Derive canonical commutation relations from Schwingers principle

Question

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

Comments

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

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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

Answer 1

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

Comments

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

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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

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I thought it through, and it somehow makes sense. But isn't there a contradiction then when I look at the case of fermions?

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

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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

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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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The commutation relations do not change when the Hamiltonian is modified (and as a consequence also the Lagrangian) because they depend only on the phase space and the phase space is basically fixed by the symplectic potential term of the action. Please see the following two articles: arxiv.org/abs/hep-th/0110114v3 arxiv.org/abs/1104.5269 Curtright and Zachos.

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

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cont. Basically, the phase space is the set of initial data that one needs in order to solve the equations of motion. In the examples of the main text, you need to know the initial position and velocity (or momentum), whatever Hamiltonian you use. Thus they are described by the same phase space; therefore they respect the same canonical commutation relations. Quantization is actually a process in which the phase space is replaced by a Hilbert space and functions over the phase space by operators.

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

Answer 2

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.)

We start with the Hamiltonian action

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

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. $\Box$

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$^1$ 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

Comments

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

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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