# A formula used in Quantum Field Theory

There is a formula: $$\left. F \left( -i \frac{\partial}{\partial \mathbf{x}} \right) G(\mathbf{x}) = G \left( -i \frac{\partial}{\partial \mathbf{y}} \right) F(\mathbf{y}) e^{i\mathbf{x} \cdot \mathbf{y}} \right|_{\mathbf{y}=0}$$ In perturbation theory this formula is used in the form: $$\left. \exp \left[ -\int d^4z\, V\left( \frac{-\delta}{\delta J(z)} \right)\right] \exp\left[{\frac{1}{2}\int d^4x d^4y\,J(x)\Delta(x-y) J(y)}\right] \\ = \exp\left[\frac{1}{2}\int d^4x d^4y\,\frac{\delta}{\delta \phi(x)}\Delta(x-y) \frac{\delta}{\delta \phi(y)}\right] \exp\left[{-\int d^4z\, V(\phi(z))-\int d^4x\,\phi(x) J(x)}\right] \right|_{\phi=0}$$
I know one way of proving it, which way is shown by Cosmas Zachos.
And I hear this formula can be proved by Fourier analysis too.
How can we prove?

• Why voted down -2? Dec 9, 2016 at 17:52
• There is nothing intrinsically bad about this question (and I didn't downvote it), but it should be asked in Math.SE, as it is devoid of physics. Dec 9, 2016 at 17:53
• well, you shouldn't've. This is a math question. If you got no luck here either, would you post it in Amateur Radio Stack Exchange? If you really want to get attention towards your question in Math.SE, add a bounty to it. If you don't have enough reputation to do so, earn some by asking good question or answering others'. Dec 9, 2016 at 18:00
• BTW I too would like to see a formal proof of the identity, but I cannot find your post in Math.SE. Would you please post a link here? Dec 9, 2016 at 18:09
• You should, in my opinion, post your question with all relevant information, saying I know one way of proving it. And I hear this formula can be proved by Fourier analysis too. is not helpful. The actual details, rather than being in a comment, should be in the post, so that other users can check it. I want you to get a proper answer, but adding elements of mystery to it (just my initial impression, no offence) does not help you get an answer. Dec 9, 2016 at 18:57

$$\left. G \left( -i \frac{\partial}{\partial \mathbf{y}} \right) F(\mathbf{y}) e^{i\mathbf{x} \cdot \mathbf{y}} \right|_{\mathbf{y}=0} =\\ =\left. G \left( -i \frac{\partial}{\partial \mathbf{y}} \right) F( -i \frac{\partial}{\partial \mathbf{x}}) e^{i\mathbf{x} \cdot \mathbf{y}} \right|_{\mathbf{y}=0} =\\ = \left. F\left( -i \frac{\partial}{\partial \mathbf{x}} \right) G \left( -i \frac{\partial}{\partial \mathbf{y}} \right) e^{i\mathbf{x} \cdot \mathbf{y}} \right|_{\mathbf{y}=0} =\\ = \left. F\left( -i \frac{\partial}{\partial \mathbf{x}} \right) G \left(\mathbf{x}\right) e^{i\mathbf{x} \cdot \mathbf{y}} \right|_{\mathbf{y}=0} =\\ =F\left( -i \frac{\partial}{\partial \mathbf{x}} \right) G(\mathbf{x}) ~.$$ The separation of conjugate variables is the same as in working in Fourier space.
Edit in response to comment. Detail of equivalent integration by parts in the end, $$F(-i\partial_x)G(x)=\int dy F(y) \tilde{G}(y) e^{ixy}/\sqrt{2\pi}= \\ =\int dy \left( \frac{1}{2\pi}\int dz e^{-izy} G(z)\right) F(y) e^{ixy}=$$ $$=\int dy \left( \frac{1}{2\pi}\int dz e^{-izy} G(i\stackrel{\leftarrow}{\partial} _y) \right) F(y) e^{ixy}=\\ =\int dy \left( \frac{1}{2\pi}\int dz e^{-izy} \right) G(-i {\partial} _y)F(y) e^{ixy}=\\ =\int dy~ \delta(y) ~ G(-i {\partial} _y)F(y) e^{ixy} ~.$$