Question 7.2: The Average Link Value Calculate the average link variable U......
The Average Link Value
Calculate the average link variable
\overline{{{U}}}=\frac{\int\mathrm{d}U U\,\mathrm{e}^{\frac{\beta}{N}}{\mathrm{Re tr}\,U W}}{\int\mathrm{d}U\,\mathrm{e}^{\frac{\beta}{N}}{\mathrm{Retr}\,U W}} (1)
for the gauge group SU(2).
Since the trace of an SU(2) matrix is real, which can be understood directly from (7.54), parameterize the SU(2) group elements with the help of the Pauli matrices:
U=\left(\begin{array}{c}{{x_{4}+\mathrm{i}x_{3}}}&{{x_{2}+\mathrm{i}x_{1}}}\\ -x_{2}+\mathrm{i}x_{1} & x_{4}-\mathrm{i}x_{3}\end{array}\right)\ , (7.54)
U=x_{1}𝟙 + i\sum\limits_{i}x_{i}\sigma_{i}\ . (2)
Exploiting this property we can write the exponent in the exponentials of (1) simply as β/2tr UW. A sum of SU(2) matrices is again proportional to a SU(2) matrix:
\sum\limits_{i}U_{i}=c{\tilde{U}}\ . (3)
This behavior becomes clear when remembering our discussion of Sect. 7.1.5. There we showed that a SU(2) matrix corresponds to a unit vector on the hypersphere {S}_{3}. Adding several SU(2) matrices together is like adding several vectors, the resulting vector pointing to the hypersurface, albeit with a different length. Note that this behavior is very specific to the SU(2) group and cannot be applied to SU(3). Using this feature we get
W=\lambda V\ ,\ \ \ V\in\mathrm{SU(2)}\ ,\ \ \ \lambda=\sqrt{\operatorname*{det}W}\ . (4)
The invariance of the measure with respect to the variable change U\to U V^{+} yields
\overline{{{U}}}=\frac{\int\mathrm{d}U U\,\mathrm{e}^{\frac{\beta}{2}\lambda\ \mathrm{tr}\,U V}}{\int\mathrm{d}U\,\mathrm{e}^{\frac{\beta}{2}\lambda\ \mathrm{tr}\,U V}}\\ =\frac{\int\mathrm{d}U U V^{\dagger}e^{\frac{\beta}{\mathrm{2}}\lambda\,\mathrm{tr}\,U}}{\int\mathrm{d}U\mathrm{e}^{\frac{\beta}{2}\lambda\,\mathrm{tr}\,U}}\\\\=V^{\dagger}\frac{\int d{U}U{\mathrm{e}}^{\frac{\beta}{2}\lambda\,\mathrm{tr}\,U}}{\int\mathrm{d}U{\mathrm{e}}^{\frac{\beta}{2}\lambda\,\mathrm{tr}\,U}}\ . (5)
Let us first evaluate the denominator of (1):
\int\mathrm{d}U\,\mathrm{e}^{{\frac{\beta}{2}}\lambda\ \mathrm{tr}\,U}={\frac{1}{4\pi^{2}}}\int\mathrm{d}\varphi\mathrm{d}\Omega(n)\sin^{2}{\frac{\varphi}{2}}\exp\left({\frac{\alpha}{2}}{\frac{\sin\varphi}{\sin{\frac{\varphi}{2}}}}\right)\equiv I\ ,\ \ \ \alpha\equiv\beta\lambda\ . (6)
The integration over the solid angle dΩ can be directly performed, yielding a value of 4π. Using the relation
\sin\varphi=2\sin{\frac{\varphi}{2}}\cos{\frac{\varphi}{2}} (7)
we get
I=\frac{1}{\pi}\int\,\mathrm{d}\varphi\,\mathrm{sin}^{2}\frac{\varphi}{2}\mathrm{e}^{\alpha\ \mathrm{cos}\frac{\varphi}{2}}\\=\frac{1}{\pi}\left(1-\frac{\partial^{2}}{\partial\alpha^{2}}\right)\underbrace{\int_{0}^{2\pi}\mathrm{d}\varphi\,\mathrm{e}^{\alpha\cos\frac{\varphi}{2}}}_{\equiv A}. (8)
We can evaluate the integral A explicitly:
A=\sum\limits_{k=0}^{\infty}{\frac{\alpha^{k}}{k!}}\int_{0}^{\;\;2\pi}\mathrm{d}\varphi\cos^{k}{\frac{\varphi}{2}}\\ =\sum\limits_{k=0}^{\infty}\frac{\alpha^{k}}{k!}2\int_{0}^{\pi}\mathrm{d}\phi\cos^{k}\phi\\ =2\sum\limits_{k=0}^{\infty}\frac{\alpha^{k}}{k!}\frac{1}{2^{k}}\sum\limits_{l=0}^{\infty}{\binom{k}{l}}\int_{0}^{\pi}\mathrm{d}\phi\cos((k-2l)\phi)\\=2\sum\limits_{k=0}^{\infty}\frac{\alpha^{k}}{k!}\frac{1}{2^{k}}\sum\limits_{l=0}^{\infty}{\binom{k}{l}}\left\{\begin{array}{l l}{{\pi}}&{{\mathrm{if}}}&{{k=2l}}\\ {{0}}&{{\mathrm{otherwise}}}\end{array}\right.\ . (9)
Thus only terms with k even contribute to the sum:
A=2\sum\limits_{\mu=0}^{\infty}\frac{\alpha^{2\mu}}{(2\mu)!}\frac{1}{2^{2\mu}}{\binom{2\mu}{\mu}}\pi \\ =2\pi\sum\limits_{\mu=0}^{\infty}\frac{\alpha^{2\mu}}{(2\mu)!}\frac{1}{2^{2\mu}}\frac{(2\mu)!}{(\mu!)^{2}}\\=2\pi\sum\limits_{\mu=0}^{\infty}\frac{\alpha^{2\mu}}{(\mu!)^{2}2^{2\mu}}=2\pi I_{0}(\alpha)\ . (10)
The sum corresponds to the series representation of the Bessel function \,I_{0}(x) of the 2nd kind and 0th order. In general, the following relation for Bessel functions holds:^{11}
I_{m}(x)=\sum\limits_{k=0}^{\infty}\frac{1}{k!(k+m)!}\left(\frac{x}{2}\right)^{2k+m}\ . (11)
For the next steps we can make use of the following recursion relations for Bessel functions:
I_{m-1}(x)-I_{m+1}(x)={\frac{2m}{x}}I_{m}(x)\,\,\,,\\ {I}_{m-1}(x)+{ I}_{m+1}(x)=2{I}_{m}^{\prime}(x)\,\,\,,{I}_{0}^{\prime}(x)={I}_{1}(x)\ . (12)
It follows that
I=2\left(1-\frac{\partial^{2}}{\partial\alpha^{2}}\right)I_{0}(\alpha)^{11} e.g. see M. Abramowitz, I.A. Stegun: Handbook of Mathematical Functions (Dover 1968) Formula (9.6.10).
=2\left(I_{0}(\alpha)-{\frac{\partial}{\partial\alpha}}I_{1}(\alpha)\right)\\ =2\left(I_{0}(\alpha)-{\frac{I_{0}(\alpha)+I_{2}(\alpha)}{2}}\right)\\ =I_{0}(\alpha)-I_{2}(\alpha)={\frac{2}{\alpha}}I_{1}(\alpha)\ . (13)
Now we have to compute the numerator of (1)
\int\,\mathrm{d}U U\,\mathrm{e}^{\frac{\beta\lambda}{2}\mathrm{tr}\,U} (14)
={\frac{1}{4\pi^{2}}}\int\mathrm{d}\varphi\mathrm{d}\Omega(n)\sin^{2}{\frac{\varphi}{2}}\left(\cos{\frac{\varphi}{2}}+{\frac{\mathrm{i}}{2}}\sin{\frac{\varphi}{2}}n\cdot\tau\right)\mathrm{exp}\left({\frac{\alpha}{2}}{\frac{\sin\varphi}{\mathrm{sin}\,{\frac{\varphi}{2}}}}\right)~.Because
\int\,\mathrm{d}\Omega(n)n=0\,\,, (15)
only the first term in the sum contributes:
\frac{1}{\pi}\int_{0}^{2\pi}\mathrm{d}\varphi\sin^{2}\frac{\varphi}{2}\cos\frac{\varphi}{2}\mathrm{e}^{\alpha\cos\frac{\varphi}{2}}\\ ={\frac{1}{\pi}}\left(\frac{\partial}{\partial\alpha}-\frac{\partial}{\partial\alpha^{3}}\right)\int_{0}^{2\pi}\mathrm{d}\varphi\mathrm{e}^{\alpha\ \mathrm{cos}\frac{\varphi}{2}}\\ ={\frac{1}{\pi}}\left({\frac{\partial}{\partial\alpha}}-{\frac{\partial}{\partial\alpha^{3}}}\right)2\pi I_{0}(\alpha)\\ =2\left(I_{1}(\alpha)-\frac{\partial}{\partial\alpha}\left\{\frac{I_{0}(\alpha)+I_{2}(\alpha)}{2}\right\}\right)\\=2I_{1}(\alpha)-I_{1}(\alpha)-\frac{I_{1}(\alpha)+I_{3}(\alpha)}{2}=\frac{2}{\alpha}I_{2}(\alpha)\,\,\,. (16)
Putting the results together we have determined the value of the average link:
\overline{{{U}}}=V^{\dagger}\frac{{\frac{2}{\alpha}}I_{2}(\alpha)}{{\frac{2}{\alpha}}I_{1}(\alpha)}=V^{\dagger}\frac{I_{2}(\alpha)}{I_{1}(\alpha)}\\=(\mathrm{det}\,W)^{-1/2}W^{\dagger}\frac{I_{2}(\beta\sqrt{\mathrm{det}\,W})}{I_{1}(\beta\sqrt{\mathrm{det}\,W})}\ . (17)