The vacuum saturation approximation

[Safinaz]

Hi all,

I'd like to ask about " the vacuum saturation approximation " in a calculation as the decay width of $B \to l \nu$ [hep-ph/0306037v2] equ. 1 till equ. 4, that why

$< 0 | \bar{u} \gamma_\mu b | B > ~ and ~ < 0 | \bar{u} b | B > = 0$ ? and for $[ \bar{l} \gamma^\mu (1-\gamma_5) \nu ]$ why didn't say:

$< l^- \bar{ \nu} | \bar{l} \gamma^\mu (1-\gamma_5) \nu | 0 >$ ?

Bests.

 

[Safinaz]

Hi, no, I still have no a concluotion yet, so should I repost the question again?

 

[Hepth]

Sorry, I totally thought I tried to answer this earlier in the week.

The way to approach the parameterization of matrix elements like those you listed is to look at a combination of things; mainly symmetries and available building blocks.

These will lead you in the right direction.

Lets say we have 4 different possible currents : $\bar{u} \left\{1, \gamma_5, \gamma_{\mu}, \gamma_{\mu} \gamma_5\right\} b$

That are SCALAR, PSEUDOSCALAR, VECTOR, and AXIAL VECTOR (I'll ignore the tensor one for now )

You're discussing a B decay, which is a pseudoscalar.

If you want parity conservation, then the parity must match on either side of the equals sign.

Scalar : $\langle 0 | \bar{u} b | B \rangle = X_S$, where $X_S$ is unknown. BUT we know it doesn't have a lorentz index, because that wouldn't make sense. It is just some CONSTANT (SCALAR).

BUT the current $\bar{u} b$is a scalar too, and the B meson is a pseudoscalar. So if you look at the parity of each side you have :

$(+1)_S (-1)_P != (+1)_S$, which is NOT possible. You would need a pseudoscalar on the right hand side, not some constant scalar. This is impossible., because we only have scalars/constants and the momentum of the B to work with (and a scalar product $P_B^2$ is a vector times a vector, so (-1)*(-1) = +1 Parity. So it must be zero, as there is no way for $X_S$ to be anything else.

Remember that the parity for each of the structures S P V A is : +1 -1 -1 +1

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Next is the Pseudoscalar P : $\langle 0 | \bar{u} \gamma_5 b | B \rangle = X_P$

This one works for Parity.

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For Vector we have to see that we have a lorentz index now. So whatever is on the right side must have an index too. Our only building block that has an index in this case is the B momentum, $P_B^{\mu}$. If you end up looking at decaying vector mesons, like $B^*$, then you also have to worry about its polarization which is an axial vector. But for now all we have is its momentum.

So we try : $\langle 0 | \bar{u} \gamma^{\mu} b | B \rangle = X_V P_B^{\mu}$

For parity we have $(-1)_V (-1)_P != (+1)_S (-1)_V$, which is NOT POSSIBLE. So its zero. $X_V$ must be zero

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But if we do the axial current :
$\langle 0 | \bar{u} \gamma^{\mu} \gamma_5 b | B \rangle = X_A P_B^{\mu}$
For parity we have $(+1)_A (-1)_P = (+1)_S (-1)_V$, which IS POSSIBLE. We here use a definition, and call (usually) :
$\langle 0 | \bar{u} \gamma^{\mu} \gamma_5 b | B \rangle = i f_B P_B^{\mu}$ , where $f_B$ is the DECAY CONSTANT of the pseudoscalar meson.

But now we want to know what $X_P$ is. So, multiply both sides of the above by $P_{B \mu}$. On the right you get $i f_B P_B^2 = i f_B M_B^2$ if its onshell.

On the left you get
$P_{B \mu} \langle 0 | \bar{u} \gamma^{\mu} \gamma_5 b | B \rangle = \langle 0 | \bar{u} \not P_B \gamma_5 b | B \rangle$

but we know that $P_B = p_u + p_b$ because the momentum is made up of the sum of the two quarks.

$\langle 0 | \bar{u} (\not p_u + \not p_b) \gamma_5 b | B \rangle$

using $\not p_b \gamma_5 = - \gamma_5 \not p_b$ the anticommutation of the gamma matrices
and $\not p_q q = m_q q$
for both quarks, you can take out a sum of the masses of the quarks and get

$- \langle 0 | \bar{u} \gamma_5 b | B \rangle (m_u + m_b) = i f_B M_B^2$

and now you can relate $X_P$ to the decay constant $f_B$.

Does that help? Sorry if its not too clear.

 

[Hepth]

As for your second question, it is common for unbound leptons to be written like this, in terms of its spinor rather than its operator. If my current acts on a state :

$\langle \ell \bar{\nu} | \bar{\ell} \, \Gamma \, \nu | 0 \rangle = [\bar{\ell} \, \Gamma \, \nu]$

where $\bar{\ell} , \nu$ are now REALLY dirac spinors, $\bar{u}_{\ell} v_{\nu}$, but we have just relabeled them because we are lazy physicists. Unless I'm mistaken in their notation.

 

[Safinaz]

Many thanks ..

 

[Safinaz]

Only some point is unclear for me, for $< 0 | \bar{u} \gamma_5 b | B >$ ,

we have: $(-1)_p (-1)_p$ on the left hand side, while only $( -1)_p$ on the right hand side, so doesn't conserve?

 

[Hepth]

No, on the right had side it should now be a constant, which is a scalar, which has positive polarity.

All of the X's wer ejust constants (scalars) so they give +1.

 

[Safinaz]

I got it, that because in parity $( -1)^n = +1$, when n is even, it's basic I know, but I just memoriz ..

For my second question, may be I were not clear enough, I meant why the vacuum saturation approximation has not been used in the evaluation is the lepton current $\bar{l} ( 1-\gamma_5 ) \nu$ as well as the quark current?

Or you just meant in your answer that the value of $< l^- \nu | \bar{l} \Gamma \nu | 0 >$ is $\bar{l} \Gamma \nu$, but why, and what does unbound leptons mean here?

Because actually I don't know were "$\bar{l} ( 1-\gamma_5 ) \nu$ " has gone in the decay width formula of B→lν [hep-ph/0306037v2] Equ. 5 after Equ. 4, that I think the factor $( 1 - \frac{m_l^2}{m_B^2})^2$ Equ. 5 comes from $\lambda$ in
$\Gamma_{B \to l \nu } = \frac{1}{16\pi m_B} \lambda^{1/2} (m_l^2,0,m_B^2) |M|^2$, also if we evaluate the lepton term by traces, 'd not we have

$| \bar{l} ( 1-\gamma_5 ) \nu |^2 = ( /\!\!\!p_l + m_l ) (1-\gamma_5) /\!\!\!p_\nu (1-\gamma_5) =! 0$ ..

 

[Hepth]

I get that $|\bar{l} (1-\gamma_5)\nu|^2 = 4 (m_B^2- m_l^2) = 4 m_B^2 (1-\frac{m_l^2}{m_B^2})$

we also have $\lambda^{1/2}(m_l^2,0,m_B^2) = \sqrt{m_l^4-2 m_l^2 m_B^2+m_B^4} = m_B^2 (1-\frac{m_l^2}{m_B^2})$

so that's where you get both powers of it from. One from the lambda, one from the amplitude squared.

Also, your last line is wrong. Remember you are taking the amplitude times its complex conjugate. So one of the Left-projections will become a right-projection operator. (pl+m)(1-g5)(pnu)(1+g5)

which is nonzero.

 

[Safinaz]

Ok, that's totaly right. Then I think $< l \nu | l \Gamma \nu | 0 > =l \Gamma \nu$ cause the vacuum approximation used only to evaluate the quarks matrix elements due to the untrivial QCD interactions and sum rules..

You know I'm still not so familiar with that approximation, that for example instead of evaluating a process as u b -> l nu according to Feynman rules , now we make some effective approximations. I know the starting of VSA was to evaluate $< \bar{k}^0 | O | k^0 >$ ..

Thanks again

 

[wzr9999]

@Hepth Hi, thank you very much for your #4 post. While it solves some confusion on my mind, i still have one further question to ask. If
$$< 0 | \bar{u}\gamma^\mu \gamma^5 b | B > =i f_B p_B^\mu$$, then what about $$ < \bar{B} | \bar{u}\gamma^\mu \gamma^5 b | 0 > $$?
Similarly, if $$ < 0 | \bar{u} \gamma^5 b | B >=-\frac{i f_B m_B^2}{m_u+m_b} $$, then what about $$< \bar{B}| \bar{u} \gamma^5 b | 0 >$$ ? Thanks!