超对称性和共形场理论 Supersymmetry and Conformal Field Theory 7CCMMS40

这是一份UCL伦敦大学 7CCMMS40作业代写的成功案例

超对称性和共形场理论 Supersymmetry and Conformal Field Theory 7CCMMS40
问题 1.

Self-duality then follows from requiring a certain coordinate dependence of the $U$ ‘s: This is fixed by giving the explicit dependence of the $v$ ‘s as
$$
v_{I i \alpha}=b_{I i \mathcal{A}} z_{\alpha}^{\mathcal{A}}, \quad v_{i^{\prime} \alpha}^{I}=b_{i^{\prime} \mathcal{A}}^{I} z^{\mathcal{A}}{ }{\alpha} $$ where the $b$ ‘s are constants. The orthonormality conditions on the $U$ ‘s then implies the constraint on the $b$ ‘s $$ b^{I}{ }{i^{\prime}(\mathcal{A}} b_{I i \mathcal{B})}=0
$$


证明 .

as well as determining the $u$ ‘s in terms of the $b$ ‘s (with much messier dependence than the $v$ ‘s), and thus $A$. Note that the $z$ dependence of $u$ can be written in terms of just $x$, as follows from rewriting the $u v$ orthogonality as (after multiplying by $z$ )
$$
u^{I}{ }{\iota} b{I i, \mathcal{A}} y^{\mathcal{A B}}=u_{I}{ }^{\iota} b^{I}{ }{i^{\prime} \mathcal{A}} y^{\mathcal{A B}}=0 $$ and noting scale invariance. Then the $x$ components of $A$ can also be written in terms of just $x$. We then can check the self-duality condition by calculating $f$ : The orthonormality condition on the $U$ ‘s can be written as $$ \delta{I}^{J}=u_{I}{ }^{\iota} u_{{ }^{J}}{ }^{J}+v_{I i}{ }^{\alpha} g^{i i^{\prime}} v_{i^{\prime} \alpha}^{J}
$$

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 7CCMMS40 COURSE NOTES :

$$
-\varphi-r^{2} \mathcal{A} \varphi=2 \mathcal{A}+r \mathcal{A}^{\prime}, \quad-r \varphi^{\prime}+r^{2} \mathcal{A} \varphi=-r \mathcal{A}^{\prime}+r^{2} \mathcal{A}^{2}
$$
After some massaging, we find the change of variables
$$
\tilde{\varphi}=\frac{1}{r}+r \varphi, \quad \tilde{\mathcal{A}}=\frac{1}{r}+r \mathcal{A}
$$
leads to the simplification
$$
\tilde{\varphi}^{\prime}=-\tilde{\mathcal{A}}^{2}, \quad \tilde{\mathcal{A}}^{\prime}=-\tilde{\mathcal{A}} \tilde{\varphi}
$$
$\tilde{\varphi}$ then can be eliminated, giving an equation for $\tilde{\mathcal{A}}$. Making a final change of variables,
$$
\psi=\tilde{\mathcal{A}}^{-1} \quad \Rightarrow \quad \psi \psi^{\prime \prime}-\left(\psi^{\prime}\right)^{2}=-1
$$







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