Cosmological constant

To incorporate a nonzero cosmological constant, use the Hamiltonian constraint

(1)
\begin{align} \mathcal{H} = e^{-3 \alpha} \left[ -p_{ \alpha } ^{2} + p_{+} ^{2} + p_{-} ^{2} \right] - \frac{2}{3} e^{ 3 \alpha } \left( ^{\left( 3 \right) } R - 2 \Lambda \right) \end{align}

instead of

(2)
\begin{align} \mathcal{H} = e^{-3 \alpha} \left[ -p_{ \alpha } ^{2} + p_{+} ^{2} + p_{-} ^{2} \right] - \frac{2}{3} e^{ 3 \alpha } \left( ^{\left( 3 \right) } R \right). \end{align}

Using the same operator ordering as in the Moncrief-Ryan case, we can now write the Wheeler-DeWitt equation as an eigenvalue equation:

(3)
\begin{align} \frac{\partial ^{2} \Psi}{\partial \alpha ^{2} } - B \frac{\partial \Psi}{\partial \alpha} - \frac{\partial ^{2} \Psi}{\partial \beta_{+}^{2}} - \frac{\partial ^{2} \Psi}{\partial \beta_{-}^{2}} + e^{4 \alpha} V \left( \beta_{\pm} \right) \Psi = - \frac{4}{3} \Lambda e^{6 \alpha} \Psi \end{align}

As in Moncrief-Ryan, we assume $\Psi = W \left( \alpha , \beta_{\pm} \right) e^{-S \left( \alpha , \beta_{\pm} \right) }$ and separate the Wheeler-DeWitt equation into the following equation for $S$:

(4)
\begin{align} \left( \frac{\partial S}{\partial \alpha} \right)^{2} - \left( \frac{\partial S}{\partial \beta_{+}} \right)^{2} - \left( \frac{\partial S}{\partial \beta_{-}} \right)^{2} + e^{4 \alpha} V \left( \beta_{\pm} \right) = 0, \end{align}

solved by Moncrief and Ryan for

(5)
\begin{align} S = \frac{1}{6}e^{2 \alpha} \left[ e^{-4 \beta_{+}} + 2 e^{2 \beta_{+} } \cosh \left( 2 \sqrt(3) \beta_{-} \right) \right], \end{align}

along with the equation

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for $W$, which simplifies using $\partial S / \partial \alpha = 2 S$ and the identity

(7)
\begin{align} \frac{\partial ^{2} S}{\partial \alpha ^{2}} - \frac{ \partial ^{2} S}{\partial \beta_{+} ^{2}} - \frac{ \partial ^{2} S}{\partial \beta_{-} ^{2}} = -12 S \end{align}

to

(8)
\begin{align} \frac{\partial ^{2} W}{\partial \alpha ^{2}} - \frac{\partial ^{2} W}{\partial \beta_{+} ^{2}} - \frac{\partial ^{2} W}{\partial \beta_{-} ^{2}} + 12S \cdot W - B \frac{\partial W}{\partial \alpha} + 2B \cdot S \cdot W - 2 \left( \frac{\partial W}{\partial \alpha} \frac{\partial S}{\partial \alpha} - \frac{\partial W}{\partial \beta_{+}} \frac{\partial S}{\partial \beta_{+}} - \frac{\partial W}{\partial \beta_{-}} \frac{\partial S}{\partial \beta_{-}} \right) = - \frac{4}{3} \Lambda e^{6 \alpha} W \left( \alpha , \beta_{\pm} \right) \end{align}

As a simplified model before approaching this equation, let us first consider the Taub model, in which $\beta_{-} = 0.$ Then we have

(9)
\begin{align} S = \frac{1}{6}e^{2 \alpha} \left( e^{-4 \beta{+}} + 2 e^{2 \beta_{+}} \right) \end{align}

as a solution to

(10)
\begin{align} \left( \frac{\partial S}{\partial \alpha} \right)^{2} - \left( \frac{\partial S}{\partial \beta_{+}} \right)^{2} + e^{4 \alpha} V \left( \beta_{+} \right) = 0, \end{align}

and $S$ satisfies

(11)
\begin{align} \frac{\partial ^{2} S}{\partial \alpha ^{2}} - \frac{\partial ^{2} S}{\partial \beta_{+} ^{2}} = -2 e^{2 \alpha} e^{-4 \beta_{+}}, \end{align}

so the equation for $W$ simplifies to

(12)
\begin{align} \frac{\partial ^{2} W}{\partial \alpha ^{2}} - \frac{\partial ^{2} W}{\partial \beta_{+} ^{2}} + 2 e^{2 \alpha} e^{-4 \beta_{+}} W - B \frac{ \partial W}{\partial \alpha} + 2B \cdot S \cdot W - 2 \left( \frac{\partial W}{\partial \alpha} \frac{\partial S}{\partial \alpha} - \frac{\partial W}{\partial \beta_{+}} \frac{\partial S}{\partial \beta_{+}} \right) = - \frac{4}{3} \Lambda e^{6 \alpha} W. \end{align}
page revision: 40, last edited: 16 Jun 2008 17:36
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