Klein Gordon 4-current from Dirac Equation

  • Thread starter shinobi20
  • Start date
  • #1
shinobi20
257
17
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?
 

Answers and Replies

  • #2
Antarres
170
81
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:
  • #3
shinobi20
257
17
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##.
 
  • #4
Antarres
170
81
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:
  • #5
shinobi20
257
17
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##.
 
  • #6
Antarres
170
81
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.
 
  • #7
shinobi20
257
17
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.
 
  • #8
Antarres
170
81
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.
 
  • #9
shinobi20
257
17
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##
 
  • #10
Antarres
170
81
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.
 
  • #11
shinobi20
257
17
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.
 
  • #12
Antarres
170
81
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.
 
  • #13
shinobi20
257
17
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)##
 
  • #14
Antarres
170
81
Yeah, that's correct.
 
  • #15
shinobi20
257
17
  • #16
Antarres
170
81
You're welcome :smile:
 

Suggested for: Klein Gordon 4-current from Dirac Equation

Replies
9
Views
641
Replies
2
Views
810
Replies
16
Views
1K
Replies
1
Views
348
  • Last Post
Replies
2
Views
876
Replies
5
Views
568
Replies
2
Views
1K
Replies
1
Views
587
  • Last Post
Replies
2
Views
501
Replies
3
Views
506
Top