Is J_z|0> equal to zero in QFT for spinor field?

In summary, the value of J_z|0> in QFT for spinor fields depends on the specific representation of the spinor field and the choice of basis. In some representations, J_z|0> may be equal to zero, while in others it may have a non-zero value. This is due to the fact that different representations of the spinor field have different transformation properties under rotations, which affect the value of J_z|0>. Therefore, the answer to whether J_z|0> is equal to zero in QFT for spinor fields cannot be generalized and must be determined based on the specific representation and basis being used.
  • #1
Heirot
151
0
Let's say that we're trying to calculate J_z|0> for spinor field. Now, one would naturally expect that it vanishes as do all of the observables in vacuum state. On the other hand, J_z isn't diagonal in the number basis, so one has J_z|0> ~ |2>, i.e. particle + antiparticle state due to the two creation operators in J_z. So, which one is it? All the books I've seen says that it vanishes. But the direct calculations says it isn't so.

Thanks
 
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  • #2
The angular momentum operators do conserve particle number. Fields are constructed to lie in a definite angular momentum representation. The vacuum state has [tex]j=m_j=0[/tex] (a scalar) so any angular momentum operator acting on it gives zero.
 
  • #3
How can one state (vaccum) have well defined momenta and angular momenta at the same time?
 
  • #4
There's no contradiction. The uncertainty relation has a bound of zero when the expectation value of one of the operators vanishes completely.
 
  • #5
What about if one uses the plane wave decomposion of spinor field psi and use it in the angular momentum operator? Then one has J_z|0> ~ |2>
 
  • #6
The angular momentum operators never change the particle number. If [tex]\vec{J}=\vec{L}+\vec{S}[/tex] then you use the differential operator representation of [tex]\vec{L}[/tex] while [tex]\vec{S}[/tex] acts on the internal spin state.
 
  • #7
Heirot said:
What about if one uses the plane wave decomposion of spinor field psi and use it in the angular momentum operator? Then one has J_z|0> ~ |2>

What expression are you trying to use for J_z ??
 
  • #8
I'm trying to derive an operator containing creation and destruction operators for electrons and positrons (b and d). I can't seem to shape it into a form that contains only bTb and dTd combinations (T - hermitian conjugation) which would preserve the particle nubmer. Is there any reference for J in terms of b and d? I could only find it in terms of Dirac field psi.
 
  • #9
Heirot said:
I'm trying to derive an operator containing creation and destruction operators for electrons and positrons (b and d). I can't seem to shape it into a form that contains only bTb and dTd combinations (T - hermitian conjugation) which would preserve the particle nubmer. Is there any reference for J in terms of b and d? I could only find it in terms of Dirac field psi.

The angular momentum operators never depend on creation and destruction operators. Field operators act on the vacuum to create a state of definite spin. The angular momentum operators are defined by

[tex]J_i = L_i + S_i. [/tex]

The first term is the spatial representation (in units where [tex]\hbar=1[/tex])

[tex] L_i = \frac{1}{2} \epsilon_{ijk} x^j \partial^k,[/tex]

while the internal spin is completely determined by its action on a general spin state:

[tex] S^2 |s,m_s\rangle = s(s+1) |s,m_s\rangle, ~ S_z |s,m_s\rangle = m_s |s,m_s\rangle , etc.[/tex]
 
  • #10
Heirot said:
I'm trying to derive an operator containing creation and destruction operators for electrons and positrons (b and d). I can't seem to shape it into a form that contains only bTb and dTd combinations (T - hermitian conjugation) which would preserve the particle nubmer. Is there any reference for J in terms of b and d? I could only find it in terms of Dirac field psi.
I don't know of an explicit reference, but you should be able to derive it from the expression in terms of the field. It should definitely come out in the form of bTb and dTd terms only.
 
  • #11
Heirot said:
I'm trying to derive an operator containing creation and destruction operators for electrons and positrons (b and d). I can't seem to shape it into a form that contains only bTb and dTd combinations (T - hermitian conjugation) which would preserve the particle nubmer. Is there any reference for J in terms of b and d? I could only find it in terms of Dirac field psi.

I remember trying to derive just such a formula a while back, but I ran into several problems -- I don't quite remember what the were. I ultimately gave up on it, and settled on whatever they have in Itzykson+Zuber p147. If you find out how to get the forumla, please let me know.
 
  • #12
The problems include the derivatives of the spinors which don't satisfy the orthonormal relation. In that way, the time dependent terms includind bTdT and db do not vanish! Apperently, no books speaks of this problem.
 
  • #13
Heirot said:
The problems include the derivatives of the spinors which don't satisfy the orthonormal relation. In that way, the time dependent terms includind bTdT and db do not vanish! Apperently, no books speaks of this problem.

Try Eugene Stefanovich's book, available as

http://arxiv.org/abs/physics/0504062

In ch7, pp246-247, you'll find a derivation of the Lorentz generators in terms of
c/a operators. E.g., eqn(7.32) gives the expression for J_z (in the noninteracting
representation).

I'm not sure if this is exactly what you want, but it might give you
some idea of how to derive this sort of stuff.
 
  • #14
Unfortunately, this book also deals with angular momentum of a scalar field which can't be easily generalized to a non zero spin.
 
  • #15
TriTertButoxy said:
I remember trying to derive just such a formula a while back, but I ran into several problems -- I don't quite remember what the were. I ultimately gave up on it, and settled on whatever they have in Itzykson+Zuber p147. If you find out how to get the forumla, please let me know.

After a long and very much tedious calculation, I've derived it! Would you like to see the derivation?
 
  • #16
Heirot said:
After a long and very much tedious calculation, I've derived it! Would you like to see the derivation?

YES YES! Yes Very MUCH SO! As I said before, I have been trying to derive it for ages, and couldn't and was forced to give up. If you could show me the derivation, I would be indebted to you.
 
  • #17
I'll post it as soon as I type it in Latex ;)
 
  • #18
Take your time but PLEASE, don't forget.

thanks.
 
  • #19
I think he forgot
 
  • #20
He didn't. Next week...
 
  • #21
Sorry for the delay... Merry Christmas!
 

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  • #22
Thanks
You too
 
  • #23
This is so awesome! Thanks. Happy new year.
 

FAQ: Is J_z|0> equal to zero in QFT for spinor field?

1. Is J_z|0> equal to zero in QFT for spinor field?

Yes, in QFT (quantum field theory), J_z|0> refers to the angular momentum operator acting on the vacuum state, which is defined as having zero particles. Since the vacuum state has no particles, it also has no angular momentum, and therefore J_z|0> is equal to zero.

2. How is the spinor field defined in QFT?

In QFT, a spinor field is a type of quantum field that describes particles with half-integer spin, such as electrons and quarks. It is represented as a complex-valued function that varies over space and time, and is subject to certain mathematical constraints and transformations.

3. What is the significance of J_z in QFT for spinor field?

J_z (or the z-component of angular momentum) is an important quantity in QFT, as it is one of the fundamental operators that describes the behavior of spinor fields. Its eigenvalues correspond to different possible orientations of a particle's spin, and its measurement can provide information about a particle's angular momentum and intrinsic properties.

4. How is J_z|0> related to the other components of angular momentum in QFT?

In QFT, J_z is just one component of the total angular momentum operator, which also includes J_x and J_y. These three operators form a set of mutually commuting observables, meaning that they can all be measured simultaneously. However, since the vacuum state has zero particles, the expectation value of J_x and J_y on the vacuum state would also be zero, while J_z|0> would still be zero.

5. Can J_z|0> ever be non-zero in QFT for spinor field?

No, according to the mathematical framework of QFT, the vacuum state is defined as having zero particles and zero angular momentum. Therefore, the expectation value of J_z|0> will always be zero, regardless of the specific spinor field being considered. However, in certain physical situations, such as in the presence of external fields or interactions, the vacuum state may be modified and J_z|0> could potentially have a non-zero expectation value.

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