# Derive canonical commutation relations from Schwingers principle

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The book of Dyson "Quantum-Field-Theory" states in section 4.4 that one can derive canonical commutation relations from Schwingers quantum action principle. However, doesn't give a calculation for the canonical momenta, which should commute.

Is it possible to derive the canonical commutation relations (for fields and momenta) just from Schwingers principle?

Edit: By canonical commutation relations I mean:

$[\Phi(x), \pi(y)]=i \hbar \delta(x-y)$

$[\Phi(x), \Phi(y)]=0$

$[\pi(x), \pi(y)] = 0$

with $x^0 = y^0$.

By Schwingers principle, I mean, that for a given Lagrangian $L$, which is an operatorvalued function on $\Phi(x)$ and $\partial_\mu \Phi(x)$, a variation $\Phi'(x')= \Phi(x) + \delta \Phi(x)= M^{-1}\Phi(x)M^{1}$, and the resulting change in the Eigenstates $|n_{new}\rangle = M |n_{old} \rangle$, the change in the transition amplitude $\langle n_{new, t_2}|m_{new, t_1} \rangle - \langle n_{old, t_2}|m_{old, t_1} = \int dx' L(\Phi'(x')) - \int dx L(\Phi(x))$ equals the change in the action.

This post imported from StackExchange Physics at 2017-09-17 13:00 (UTC), posted by SE-user Quantumwhisp
I think there is some discussion of this in the book by Manoukian (https://www.springer.com/gp/book/9783319309385), esp. chapter 4.3 & 4.6.

This post imported from StackExchange Physics at 2017-09-17 13:00 (UTC), posted by SE-user To Chin Yu
I read it. And it basically does the same things that Dysons book does. They both just derive the commutation relations for $[\Phi, \pi]$, but not for $[\Phi, \Phi]$ or $[\pi, \pi]$

This post imported from StackExchange Physics at 2017-09-17 13:00 (UTC), posted by SE-user Quantumwhisp

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Yes, it is possible. I'll demonstrate it with a very simple quantum mechanical example. The generalization to quantum field theory is straightforward. This derivation is more or less along the lines of Dyson's book, with a special choice of $t=0$ as the spacelike hypersurface.

Starting from a simple free particle Action: $$I = \int dt L = \int dt\frac{m}{2}\dot{x}^2$$ The canonical momentum $$p = \frac{\partial L}{\partial \dot{x}}$$ Performing the Legendre transformation, we obtain the action in the Hamiltonian form $$I = \int dt (p \dot{x} - H)$$ With the Hamiltonian $$H = \frac{p^2}{2 m}$$ In order to derive the commutation relations of an operator with the phase space variables, we replace the existing Hamiltonian with this operator as the new Hamiltonian $H'$.

In our case, we choose a new Hamiltonian is: $$H' = p$$ This will allow us to find the commutation relations of the momentum operator with the other operators. The ation: $$I' = \int dt (p \dot{x} - H') = \int dt (p \dot{x} - p)$$

We derive the equations of motion using the variation principle on the new action \begin{align*} \delta I' &= \int dt ( \delta p \dot{x} + p \dot{\delta x} - \delta p ) \\ &= \int dt ( \delta p \dot{x} -\dot{ p} \delta x - \delta p ) + \mathrm{boundary \quad terms} \\ &= \int dt ( \delta p (\dot{x}-1) -\dot{ p} \delta x ) + \mathrm{boundary \quad terms} \end {align*}

Thus the equations of motion $$\dot{x} = 1, \quad \dot{p} = 0$$ But according to Heisenberg equations of motion, for any operator $O$, we must have: $$\dot{O} = \frac{i}{\hbar}[H', O]$$ Thus, in our case $$\dot{x} = 1 \implies \frac{i}{\hbar}[p, x] = 1 \implies [x, p] = i \hbar$$ $$\dot{p} = 0 \implies \frac{i}{\hbar}[p, p] = 0 \implies [p, p] = 0$$

This post imported from StackExchange Physics at 2017-09-17 13:00 (UTC), posted by SE-user David Bar Moshe
answered Aug 29, 2017 by (4,095 points)
I'm sorry, but neither is this proof based on schwingers principle, nor do I understand it fully: Why would you be allowed to choose the Hamiltonian to be $P$?

This post imported from StackExchange Physics at 2017-09-17 13:00 (UTC), posted by SE-user Quantumwhisp
To be more specific: Your derivation holds for a special Hamiltonian <-> a special lagrangian. but the commutation relations are imposed in every theory, independent of the choice of the lagrangian.

This post imported from StackExchange Physics at 2017-09-17 13:00 (UTC), posted by SE-user Quantumwhisp
The canonical commutation relations do not depend on the Hamiltonian. The depend only on the first term of the action not containing the Hamiltonian (in our case the term $p\dot{x}$). (This term is customarily called the symplectic potential). For example, the canonical commutation relations of free particle, harmonic oscillator and a particle in a potential well are the same despite the fact that Hamiltonians are different. This is the reason why the commutation relations were computed from a theory with the same symplectic potential but with a different but appropriately chosen Hamiltonian.

This post imported from StackExchange Physics at 2017-09-17 13:00 (UTC), posted by SE-user David Bar Moshe
Cont. Secondly, it is not the question of being allowed to change the Hamiltonian. You are not using the theory with the modified Hamiltonian for anything else besides the derivation of the commutation relations. Your theory is still the original theory, but after the derivation of the commutation relations, you can apply them to the original theory. The trick in replacing the Hamiltonian is that the Hamiltonian is the generator of the time evolution, thus you can derive its infinitesimal action from the equations of motion using Heisenbeg's principle.

This post imported from StackExchange Physics at 2017-09-17 13:00 (UTC), posted by SE-user David Bar Moshe
Thirdly, as I wrote in the beginning of the answer, I am using the time $t=0$ as the spacelike hypersurface. In this choice Dyson's equation (188) derived from Schwinger's principle, becomes equation (190) (Heisenberg) which I actually used. Please, try to take as an exercise a theory of a particle minimally coupled to an electromagnetic potential and derive the generalized commutation relations. (This case is different than the above because the symplectic potential is modified by the interaction term).

This post imported from StackExchange Physics at 2017-09-17 13:00 (UTC), posted by SE-user David Bar Moshe
For fermions I could do the same things you did, and it would yield the "wrong" commutiation relations, because it would be commutation, and not anticommutation relations. What's wrong in this reasoning?

This post imported from StackExchange Physics at 2017-09-17 13:00 (UTC), posted by SE-user Quantumwhisp
As was mentioned briefly, this method is basically classical. It actually yields the Poisson brackets, on which we perform the Dirac canonical quantization and replace by $i\hbar$ times the commutator. Actually the last step is done by hand. In the fermionic case, this method still gives the Poisson brackets, but we replace it by hand with the commutator when the Hamiltonian is even and the anticommutator when the Hamiltonian is odd.

This post imported from StackExchange Physics at 2017-09-17 13:00 (UTC), posted by SE-user David Bar Moshe
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In this answer we shall use the Schwinger action principle to prove the CCRs, as OP requested. Let us for simplicity consider the Hamiltonian formulation of bosonic point mechanics. (It is in principle possible to generalize to the Lagrangian formulation, field theory, & fermionic variables.)

$$S_H(t_i,t_f)~=~\int_{t_i}^{t_f}\!dt~ L_H, \qquad L_H~=~\sum_{k=1}^np_k \dot{q}^k - H . \tag{1}$$

The Hamiltonian phase space path integral reads$^1$

$$\langle q_f,t_f | q_i, t_i \rangle ~=~\langle q_f,t |\exp\left\{-\frac{i}{\hbar} \hat{H} \Delta t \right\}| q_i, t \rangle ~=~\int_{q(t_i)=q_i}^{q(t_f)=q_f} \!{\cal D}q~{\cal D}p~\exp\left\{\frac{i}{\hbar} S_H(t_i,t_f)\right\}, \tag{2}$$

$$\langle q_f,t_f | \hat{F}|q_i, t_i \rangle ~=~ \int_{q(t_i)=q_i}^{q(t_f)=q_f} \!{\cal D}q~{\cal D}p~F~\exp\left\{\frac{i}{\hbar} S_H(t_i,t_f)\right\} . \tag{3}$$

If we change infinitesimally the action (1), we derive the Schwinger action principle: $$\frac{\hbar}{i}\delta\langle q_f,t_f | q_i, t_i \rangle ~\stackrel{(2)}{=}~ \int_{q(t_i)=q_i}^{q(t_f)=q_f} \!{\cal D}q~{\cal D}p~ \delta_0 S_H(t_i,t_f)~\exp\left\{\frac{i}{\hbar} S_H(t_i,t_f)\right\} ~\stackrel{(3)}{=}~\langle q_f,t_f | \widehat{\delta_0 S}_H(t_i,t_f)|q_i, t_i \rangle . \tag{4}$$

Similarly, we are interested in calculating the change to $$\langle q^{\prime}_i,t_i | \hat{F}(t_f)|q^{\prime\prime}_i, t_i \rangle ~=~ \int \! dq^{\prime}_f~dq^{\prime\prime}_f~ \langle q^{\prime}_i,t_i |q^{\prime}_f,t_f \rangle ~ \langle q^{\prime}_f,t_f | \hat{F}(t_f)|q^{\prime\prime}_f, t_f \rangle~\langle q^{\prime\prime}_f,t_f |q^{\prime\prime}_i,t_i \rangle , \tag{5}$$ where the initial states $|q, t_i \rangle$ and the middle bracket on the rhs. of eq. (5) are unaffected by the (later) change of action in the (later) time interval $[t_i,t_f]$. We calculate the change $$\langle q^{\prime}_i,t_i | \delta\hat{F}(t_f)|q^{\prime\prime}_i,t_i\rangle ~=~ \delta\langle q^{\prime}_i,t_i | \hat{F}(t_f)|q^{\prime\prime}_i,t_i\rangle$$ $$~\stackrel{(5)}{=}~ \int \! dq^{\prime}_f~dq^{\prime\prime}_f~ \delta\langle q^{\prime}_i,t_i |q^{\prime}_f,t_f \rangle ~ \langle q^{\prime}_f,t_f | \hat{F}(t_f)|q^{\prime\prime}_f, t_f \rangle~\langle q^{\prime\prime}_f,t_f |q^{\prime\prime}_i,t_i \rangle$$ $$+\int \! dq^{\prime}_f~dq^{\prime\prime}_f~ \langle q^{\prime}_i,t_i |q^{\prime}_f,t_f \rangle ~ \langle q^{\prime}_f,t_f | \hat{F}(t_f)|q^{\prime\prime}_f, t_f \rangle~\delta\langle q^{\prime\prime}_f,t_f |q^{\prime\prime}_i,t_i \rangle$$ $$~\stackrel{(4)}{=}~\frac{i}{\hbar}\langle q^{\prime}_i,t_i | [\hat{F}(t_f),\widehat{\delta_0 S}_H(t_i,t_f)]|q^{\prime\prime}_i,t_i\rangle , \tag{6}$$ or equivalently, as an operator identity $$\delta\hat{F}(t_f)~\stackrel{(6)}{=}~\frac{i}{\hbar} [\hat{F}(t_f),\widehat{\delta_0 S}_H(t_i,t_f)]. \tag{7}$$ In other words, the cause $\delta_0$ leads to the effect $\delta$. We next go to the adiabatic limit $\Delta t:=t_f-t_i\to 0$ where the symplectic term of the action (1) dominates over the Hamiltonian term, i.e. we can effectively remove the Hamiltonian $H$ from the calculation, i.e. operators in the Heisenberg picture can be treated as time-independent in this limit.$^2$ $$\frac{\hbar}{i} \delta\hat{q}^k(t_f) ~\stackrel{(7)}{\simeq}~ [\hat{q}^k(t_f),\widehat{\delta_0 S}_H(t_i,t_f)]$$ $$~\stackrel{(1)}{\simeq}~ \left[\hat{q}^k(t_f), \sum_{\ell=1}^n \hat{p}_{\ell}(t_f)~ \delta_0\{ \hat{q}^{\ell}(t_f) - \hat{q}^{\ell}(t_i)\} \right] ~=~\sum_{\ell=1}^n [\hat{q}^k(t_f), \hat{p}_{\ell}(t_f)]~\delta_0\hat{q}^{\ell}(t_f) . \tag{8}$$ Since there is effectively no Hamiltonian, the cause & effect must cancel: $$\delta_0\hat{q}^{\ell}(t_f) + \delta\hat{q}^{\ell}(t_f)~=~0. \tag{9}$$ Eq. (8) is only possible if we have the equal-time CCR $$[\hat{q}^k(t), \hat{p}_{\ell}(t)]~=~i\hbar \delta_{\ell}^k~{\bf 1} , \qquad k,\ell~\in~\{1,\ldots, n\}, \tag{10}$$ as OP wanted to show. $\Box$

--

$^1$ Here we will work in the Heisenberg picture with time-independent ket and bra states, and time-dependent operators $$\hat{F}(t_f)~=~\exp\left\{\frac{i}{\hbar} \hat{H} \Delta t \right\} \hat{F}(t_f)\exp\left\{-\frac{i}{\hbar} \hat{H} \Delta t \right\}, \qquad \Delta t~:=~t_f-t_i. \tag{11}$$ Moreover, $| q, t \rangle$ are instantaneous position eigenstates in the Heisenberg picture, $$\hat{q}^k(t)| q, t \rangle~=~q^k| q, t \rangle, \qquad k~\in~\{1,\ldots, n\}. \tag{12}$$ see e.g. J.J. Sakurai, Modern Quantum Mechanics, Section 2.5. Such states (12) are only well-defined for commuting observables $$[\hat{q}^k(t), \hat{q}^{\ell}(t)]~=~0 , \qquad k,\ell~\in~\{1,\ldots, n\}, \tag{13}$$ so we are not going to derive the CCR (13). Rather (13) is an assumption with this proof.

Finally for completeness, let us mention that instead of instantaneous position eigenstates $| q, t \rangle$, we could use instantaneous momentum eigenstates $| p, t \rangle$, but again we would have to assume the corresponding CCR $$[\hat{p}_k(t), \hat{p}_{\ell}(t)]~=~0 , \qquad k,\ell~\in~\{1,\ldots, n\}. \tag{14}$$ Assumptions (13) & (14) can be avoided by using other methods. E.g. the CCRs (10), (13) & (14) also follows from the Peierls bracket.$^3$

$^2$ Notation: The $\approx$ symbol means equality modulo eqs. of motion. The $\sim$ symbol means equality modulo boundary terms. The $\simeq$ symbol means equality in the adiabatic limit, where the Hamiltonian $H$ can be ignored.

$^3$ The correspondence principle between quantum mechanics and classical mechanics states that the equal-time CCRs $$[\hat{z}^I(t), \hat{z}^K(t)] ~\stackrel{(16)}{=}~i\hbar\omega^{IK}~{\bf 1} , \qquad I,K~\in~\{1,\ldots, 2n\}, \tag{15}$$ is $i\hbar$ times the equal-time Peierls bracket $$\{z^I(t), z^K(t^{\prime})\} ~\stackrel{(17)}{\simeq}~\omega^{IK} , \qquad I,K~\in~\{1,\ldots, 2n\}. \tag{16}$$ The Peierls bracket is defined as $$\{ F,G \}~:=~\iint_{[t_i,t_f]^2}\!dt~dt^{\prime}~\sum_{I,K=1}^{2n} \frac{\delta F }{\delta z^I(t)}~G^{IK}_{\rm ret}(t,t^{\prime})~\frac{\delta G }{\delta z^K(t^{\prime})} - (F\leftrightarrow G)$$ $$~\stackrel{(22)}{\simeq}~\iint_{[t_i,t_f]^2}\!dt~dt^{\prime}~\sum_{I,K=1}^{2n} \frac{\delta F }{\delta z^I(t)}~\omega^{IK}~\frac{\delta G }{\delta z^K(t^{\prime})}. \tag{17}$$ Classically, if the Hamiltonian action (1) is recast in symplectic notation $$S_H(t_i,t_f)~\sim~\int_{[t_i,t_f]}\!dt\left(\frac{1}{2}\sum_{I,K=1}^{2n} z^I ~\omega_{IK}~ \dot{z}^K-H\right)$$ $$~\sim~\frac{1}{2} \iint_{[t_i,t_f]^2}\!dt~dt^{\prime}\sum_{I,K=1}^{2n} z^I(t)~\omega_{IK}~ \delta^{\prime}(t\!-\!t^{\prime}) ~z^K(t^{\prime}) -\int_{[t_i,t_f]}\!dt~H, \tag{18}$$ and changed infinitesimally, then the classical solution $z^I(t)$ is also changed infinitesimally $\delta z^I(t)$, to ensure that the deformed EL eqs. $$0~\approx~ \frac{\delta}{\delta z^I(t)} (S_H+\delta_0 S_H)[z+\delta z] \tag{19}$$ are satisfied. (We are ignoring boundary terms everywhere in this calculation.) In other words $$\frac{\delta ~\delta_0 S_H(t_i,t_f)}{\delta z^I(t)}~\approx~-\int_{[t_i,t_f]}\! dt^{\prime} \sum_{K=1}^{2n}H_{IK}(t,t^{\prime})~\delta z^K(t^{\prime}), \tag{20}$$ where the Hessian is $$H_{IK}(t,t^{\prime}) ~:=~ \frac{\delta^2 S_H(t_i,t_f)}{\delta z^I(t) \delta z^K(t^{\prime})} ~=~\omega_{IK}~\delta^{\prime}(t\!-\!t^{\prime}) -\partial_I\partial_K H ~\delta(t\!-\!t^{\prime})~\simeq~\omega_{IK}~\delta^{\prime}(t\!-\!t^{\prime}). \tag{21}$$ Then the retarded Green's function simplifies to $$G^{IK}_{\rm ret}(t,t^{\prime}) ~\stackrel{(21)+(23)}{\simeq}~\omega^{IK}~\theta(t\!-\!t^{\prime}),\tag{22}$$ $$\int_{[t_i,t_f]}\! dt^{\prime} \sum_{J=1}^{2n}H_{IJ}(t,t^{\prime})~G^{JK}_{\rm ret}(t^{\prime},t^{\prime\prime}) ~=~\delta_I^K~\delta^{\prime}(t\!-\!t^{\prime\prime}). \tag{23}$$ Therefore we derive the classical analogue of the operator identity (7) $$\delta F(t_f)~\stackrel{(25)}{=}~\{\delta_0 S_H(t_i,t_f), F(t_f)\} \tag{24}$$ from $$\delta z^I(t_f) ~\stackrel{(20)+(23)}{=}~-\int_{[t_i,t_f]}\! dt^{\prime} \sum_{K=1}^{2n} G^{IK}_{\rm ret}(t_f,t^{\prime})~\frac{\delta ~\delta_0 S_H(t_i,t_f)}{\delta z^K(t^{\prime})} ~\stackrel{(17)}{=}~\{\delta_0 S_H(t_i,t_f), z^I(t_f)\}$$ $$~\simeq~\left\{ \sum_{J=1}^{2n} z^J(t_f)~\omega_{JK} \delta_0z^K(t_f) , z^I(t_f)\right\}~=~-\sum_{J=1}^{2n} \delta_0 z^J(t_f)~\omega_{JK} \{z^K(t_f) , z^I(t_f)\}~\stackrel{(16)}{\simeq}~-\delta_0z^I(t_f), \tag{25}$$ which in turn is consistent with the fact that cause & effect must cancel: $$\delta_0z^I(t_f) + \delta z^I(t_f)~=~0, \tag{26}$$ since there is effectively no Hamiltonian.

This post imported from StackExchange Physics at 2017-09-17 13:00 (UTC), posted by SE-user Qmechanic
answered Sep 3, 2017 by (2,860 points)
Could one do the same thing for Fermions? From what I understand, the CCRS for position and momentum (13) and (14) are an assumption that guarantees the full freedom to a field configuration at one time. Could I assume the anticommutation-relation in the same way?

This post imported from StackExchange Physics at 2017-09-17 13:00 (UTC), posted by SE-user Quantumwhisp
Furthermore, what do you do if you assume the lagrangian to be a function of $\dot{\phi}$ and $\phi$, not of $\phi$ and $\pi$ (as you did, you called them $q$ and $p$, but it's the same). I know how to derive the commutation relations (10) for this case, but I fail to "assume" (14) in this case. I can understand that assuming (13) is needed for independency of fields (that would be $q$), but where do I know that the same is true for $\hat{p} = \frac{\partial L(\hat{q}, \hat{\dot{q}}) }{\partial \dot{q}}$?

This post imported from StackExchange Physics at 2017-09-17 13:00 (UTC), posted by SE-user Quantumwhisp

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