could be introduced in some further step.

Antarres said:
Hmm, well that stems from the relation ##[\gamma^\mu,\gamma^\nu] = -2i\sigma^{\mu\nu}##. So it should be correct from the definition of ##\sigma^{\mu\nu}##.
I was somehow suspicious that there might be some error in the question, what do you think? Also, how did you move ##{\not}a## to the right beside ##\psi_1?##

What I have done is

##\bar{\psi}_2\not{a}(\gamma^\mu p_\mu - m)\psi_1 = \bar{\psi}_2\not{a} \not{p} \psi_1 - m \bar{\psi}_2\not{a} \psi_1 = \bar{\psi}_2 (\vec{a} \cdot \vec{p} - i a^\mu \hat{\sigma}_{\mu \nu} p^\nu ) \psi_1 - m \bar{\psi}_2\not{a} \psi_1##

The way I did it was just commuting gamma matrices, introducing the commutator via ##\sigma^{\mu\nu}##. However, the derivation seems longer that way, since some terms don't take on the form we wish them to. Applying the identity you applied to both terms, gives the correct formula as I checked right now.

You just have to know that ##p_\mu## doesn't commute with anything except ##\gamma## matrices and ##a##. So you can't put it before or after wave functions without thinking, because it's a derivative operator. So for the first term, you would use the identity you mentioned in the following way.

$$\not{p}\not{a} = p \cdot a -i p^\mu \sigma_{\mu\nu} a^\nu = a \cdot p + i a^\mu\sigma_{\mu\nu}p^\nu$$

Where we used antisymmetry of ##\sigma_{\mu\nu}##. Therefore we get in the first term without mass(the mass term is the same):

$$-(p_\mu\bar{\psi}_2)\gamma^\mu\not{a}\psi_1 = -p_\mu\bar{\psi}_2 a^\mu\psi_1 - ia^\mu(p^\nu\bar{\psi}_2)\sigma_{\mu\nu}\psi_1$$

Substituting this and then using the product rule to pull out the momentum operator will give you the right relation.

Antarres said:
The way I did it was just commuting gamma matrices, introducing the commutator via ##\sigma^{\mu\nu}##. However, the derivation seems longer that way, since some terms don't take on the form we wish them to. Applying the identity you applied to both terms, gives the correct formula as I checked right now.

You just have to know that ##p_\mu## doesn't commute with anything except ##\gamma## matrices and ##a##. So you can't put it before or after wave functions without thinking, because it's a derivative operator. So for the first term, you would use the identity you mentioned in the following way.

$$\not{p}\not{a} = p \cdot a -i p^\mu \sigma_{\mu\nu} a^\nu = a \cdot p + i a^\mu\sigma_{\mu\nu}p^\nu$$

Where we used antisymmetry of ##\sigma_{\mu\nu}##. Therefore we get in the first term without mass(the mass term is the same):

$$-(p_\mu\bar{\psi}_2)\gamma^\mu\not{a}\psi_1 = -p_\mu\bar{\psi}_2 a^\mu\psi_1 - ia^\mu(p^\nu\bar{\psi}_2)\sigma_{\mu\nu}\psi_1$$

Substituting this and then using the product rule to pull out the momentum operator will give you the right relation.
Wheeew, I just realized now that what got me stuck is not thinking right away that ##p \cdot a = p^\mu a_\mu##, I guess I'm just bombarded with too many things now and many things doesn't ring to me although I'm supposed to know it. So by using the identity that I showed you I don't have to bother so much with how ##p## acts. Anyways, I will summarize it below.

From equation ##4## (disregarding the mass term), the first term is,

##-\bar{\psi_2} {\not}p {\not}a \psi_1 = -\bar{\psi_2} p \cdot a \psi_1 + \bar{\psi_2} i p^\mu \hat{\sigma}_{\mu\nu} a^\nu \psi_1 = -\bar{\psi_2} \overleftarrow{p_\mu} a^\mu \psi_1 - \bar{\psi_2} i a^\mu \hat{\sigma}_{\mu\nu} p^\nu \psi_1##

notice the left arrow on top of ##p\_mu## since it acts on the left wave function

the second term is,

##\bar{\psi_2} {\not}a {\not}p \psi_1 = \bar{\psi_2} a \cdot p \psi_1 - \bar{\psi_2} i a^\mu \hat{\sigma}_{\mu\nu} p^\nu \psi_1 = \bar{\psi_2} a^\mu p_\mu \psi_1 - \bar{\psi_2} i a^\mu \hat{\sigma}_{\mu\nu} p^\nu \psi_1##.

By adding the two and the mass term, I will arrive at equation ##5## WITH and extra factor of 2 beside the quantity ##\hat{\sigma}_{\mu\nu}##. I hope this is what you are implying.

Well no, that's the point. You're close to the solution though, there is no extra factor, actually.
So in your first equation, you must pay attention that ##p## is acting to the left, that's why in my post above yours you have ##p## acting on the left in both terms, not just the first one. That way, you will get two terms, one in which ##p## is acting on ##\bar{\psi}_2## and the second term gives terms in which ##p## is acting on ##\psi_1##.

Therefore, if we look at terms with ##\sigma_{\mu\nu}##, we get, similar to you(just added my example from the above with your second term):
$$-ia^\mu(p^\nu\bar{\psi}_2)\sigma_{\mu\nu}\psi_1 - ia^\mu\bar{\psi}_2\sigma_{\mu\nu}p^\nu\psi_1$$

So we have to pay attention, as I said before, that ##p^\mu## is not a simple vector, but is a derivative operator, that is ##p_\mu \equiv i\partial_\mu##. So if we look at what we got above, by product rule, we get precisely the last term in your equation 5, without extra factor. Hope that makes it clearer. You have to stick ##p## to the function it's acting on, that way you don't make mistakes when transforming. So it would be easier if you just substitute ##p## with that left arrow with it's definition, that is, put it in front of the function it's acting on, and then do transformations, like I did.

Antarres said:
Well no, that's the point. You're close to the solution though, there is no extra factor, actually.
So in your first equation, you must pay attention that ##p## is acting to the left, that's why in my post above yours you have ##p## acting on the left in both terms, not just the first one. That way, you will get two terms, one in which ##p## is acting on ##\bar{\psi}_2## and the second term gives terms in which ##p## is acting on ##\psi_1##.

Therefore, if we look at terms with ##\sigma_{\mu\nu}##, we get, similar to you(just added my example from the above with your second term):
$$-ia^\mu(p^\nu\bar{\psi}_2)\sigma_{\mu\nu}\psi_1 - ia^\mu\bar{\psi}_2\sigma_{\mu\nu}p^\nu\psi_1$$

So we have to pay attention, as I said before, that ##p^\mu## is not a simple vector, but is a derivative operator, that is ##p_\mu \equiv i\partial_\mu##. So if we look at what we got above, by product rule, we get precisely the last term in your equation 5, without extra factor. Hope that makes it clearer. You have to stick ##p## to the function it's acting on, that way you don't make mistakes when transforming. So it would be easier if you just substitute ##p## with that left arrow with it's definition, that is, put it in front of the function it's acting on, and then do transformations, like I did.
So to summarize,

##\bar{\psi_2} \Big( -\overleftarrow{p_\mu} \gamma^\mu - m \Big) {\not}a \psi_1 + \bar{\psi_2} {\not}a \Big(\gamma^\mu p_\mu - m \Big) \psi_1 = 0##

## 2 m \bar{\psi_2} \gamma_\mu a^\mu \psi_1 = -\bar{\psi_2} \overleftarrow{p_\mu} \gamma^\mu {\not}a \psi_1 + \bar{\psi_2} {\not}a \gamma^\mu p_\mu \psi_1##

## 2 m \bar{\psi_2} \gamma_\mu a^\mu \psi_1 = -\bar{\psi_2} {\not}p {\not}a \psi_1 + \bar{\psi_2} {\not}a {\not}p \psi_1##

## 2 m \bar{\psi_2} \gamma_\mu a^\mu \psi_1 = \bar{\psi_2} \Big( - {\not}p {\not}a + {\not}a {\not}p \Big) \psi_1##

## 2 m \bar{\psi_2} \gamma_\mu a^\mu \psi_1 = \bar{\psi_2} \Big( - \overleftarrow{p} \cdot a + i \overleftarrow{p^\mu} \hat{\sigma}_{\mu \nu} a^\nu + a \cdot p - i \hat{\sigma}_{\mu \nu} a^\mu p^\nu \Big) \psi_1##

because of the anti-symmetry of ##\hat{\sigma}_{\mu \nu}## and changing indices,

## 2 m \bar{\psi_2} \gamma_\mu a^\mu \psi_1 = \bar{\psi_2} \Big( - \overleftarrow{p} \cdot a - i \overleftarrow{p^\nu} a^\mu \hat{\sigma}_{\mu \nu} + a \cdot p - i \hat{\sigma}_{\mu \nu} a^\mu p^\nu \Big) \psi_1##

## 2 m \bar{\psi_2} \gamma_\mu a^\mu \psi_1 = -p_\mu \bar{\psi_2} a^\mu \psi_1 + \bar{\psi_2} a^\mu p_\mu \psi_1 - i p^\nu \bar{\psi_2} a^\mu \hat{\sigma}_{\mu \nu} \psi_1 - i \bar{\psi_2} \hat{\sigma}_{\mu \nu} a^\mu p^\nu \psi_1##

## 2 m \bar{\psi_2} \gamma_\mu \psi_1 = \bar{\psi_2} p_\mu \psi_1 - (p_\mu \bar{\psi_2}) \psi_1 - i p^\nu \Big( \bar{\psi_2} \hat{\sigma}_{\mu \nu} \psi_1 \Big)##

where ##a^\mu## was factored out since it is an arbitrary four-vector and the terms with ##\hat{\sigma}_{\mu \nu}## were combined into one due to the product rule. Thus,

##\bar{\psi_2} \gamma_\mu \psi_1 = \frac{1}{2m} \Big( \bar{\psi_2} p_\mu \psi_1 - (p_\mu \bar{\psi_2}) \psi_1 \Big) - \frac{i}{2m} p^\nu \Big(\bar{\psi_2} \hat{\sigma}_{\mu\nu} \psi_1 \Big)##

Yeah, that's correct.

You're welcome

## 1. What is the Klein Gordon 4-current?

The Klein Gordon 4-current is a mathematical expression that describes the flow of charge and current in a relativistic system. It is derived from the Dirac equation, which is a relativistic wave equation that describes the behavior of spin-1/2 particles.

## 2. How is the Klein Gordon 4-current related to the Dirac equation?

The Klein Gordon 4-current is derived from the Dirac equation by taking the product of the conjugate wave function and the Dirac equation itself. This results in a four-component vector that represents the flow of charge and current in a relativistic system.

## 3. What are the components of the Klein Gordon 4-current?

The four components of the Klein Gordon 4-current are the charge density, spatial current components, and time current component. These components represent the flow of charge and current in the x, y, z, and t directions, respectively.

## 4. How is the Klein Gordon 4-current used in physics?

The Klein Gordon 4-current is used in the study of relativistic systems, such as particles with high energies or moving at high speeds. It is also used in quantum field theory to describe the behavior of particles and their interactions.

## 5. What are the implications of the Klein Gordon 4-current in physics?

The Klein Gordon 4-current has important implications in physics, such as the conservation of charge and the Lorentz invariance of the Dirac equation. It also helps to explain the behavior of particles in relativistic systems and is a fundamental concept in quantum field theory.

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