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Is it possible to calculate in a general way the commutator of an operator $O$ which depends on some variable $x$ and the derivative of this $O$ with respect to $x$? $${O}={O}(x)\\ \left[\partial_x{O}(x), O(x^{\prime})\right]=? $$ To be honest I don't quite understand, what the derivative of an operator is. When one plugs in the standard definition of the derivative of a function, it looks as though the operator commutes with its derivative, but I don't quite know how to feel about that "proof".

In practice I need this to calculate the commutator of the field operator of a free scalar field and any of its four derivatives: $$\left[\partial_\mu\phi(\mathbf{x},t),\phi(\mathbf{x}^{\prime},t)\right]$$ (I'm looking at scalar field theory described by a Lagrangian density $\mathcal{L}=\partial_\mu \phi^*\partial^\mu\phi-m^2\phi\phi^*$)

One can of course simply compute this by plugging in the field operator, but I was wondering about the general situation.

And what about the more general case of two operators, whose commutator is known? $$\left[O(x),U(x^{\prime})\right]=A(x,x^{\prime})\\ \left[\partial_x{O}(x),U(x^{\prime})\right]=? $$

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  • $\begingroup$ Comment to the question (v1): Which theory are you considering? $\endgroup$ – Qmechanic Mar 14 '14 at 12:44
  • $\begingroup$ I'm hearing an introductory lecture on QFT, so I'd guess standard QFT? The field I'm looking at is described by $\mathcal{L}=\partial_\mu \phi^*\partial^\mu\phi-m^2\phi\phi^*$ $\endgroup$ – user35915 Mar 15 '14 at 10:03
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The derivative of an operator: Let $X(t),\;\mathbb{R}\rightarrow \mathcal{X}$, where $\mathcal{X}$ is some normed linear space, say a Banach or Hilbert space. Then we can define the derivative in the usual way: $$\partial _{t}X(t)=\lim_{\delta \rightarrow 0}\frac{X(t+\delta )-X(t)}{\delta}.$$ However, on $\mathcal{X}$ different topologies exist, strong, weak, uniform, etc. Hille and Phillips, "Functional Analysis and Semi-Groups" contains a quite readable discussion of these matters.

In general the commutator does not vanish. Consider

$$X\left(\xi\right)=U\cos\xi+V\sin\xi\Rightarrow\dfrac{dX}{d\xi}=-U\sin\xi+V\cos\xi$$ then one gets, after some algebras

$$\left[X\left(\xi\right),\dfrac{dX}{d\xi}\right]=\left[U,V\right]$$

Then it is clear that one recovers the usual result for commuting $U$ and $V$ operators.

Edit: the result is correct ! Thanks to the comments.

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    $\begingroup$ This can't be quite right. If $ X $ is scalar then those two must commute. Aren't there more terms in that commutator? $\endgroup$ – Emilio Pisanty Mar 14 '14 at 22:18
  • $\begingroup$ @Urgje: Are you sure about your calculation? I got the answer $\left[X,\partial_\xi X\right]=\left[U,V\right]$. But I guess that this does not change the statement, that one can not answer my question in a general way? $\endgroup$ – user35915 Mar 15 '14 at 10:19
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    $\begingroup$ In the last expression UV should be [U,V] and VU should be [V,U] so we obtain the result noted by user35915. Indeed, this does not change the statement. $\endgroup$ – Urgje Mar 15 '14 at 10:52
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    $\begingroup$ @Urgje You mention the result is wrong instead of simply corrected it ... you don't feel a bit uneasy with that ?!? I wasn't, so I did the 2 lines calculations and edit your post. $\endgroup$ – FraSchelle Oct 21 '14 at 13:42
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    $\begingroup$ Oh, I see, but the definition $[X(\xi_1),X(\xi_2)]=X(\xi_1)X(\xi_2)-X(\xi_2)X(\xi_1)$ does vanish, right? Which would mean that $[U,V]\sin(\xi_2-\xi_1)$ would also vanish, right? $\endgroup$ – user24999 Oct 22 '14 at 13:22
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Notice that in $[\partial_xO(x),O(x')]$, the partial derivative only acts on $x$, not on $x'$. So we can pull the partial derivative operator out of the bracket and get $\partial_x[O(x),O(x')]$.

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  • $\begingroup$ but $x$ may be equal to $x'$ $\endgroup$ – Soap Nov 22 '17 at 17:02
  • $\begingroup$ @Soap Hi, if they are the same, the commutator vanishes identically. This is consistent with what I wrote. $\endgroup$ – Drake Marquis Nov 23 '17 at 1:35
  • $\begingroup$ The last commutator you wrote vanishes identically. However, it doesn't help to solve $[\partial_x O(x),O(x)]\neq\partial_x[O(x),O(x)]$. In particular, there is no reason why $[\partial_x O(x),O(x)]$ vanishes in general. Rather, One should evaluate $[\partial_x O(x),O(x')]$ in the fashion you showed and then evaluate at $x'=x$. $\endgroup$ – Iván Mauricio Burbano May 21 at 2:11
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I came here before asking myself the same question and, as I figured it out, I'd like to answer just for the record.

Respecting the former answer, with plain and usual commutators (e.g. see this question) in the Schrödinger picture with locality and causality, if $X(\xi)=\sum_i\left(\alpha_i(\xi){U}+b_i(\xi){V}\right)$, the condition $[X(\xi),X(\xi^\prime)]=0$ (which comes from definition of the usual commutator) implies $[U,V]=0$ (yes, explicitly you will get something non-zero, but this must be equated to the zero that comes from the definition of the commutator of the operator with itself) and thus also $[\partial_\xi{X}(\xi),X(\xi^\prime)]=0$.

Now, in general for any field $\phi$ in the Heisenberg picture, $$[\partial_i\phi(x),\phi(y)]=\partial_i[\phi(x),\phi(y)]+[\partial_i,\phi(y)]\phi(x)\tag{1}$$ where $x$ and $y$ are 4-vectors and $i$ stands for $x^i$ derivatives in the spatial components $i=1,2,3$. Now, $[\partial_i,\phi(y)]\phi(x)=0$ as all derivatives are in $x^i$, and thus, simply $$[\partial_i\phi(x),\phi(y)]=\partial_i[\phi(x),\phi(y)]=0\tag{2}$$ as $[\phi(x),\phi(y)]=0$ for spacelike separation and particularly for equal times $x^0=y^0$, which as far as I've seen are usually assumed in the Heisenberg picture unless said otherwise. As user Pisanty commented in the former answer, at least in the Schrödinger picture, scalars always commute with themselves and with their derivatives. The case of the time component is different depending on spacetime separation; if you want to compute $[\dot\phi,\phi]$ directly (at equal times) I guess you should either know the Fourier decomposition of $\dot\phi$ or compute $\dot\phi=i[H,\phi]$ using the Hamiltonian $H$, at least you get from $[\partial_0,\phi(y)]\phi(x)$ that $[\dot\phi(x),\phi(y)]$ should be an even operator function in space.

And so, if you know that $[A(x),B(y)]=C(x,y)$, then $[\partial_xA(x),B(y)]=\partial_xC(x,y)$.

Let me exemplify this with the equations of motion of the real scalar theory (you should be able to extend it for your complex scalar one): $$\mathcal{L}=\frac{1}{2}\partial_\mu\phi\partial^\mu\phi-\frac{1}{2}m^2\phi^2\,\Longrightarrow\,H=\frac{1}{2}\int{d^3x}\left(\pi^2+(\nabla\phi)^2+m^2\phi^2\right)\tag{3}$$ Then you get \begin{align}\dot\phi(x)=i[H,\phi(x)]&=\frac{i}{2}\int{d^3y}[\pi^2(y)+(\nabla\phi)^2(y)+m^2\phi^2(y),\phi(x)]\\&=\frac{i}{2}\int{d^3y}[\pi^2(y),\phi(x)]=\pi(x)\tag{4}\end{align} where I used $[(\nabla\phi)^2(y),\phi(x)]=0$ because of the reasons above. From here you also get in particular for this theory $[\dot\phi,\phi]=[\pi,\phi]=-i\delta(\vec{x}-\vec{y})$. And for $\pi$, \begin{align}\dot\pi(x)=i[H,\pi(x)]&=\frac{i}{2}\int{d^3y}[\pi^2(y)+(\nabla\phi)^2(y)+m^2\phi^2(y),\pi(x)]\\&=\frac{i}{2}\int{d^3y}[(\nabla\phi)^2(y)+m^2\phi^2(y),\pi(x)]\tag{5}\end{align} from where, using the above statements, \begin{align}\int{d^3y}[(\nabla\phi)^2(y),\pi(x)]&=\int{d^3y}\left(\nabla\phi\cdot[\nabla\phi,\pi]+[\nabla\phi,\pi]\cdot\nabla\phi\right)\\ &=2\int{d^3y}(\nabla\phi\cdot\nabla[\phi,\pi])\\ &=2i\int{d^3y}\left(\nabla\phi\cdot\nabla\delta(\vec{x}-\vec{y})\right)\\ &=2i\nabla\phi\cdot\vec\delta(\vec{x}-\vec{y})\bigg|_{\pm\infty}-2i\int{d^3y}\nabla^2\phi\,\delta(\vec{x}-\vec{y})\\ &=-2i\nabla^2\phi(x)\tag{6}\end{align} where I used the commutation of the Euclidean dot product and good old integration by parts, I also stopped writing the variable dependence but just remember that $\nabla$ acts on $y$'s only. So this way one gets $\dot\pi=\nabla^2\phi-m^2\phi^2$ which using $(4)$ gives the Klein-Gordon equation for $\phi$ as expected.

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