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Hi,

I'm reading www.phys.ethz.ch/~babis/Teaching/QFTI/qft1.pdf [Broken] and trying to understand the canonical quantization of the Schrodinger field. In particular, the Lagrangian:

\begin{equation}

\mathcal{L} = \frac{i}{2}\psi^* \partial_0 \psi - \frac{i}{2}\psi \partial_0 \psi^* + \frac{\left(\partial_i \psi^*\right) \left(\partial^i \psi\right)}{2m}

\end{equation}

results in the Schrodinger equation for a free particle. Now, the canonical momentum is given by:

\begin{equation}

\Pi = \frac{\partial \mathcal{L}}{\partial \left(\partial_0 \psi\right)}= \frac{i}{2} \psi^*

\end{equation}

and:

\begin{equation}

\Pi^* = \frac{\partial \mathcal{L}}{\partial \left(\partial_0 \psi^* \right)}= - \frac{i}{2} \psi

\end{equation}

Subsequently, imposing the canonical commutation relation, I obtain:

\begin{equation}

[ \psi(\mathbf{x}_1,t),\Pi(\mathbf{x}_2,t)] = [ \psi(\mathbf{x}_1,t),\frac{i}{2}\psi^*(\mathbf{x}_2,t)] = i \delta^3(\mathbf{x}_1 - \mathbf{x}_2)

\end{equation}

and therefore:

\begin{equation}

[ \psi(\mathbf{x}_1,t),\psi^*(\mathbf{x}_2,t)] = 2 \delta^3(\mathbf{x}_1 - \mathbf{x}_2)

\end{equation}

However, it turns out I'm wrong by a factor of 2, because according to the link I provided (see equation (3.22)), the canonical commutation relation should be:

\begin{equation}

[ \psi(\mathbf{x}_1,t),\psi^*(\mathbf{x}_2,t)] = \delta^3(\mathbf{x}_1 - \mathbf{x}_2)

\end{equation}

Have I done something wrong in my calculations? Or am I supposed to renormalize the wave function such that the factor of 2 disappears? Or is there something else I am not understanding?

Any help would be much appreciated.

I'm reading www.phys.ethz.ch/~babis/Teaching/QFTI/qft1.pdf [Broken] and trying to understand the canonical quantization of the Schrodinger field. In particular, the Lagrangian:

\begin{equation}

\mathcal{L} = \frac{i}{2}\psi^* \partial_0 \psi - \frac{i}{2}\psi \partial_0 \psi^* + \frac{\left(\partial_i \psi^*\right) \left(\partial^i \psi\right)}{2m}

\end{equation}

results in the Schrodinger equation for a free particle. Now, the canonical momentum is given by:

\begin{equation}

\Pi = \frac{\partial \mathcal{L}}{\partial \left(\partial_0 \psi\right)}= \frac{i}{2} \psi^*

\end{equation}

and:

\begin{equation}

\Pi^* = \frac{\partial \mathcal{L}}{\partial \left(\partial_0 \psi^* \right)}= - \frac{i}{2} \psi

\end{equation}

Subsequently, imposing the canonical commutation relation, I obtain:

\begin{equation}

[ \psi(\mathbf{x}_1,t),\Pi(\mathbf{x}_2,t)] = [ \psi(\mathbf{x}_1,t),\frac{i}{2}\psi^*(\mathbf{x}_2,t)] = i \delta^3(\mathbf{x}_1 - \mathbf{x}_2)

\end{equation}

and therefore:

\begin{equation}

[ \psi(\mathbf{x}_1,t),\psi^*(\mathbf{x}_2,t)] = 2 \delta^3(\mathbf{x}_1 - \mathbf{x}_2)

\end{equation}

However, it turns out I'm wrong by a factor of 2, because according to the link I provided (see equation (3.22)), the canonical commutation relation should be:

\begin{equation}

[ \psi(\mathbf{x}_1,t),\psi^*(\mathbf{x}_2,t)] = \delta^3(\mathbf{x}_1 - \mathbf{x}_2)

\end{equation}

Have I done something wrong in my calculations? Or am I supposed to renormalize the wave function such that the factor of 2 disappears? Or is there something else I am not understanding?

Any help would be much appreciated.

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