# A contradiction in Nonrelativistic Quantum Field Theory

Reference : "Field Quantization" by W.Greiner & J.Reinhardt, Edition 1996.

In the above reference as concerns the Hamilton density $$\:\mathcal H\:$$ and the Hamiltonian $$\:H\:$$ of the Schrodinger field, we read :

The Hamilton density is $$$$\mathcal H = \pi\dfrac{\partial\psi}{\partial t}-\mathcal L = \dfrac{\hbar^2}{2m}\boldsymbol\nabla\psi^*\boldsymbol{\cdot\nabla}\psi+V\left(\boldsymbol x,t\right)\psi^* \psi\,, \tag{3.6}\label{3.6}$$$$ which, after an integration by parts, leads to the Hamiltonian $$$$H=\int\mathrm d^3\boldsymbol x\,\mathcal H\left(\boldsymbol x\right) =\int\mathrm d^3\boldsymbol x\,\psi^*\left(\boldsymbol x\right)\left(-\dfrac{\hbar^2}{2m}\nabla^2+V\left(\boldsymbol x,t\right)\right)\psi\left(\boldsymbol x\right) \tag{3.7}\label{3.7}$$$$

The path from expression \eqref{3.6} to \eqref{3.7} is based on the relation

$$$$\int\mathrm d^3\boldsymbol x\left(\boldsymbol\nabla\psi^*\boldsymbol{\cdot\nabla}\psi\right) = -\int\mathrm d^3\boldsymbol x\left(\psi^*\nabla^2 \psi\right) \tag{01}\label{01}$$$$

I didn't have trouble proving the above equation. The trouble here is the contradiction between equating a real number (left hand side) to a complex in general number (right hand side).

So, the question is if equation \eqref{01} is wrong or is a priori accepted even with this contradiction.

• Parts twice gives $\int\psi^\ast\nabla^2\psi d^3\boldsymbol x=\int\psi\nabla^2\psi^\ast d^3\boldsymbol x$ (relating the boundary assumptions this requires to unitarity et al is a valuable exercise), which implies $\int\psi^\ast\nabla^2\psi d^3\boldsymbol x\in\Bbb R$.
– J.G.
Commented Jun 24, 2022 at 8:01

There is a seeming contradiction as the integrands are in fact as you described them ($$\boldsymbol{\nabla} \psi ^{*} \cdot \boldsymbol{\nabla} \psi \in \mathbb{R}$$ while $$\psi ^{*} \nabla ^{2} \psi \in \mathbb{C}$$). However, the equality is in regards to their volume integrals which can be shown to be strictly real. That is because the resulting surface terms from the integration by parts are assumed to vanish at the boundaries of the integral (basically at infinity).

Let's write the complex $$\psi (x)$$ as:

$$$$\psi (x) = \phi (x) + i \, \chi (x) \, , \quad \phi (x) , \chi (x) \in \mathbb{R}$$$$

The real quantity you pointed out shall be:

$$$$\boldsymbol{\nabla} \psi ^{*} \cdot \boldsymbol{\nabla} \psi = (\boldsymbol{\nabla} \phi )^{2} + (\boldsymbol{\nabla} \chi )^{2}$$$$

While the complex quantity is:

$$$$\psi ^{*} \nabla ^{2} \psi = \phi \nabla ^{2} \phi +i \, (\phi \nabla ^{2} \chi - \chi \nabla ^{2} \phi) + \chi \nabla ^{2} \chi$$$$

The imaginary part of the latter equality can be integrated by parts:

$$$$\int i \, (\phi \nabla ^{2} \chi - \chi \nabla ^{2} \phi) \, d^{3}x = i\, \Big[\phi \mathbf{\hat{r}} \cdot \boldsymbol{\nabla} \chi - \chi \mathbf{\hat{r}} \cdot \boldsymbol{\nabla} \phi \Big]_{\mathcal{V} } + \int i \, (\boldsymbol{\nabla} \chi \cdot \boldsymbol{\nabla} \phi - \boldsymbol{\nabla} \phi \cdot \boldsymbol{\nabla} \chi) \, d^{3}x$$$$

where $$\Big[...\Big]_{\mathcal{V} }$$ implies evaluating the argument in the brackets at the limits if the integral and $$\mathbf{\hat{r}}$$ the unit vector of the gradient. The integral is identically 0, so the only complex part remaining are the surface terms in the bracket. But since in your proof you implicitly assumed that either $$\psi$$ or $$\boldsymbol{\nabla} \psi$$ (or both) vanish at infinity, this follows trivially for its constituent parts $$\phi$$ and $$\chi$$. Thus the complex surface term is also 0.

EDIT:

A minor addition, we may also show the vanishing of the imaginary part of the integral using Green's theorem as mentioned in the comments. Like you said:

$$$$\int _{\mathcal{V}} (\phi \nabla ^{2} \chi - \chi \nabla ^{2} \phi) \, d^{3}x = \int _{\mathcal{V}} \boldsymbol{\nabla} \cdot (\phi \boldsymbol{\nabla} \chi - \chi \boldsymbol{\nabla} \phi) \, d^{3}x = \int _{\partial \mathcal{V}} (\phi \boldsymbol{\nabla} \chi - \chi \boldsymbol{\nabla} \phi) \cdot d\mathbf{S}$$$$

In order to perform this, we initially assume a finite volume $$\mathcal{V}$$ so that a surface boundary $$\partial \mathcal{V}$$ may be defined which, again as you said, we shall send to infinity i.e. $$\mathcal{V}$$ becomes all of space. We therefore assume the finite $$\mathcal{V}$$ is a sphere with a boundary being its surface at the maximum radial distance $$R$$. Hence the final integral is:

$$$$\int _{0} ^{2\pi} d\varphi \int _{0} ^{\pi} \sin{\theta} \, d\theta \, (\phi \boldsymbol{\nabla} \chi - \chi \boldsymbol{\nabla} \phi) \Big |_{\rho = R} \cdot \boldsymbol{\hat{\rho}} = \int _{0} ^{2\pi} d\varphi \int _{0} ^{\pi} \sin{\theta} \, d\theta \, \left (\phi \frac{\partial \chi}{\partial \rho} - \chi \frac{\partial \phi}{\partial \rho} \right ) \Bigg |_{\rho = R}$$$$

Now for all of space, send $$R \rightarrow +\infty$$. Thus you once again get the surface terms at infinity where either the functions themselves or their derivatives (or both) vanish:

$$$$\left (\phi \frac{\partial \chi}{\partial \rho} - \chi \frac{\partial \phi}{\partial \rho} \right ) \Bigg |_{\rho = +\infty} = 0$$$$

• Thanks for your answer. I don't understand why I stuck to such a simple problem , solved by the analysis of an item to its real and imaginary parts. So, there is no contradiction. As in a comment by @J.B. $$\left(\psi^*\nabla^2 \psi\right) \in \mathbb C \quad\text{but}\quad \int\mathrm d^3\boldsymbol x\left(\psi^*\nabla^2 \psi\right) \in \mathbb R \nonumber$$ Commented Jun 24, 2022 at 15:27
• For the final step of your answer to be more clear I would like to use this $$\int\limits_{\mathcal V}\left(\phi\nabla^2\chi-\chi\nabla^2\phi\right)\mathrm d^3\boldsymbol x = \int\limits_{\mathcal V}\boldsymbol{\nabla\cdot}\left(\phi\boldsymbol\nabla\chi-\chi\boldsymbol\nabla\phi\right)\mathrm d^3\boldsymbol x=\int\limits_{\partial\mathcal V}\left(\phi\boldsymbol\nabla\chi-\chi\boldsymbol\nabla\phi\right)\boldsymbol{\cdot}\mathrm d\mathbf S \nonumber$$ (aka Green's theorem) and take the limit $\,\mathcal V\rightarrow \texttt{all space}$. Commented Jun 24, 2022 at 15:28
• ...by the way, Welcome to PSE. Commented Jun 24, 2022 at 15:32
• Thank you! As for proving the equality using Green's theorem, I can make an edit to my post explaining it via that method as well. Commented Jun 24, 2022 at 22:04