Question 3.11: The Bag Wave Function for Massive Quarks Show that the bag w......
The Bag Wave Function for Massive Quarks
Show that the bag wave function of massive quarks is
\Psi=N\left(\begin{array}{c} j_{\ell_{\kappa}}(p r) \chi_{\kappa}^{\mu} \\ \mathrm{i} \frac{p}{E+m} \operatorname{sgn}(\kappa) j_{\bar{\ell}_{\kappa}}(p r) \chi_{-\kappa}^{\mu} \end{array}\right) \mathrm{e}^{-\mathrm{i} E t} (1)
with
\begin{aligned} E & =\sqrt{p^{2}+m^{2}}, \\ \ell_{\kappa} & =\left\{\begin{array}{cc} -\kappa-1 & \text { for } \kappa<0 \\ \kappa & \text { for } \kappa>0 \end{array},\right. \\ \bar{\ell}_{\kappa} & =\left\{\begin{array}{cc} -\kappa & \text { for } \kappa<0 \\ \kappa-1 & \text { for } \kappa>0 \end{array} .\right. \end{aligned}
Show, furthermore, that the normalization condition
\int\limits_{\text {bag }} \mathrm{d}^{3} r \Psi^{\dagger} \Psi=1 (2)
leads to
N=\frac{p}{\sqrt{2 E(E R+\kappa)+m}} \frac{1}{\left|j \ell_{\kappa}(p R)\right| R}, (3)
R being the bag radius.
Adding the mass term to (3.113)
(\not p) \Psi=0 (3.113)
yields
(\not p-m) \Psi=0 (4)
Correspondingly (3.119) becomes
\left[-\mathrm{i}\left(\begin{array}{cc} 0 & \sigma_{r} \\ \sigma_{r} & 0 \end{array}\right)\left(\frac{\partial}{\partial r}+\frac{1}{r}-\frac{\beta}{r} \hat{K}\right)+\beta m\right] \Psi=E \Psi (5)
and (3.121)
\left(\frac{\mathrm{d}}{\mathrm{d} r}+\frac{1}{r}-\frac{\kappa}{r}\right) f(r)=E g(r) (3.121a)
\left(\frac{\mathrm{d}}{\mathrm{d} r}+\frac{1}{r}+\frac{\kappa}{r}\right) g(r)=-E f(r) (3.121b)
then becomes
\left(\frac{\mathrm{d}}{\mathrm{d} r}+\frac{1}{r}-\frac{\kappa}{r}\right) f(r)-(E-m) g(r)=0 (6a)
and
\left(\frac{\mathrm{d}}{\mathrm{d} r}+\frac{1}{r}+\frac{\kappa}{r}\right) g(r)+(E+m) f(r)=0 . (6b)
Now we insert g(r)=j_{\ell_{\kappa}}(p r) and f(r)=-[p /(E+m)] \operatorname{sgn}(\kappa) j_{\bar{\ell}_{\kappa}}(p r) and evaluate the left-hand sides of (6a) and (6b), respectively:
\frac{-p^{2}}{E+m} \operatorname{sgn}(\kappa)\left(\frac{\mathrm{d}}{\mathrm{d}(p r)}+\frac{1-\kappa}{p r}\right) j_{\bar{\ell}_{\kappa}}(p r)-(E-m) j \ell_{\kappa}(p r)=0 (7a)
and
p\left(\frac{\mathrm{d}}{\mathrm{d}(p r)}+\frac{1+\kappa}{p r}\right) j_{\ell_{\kappa}}(p r)-p \operatorname{sgn}(\kappa) j_{\bar{\ell}_{\kappa}}(p r)=0 . (7b)
With the help of recursion relations (3.127)
\begin{aligned} \left(\frac{\mathrm{d}}{\mathrm{d} z}+\frac{1-\kappa}{z}\right) j_{\bar{\ell}}(z) & =A j_{\ell}(z), & (3.127a) \\ \left(\frac{\mathrm{d}}{\mathrm{d} z}+\frac{1+\kappa}{z}\right) A j_{\ell}(z) & =-j_{\bar{\ell}}(z) . & (3.127b) \end{aligned}
(c=-\operatorname{sgn}(\kappa)) we then obtain
\frac{p^{2}}{E+m} j_{\ell_{\kappa}}(p r)-\frac{p^{2}}{E+m} j_{\ell_{\kappa}}(p r)=0 (8a)
and
p \operatorname{sgn}(\kappa) j_{\bar{\ell}_{\kappa}}(p r)-p \operatorname{sgn}(\kappa) j_{\bar{\ell}_{\kappa}}(p r)=0 (8b)
This yields the wave function of a massive quark in the MIT bag
\Psi=N\left(\begin{array}{c} j_{l \kappa}(p r) \chi_{\kappa}^{\mu} \\ i \frac{p}{E+m} \operatorname{sgn}(\kappa) j_{\bar{l}_{\kappa}}(p r) \chi_{-\kappa}^{\mu} \end{array}\right) e^{-i E t}
which transforms into (3.128) for m \rightarrow 0. Accordingly we obtain from (3.129) and (3.130)
ψ=N\left(\begin{array}{c}{{j_{\ell}(E r)\;\chi_{\kappa}^{\mu}(\theta,\phi)}}\\ {{\mathrm{i\:sgn}(\kappa)\:j_{\bar{\ell}}(E r)\:\chi_{-\kappa}^{\mu}(\theta,\phi)}}\end{array}\right)\,\mathrm{e}^{-\mathrm{i}E t}\ . (3.128)
\left(\begin{array}{c} -\operatorname{sgn}(\kappa) j_{\bar{\ell}}(E R) \chi_\kappa^\mu(\theta, \phi) \\ -\mathrm{i} j_{\ell}(E R) \chi_{-\kappa}^\mu(\theta, \phi) \end{array}\right)=\left(\begin{array}{c} j_{\ell}(E R) \chi_\kappa^\mu(\theta, \phi) \\ \mathrm{i} \operatorname{sgn}(\kappa) j_{\bar{\ell}}(E R) \chi_{-\kappa}^\mu(\theta, \phi) \end{array}\right) (3.129)
j_{\ell}(E R)=-\operatorname{sgn}(\kappa) j_{\bar{\ell}}(E R) . (3.130)
by substituting \operatorname{sgn}(\kappa) \rightarrow[p /(E+m)] \operatorname{sgn}(\kappa) the modified linear boundary condition
j_{l \kappa}(p r)=-\operatorname{sgn}(\kappa) \frac{p}{E+m} j_{\bar{l}_{\kappa}}(p r). (9)
By means of (9) we then evaluate the normalization integral:
\begin{aligned} 1 & =\int_{\text {bag }} \mathrm{d}^{3} r \Psi^{\dagger} \Psi \\ & =\int_{0}^{R} \mathrm{~d} r r^{2}\left[j_{\ell_{\kappa}}^{2}(p r)+\frac{p^{2}}{(E+m)^{2}} j_{\bar{\ell}_{\kappa}}^{2}(p r)\right] N^{2} . & (10) \end{aligned}
Employing the differentiation and recursion formulas of the Bessel functions (see (3.139))
\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}(z)} j_l(z) & =\frac{1}{2 l+1}\left[l j_{l-1}(z)-(l+1) j_{l+1}(z)\right], \\ j_{l+1}(z)+j_{l-1}(z) & =\frac{2 l+1}{z} j_l(z), & (3.139)\end{aligned}
\begin{aligned} j_{n}^{\prime}(z) & =\frac{n}{z} j_{n}(z)-j_{n+1}(z), \\ j_{n-1}^{\prime}(z) & =\frac{n-1}{z} j_{n-1}(z)-j_{n}(z), \\ j_{n+1}^{\prime}(z) & =-\frac{n+2}{z} j_{n+1}(z)+j_{n}(z), \\ j_{n+1}(z)+j_{n-1}(z) & =\frac{2 n+1}{z} j_{n}(z) & (12) \end{aligned}
we get
\begin{aligned} \frac{\mathrm{d}}{\mathrm{d} z} & \left.z^{3}\left(j_{n}^{2}(z)-j_{n-1}(z) j_{n+1}(z)\right)\right] \\ = & 3 z^{2}\left(j_{n}^{2}(z)-j_{n-1}(z) j_{n+1}(z)\right)+z^{3}\left[2 j_{n}(z)\left(-j_{n+1}(z)+\frac{n}{z} j_{n}(z)\right)\right. \\ & -j_{n-1}(z)\left(-\frac{n+2}{z} j_{n+1}(z)+j_{n}(z)\right) \\ & \left.-\left(\frac{n-1}{z} j_{n-1}(z)-j_{n}(z)\right) j_{n+1}(z)\right] \\ = & 3 z^{2}\left(j_{n}^{2}(z)-j_{n-1}(z) j_{n+1}(z)\right) \\ & +z^{3}\left[\frac{2 n}{z} j_{n}^{2}(z)-j_{n}(z)\left(j_{n+1}(z)+j_{n-1}(z)\right)+\frac{3}{z} j_{n-1}(z) j_{n+1}(z)\right] \\ = & 3 z^{2} j_{n}^{2}(z)+z^{3}\left[\frac{2 n}{z} j_{n}^{2}(z)-\frac{2 n+1}{z} j_{n}^{2}(z)\right] \\ = & 2 z^{2} j_{n}^{2}(z) . & (13) \end{aligned}
Terms of the form z^{2} j_{n}^{2}(z) appear in the integral (10). Therefore, relation (13) allows a direct evaluation of that integral:
\begin{aligned} 1= & N^{2} \frac{R^{3}}{2}\left[j_{\ell_{\kappa}}^{2}(p R)-j_{\ell_{\kappa}-1}(p R) j_{\ell_{\kappa}+1}(p R)\right. \\ & \left.+\frac{p^{2}}{(E+m)^{2}}\left(j_{\bar{\ell}_{\kappa}}^{2}(p R)-j_{\bar{\ell}_{\kappa}-1}(p R) j_{\bar{\ell}_{\kappa}+1}(p R)\right)\right] . & (14) \end{aligned}
Equation (14) can be simplified using (9). For \kappa<0 we have (see (12))
\begin{aligned} \bar{\ell}_{\kappa} & =\ell_{\kappa}+1, \quad \ell_{\kappa}=-\kappa-1 \\ j_{\ell_{\kappa}+1}(p R) & =\frac{E+m}{p} j \ell_{\kappa}(p R) \\ j_{\ell_{\kappa}-1}(p R) & =\frac{2 \ell_{\kappa}+1}{p R} j \ell_{\ell_{\kappa}}(p R)-j \ell_{\ell_{\kappa}+1}(p R) \\ & =\left(\frac{2 \ell_{\kappa}+1}{p R}-\frac{E+m}{p}\right) j_{\ell_{\kappa}}(p R) \\ j_{\ell_{\kappa}+2}(p R) & =\frac{2 \ell_{\kappa}+3}{p R} j \ell_{\ell_{\kappa}+1}(p R)-j \ell_{\kappa}(p R) \\ & =\left(\frac{2 \ell_{\kappa}+3}{p R} \frac{E+m}{p}-1\right) j_{\ell_{\kappa}}(p R) & (15) \end{aligned}
Equation (14) therefore becomes
\begin{aligned} 1= & N^{2} \frac{R^{3}}{2} j_{\ell_{\kappa}}^{2}(p R)\left[2-\left(\frac{2 \ell_{\kappa}+1}{p R}-\frac{E+m}{p}\right) \frac{E+m}{p}\right. \\ & \left.+\frac{p^{2}}{(E+m)^{2}}\left(1-\frac{2 \ell_{\kappa}+3}{p R} \frac{E+m}{p}\right)\right] \end{aligned}
\begin{aligned} = & N^{2} \frac{R^{3}}{2} j_{\ell_{\kappa}}^{2}(p R)\left[2-2 \ell_{\kappa}\left(\frac{E+m}{p^{2} R}+\frac{1}{(E+m) R}\right)\right. \\ & \left.-\frac{E+m}{p^{2} R}-\frac{3}{(E+m) R}+\frac{E+m}{E-m}+\frac{E-m}{E+m}\right] \\ = & N^{2} \frac{R^{3}}{2} j_{\ell_{\kappa}}^{2}(p R)\left(-2 \ell_{\kappa} \frac{2 E}{p^{2} R}-\frac{4 E-2 m}{p^{2} R}+\frac{4 E^{2}}{p^{2}}\right) \\ = & N^{2} \frac{R^{2}}{p^{2}} j_{\ell_{\kappa}}^{2}(p R)\left(2 \kappa E+m+2 E^{2} R\right) . & (16) \end{aligned}
Hence
N=\frac{p}{R\left|j_{\ell_{\kappa}}(p r)\right| \sqrt{2 \kappa E+m+2 E^{2} R}} . (17)
For \kappa>0 an analogous calculation yields
\begin{aligned} & \bar{\ell}_{\kappa}=\ell_{\kappa}-1, \quad \ell_{\kappa}=\kappa, \\ & j_{\ell_{\kappa}-1}(p R)=-\frac{E+m}{p} j_{\ell_{\kappa}}(p R) \\ & j \ell_{\kappa}+1(p R)=\frac{2 \ell_{\kappa}+1}{p R} j \ell_{\ell_{\kappa}}(p R)-j \ell_{\ell_{\kappa}-1}(p R) \\ & =\left(\frac{2 \ell_{\kappa}+1}{p R}+\frac{E+m}{p}\right) j \ell_{\kappa}(p R) \text {, } \\ & j \ell_{\kappa}-2(p R)=\frac{2 \ell_{\kappa}-1}{p R} j \ell_{\ell_{\kappa}-1}(p R)-j \ell_{\kappa}(p R) \\ & =-\left(\frac{E+m}{p} \frac{2 \ell_{\kappa}-1}{p R}+1\right) j \ell_{\kappa}(p R) . & (18) \end{aligned}
Equation (14) now has the form
\begin{aligned} 1= & N^{2} \frac{R^{3}}{2} j_{\ell_{\kappa}}^{2}(p r)\left[2+\left(\frac{2 \ell_{\kappa}+1}{p R}+\frac{E+m}{p}\right) \frac{E+m}{p}\right. \\ & \left.+\frac{p^{2}}{(E+m)^{2}}\left(\frac{E+m}{p} \frac{2 \ell_{\kappa}-1}{p R}+1\right)\right] \\ = & N^{2} \frac{R^{3}}{2} j_{\ell_{\kappa}}^{2}(p r)\left(2 \ell_{\kappa} \frac{2 E}{p^{2} R}+\frac{2 m}{p^{2} R}+\frac{4 E^{2}}{p^{2}}\right) \\ = & N^{2} \frac{R^{2}}{p^{2}} j_{\ell_{\kappa}}^{2}(p r)\left(2 \kappa E+m+2 E^{2} R\right) .& (19) \end{aligned}
Therefore N is given by (17) also for \kappa>0. The corresponding normalization factors for massless quarks are obtained by setting p=E and m=0, i.e. N_{\kappa}(m=0)=\frac{E}{R\left|j_{l k}(E R)\right|} \frac{1}{\sqrt{2 E(\kappa+E R)}}( see also (3.142)).
N_{\kappa=-1}^2=\frac{E R}{2 R^3(E R-1) j_0^2(E R)}, (3.142)