[Physics] Klein-Gordon Canonical Commutation Relation (CCR)

commutatorklein-gordon-equationoperatorsquantum-field-theory

In the complex Klein-Gordon field we regard as dynamical variables the field $\phi$, the complex conjugate of the field $\phi^*$, and the momenta $\pi$, $\pi^*$. I can't see how should arise the (equal time) canonical commutation relations, in particular:

$$ \left[\phi\left(t, {\bf x}\right),\ \pi\left(t,{\bf y} \right)\right]=i\hbar\delta^3\left({\bf x}-{\bf y}\right)$$

because the momenta are time derivatives of the field, and hence a function of $(ct,{\bf x})$, so the commutator above should be zero because both are functions of the space time coordinates, $\pi$ is not a differential operator.

Where am I wrong?

Best Answer

The field and its conjugate momentum are operators. They act on the Hilbert space of the theory, which is not the space of square-integrable functions of position, as it is in single particle quantum mechanics. Rather, the configuration space of the theory is the set of all field configurations and the wavefunction is actually a functional, which ascribes an amplitude $\Psi[\phi(x^\mu)]$ for the field to be in a configuration $\phi(x^\mu)$.

QFT is just quantum mechanics applied to an infinite collection of degrees of freedom labeled by positions $x$. Once you get used to the switch it comes to seem pretty simple, but the set of all field configurations makes a huge space so naturally it is very difficult to make things mathematically rigorous. For the most part physicists can live without rigourously defining the configuration space.

The dictionary from ordinary quantum mechanics to quantum field theory is

$$ \begin{array}{lcl} \mathrm{QM} && \mathrm{QFT}\\ t,i &\to& \left(t,x,y,z\right)\equiv\left(t,{\bf x}\right)\equiv x^{\mu}\\ q_{i}\left(t\right) &\to& \phi\left(x^{\mu}\right)\\ p_{i}\left(t\right) &\to& \pi\left(x^{\mu}\right)\\ \psi\left(t,q_{1},q_{2},\cdots,q_{N}\right) &\to& \Psi\left[\phi\left(x^{\mu}\right)\right]\\ \left[q_{i}\left(t\right),\ p_{j}\left(t\right)\right]~=~i\hbar\delta_{ij} &\to& \left[\phi\left(t, {\bf x}\right),\ \pi\left(t,{\bf y} \right)\right]=i\hbar\delta^3\left({\bf x}-{\bf y}\right)\\ \hat{H}\left(t\right) &\to& \hat{H}\left(t\right)=\int\mathrm{d}^{3}x\ \hat{\mathcal{H}}\left(x^{\mu}\right) \end{array} $$

You can represent the field momenta explicitly by functional derivatives:

$$ \pi(x^\mu) ~=~ - i\hbar \frac{\delta}{\delta \phi(x^\mu)}, $$

and check that this satisfies the commutation relationship.

Related Solutions

Quantum Field Theory – Quantizing a Complex Klein-Gordon Field and Its Two Types of Excitations

There is a conceptually simple (but fiddly) way of relating this expansion to the usual Fourier expansion. TL;DR: Requiring $\phi$ to satisfy the Klein-Gordon equation divides the nonzero Fourier components into two classes, corresponding to particles and antiparticles.

For a general complex scalar field defined on spacetime, $$ \phi(x) = \int\frac{d^4 p}{(2\pi)^4} \hat{\phi}(p) e^{-ip \cdot x}. \label{fourier}\tag{1} $$ This is the ordinary four-dimensional Fourier expansion. Now, let us impose the Klein-Gordon equation $$ 0 = (\partial^2 + m^2) \phi = \int \frac{d^4 p}{(2\pi)^4} (-p^2 + m^2) \hat{\phi}(p) e^{-ip \cdot x}. $$ A general field satisfying this must consist of only modes where $p^2 = m^2$, i.e. on-shell modes. Thus, $$ \hat{\phi}(p) = 2\pi \delta(p^2 - m^2) f(p) $$ for some function $f$ (the factor $2\pi$ is convenient for comparing with the standard expansion). Writing $p = (p^0, \mathbf{p})$, $$ \delta(p^2 - m^2) = \delta\left((p^0)^2 - (\mathbf{p}^2 + m^2)\right). $$ The argument of the $\delta$-function has two zeros $p^0 = \pm \sqrt{\mathbf{p}^2 + m^2}$ (for fixed $\mathbf{p}$), so we use the general rule $$ \delta(f(x)) = \sum_{f(x_i) = 0} \frac{1}{|f'(x_i)|} \delta(x - x_i) $$ to find $$ \delta(p^2 - m^2) = \frac{1}{|2 p^0|} \left[ \delta\left(p^0 - \sqrt{\mathbf{p}^2 + m^2}\right) + \delta\left(p^0 + \sqrt{\mathbf{p}^2 + m^2}\right) \right]. $$ Putting this back into (\ref{fourier}) and performing the $p^0$ integral (abbreviating $\sqrt{\mathbf{p}^2 + m^2} = E_\mathbf{p}$): $$ \begin{align} \phi(x) &= \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 E_\mathbf{p}} \left[ f(E_\mathbf{p}, \mathbf{p}) e^{-i(E_\mathbf{p}, \mathbf{p}) \cdot x} + f(-E_\mathbf{p}, \mathbf{p}) e^{-i(-E_\mathbf{p}, \mathbf{p}) \cdot x} \right] \\ &= \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 E_\mathbf{p}} \left[ f(E_\mathbf{p}, \mathbf{p}) e^{-i(E_\mathbf{p}, \mathbf{p}) \cdot x} + f(-E_\mathbf{p}, \mathbf{p}) e^{i(E_\mathbf{p}, -\mathbf{p}) \cdot x} \right] \\ &= \int \frac{d^3 p}{(2\pi)^3} \frac{1}{2 E_\mathbf{p}} \left[ f(E_\mathbf{p}, \mathbf{p}) e^{-i(E_\mathbf{p}, \mathbf{p}) \cdot x} + f(-E_\mathbf{p}, -\mathbf{p}) e^{i(E_\mathbf{p}, \mathbf{p}) \cdot x} \right] \\ \end{align} $$ where we have swapped $\mathbf{p} \mapsto -\mathbf{p}$ in the second term. We identify the usual (Heisenberg picture) expansion $$ \phi(x) = \int \frac{d^3 p}{(2\pi)^3} \frac{1}{\sqrt{2 E_\mathbf{p}}} \left[ a_\mathbf{p} e^{-ip \cdot x} + b^\dagger_\mathbf{p} e^{ip \cdot x} \right] $$ with $$ \begin{align} a_\mathbf{p} &= \frac{1}{\sqrt{2 E_\mathbf{p}}} f(E_\mathbf{p}, \mathbf{p}) \\ b_\mathbf{p}^\dagger &= \frac{1}{\sqrt{2 E_\mathbf{p}}} f(-E_\mathbf{p}, -\mathbf{p}) \end{align} $$ ($f(E_\mathbf{p}, \mathbf{p})$ and $f(-E_\mathbf{p}, -\mathbf{p})$ are the relativistically normalized annihilation and creation operators). If $\phi$ is real, we know from Fourier analysis that $\hat{\phi}(p) = \hat{\phi}(-p)^\dagger$, which immediately translates to $a_\mathbf{p} = b_\mathbf{p}$.

Up until now, the field is entirely classical ($a_\mathbf{p}$ and $b_\mathbf{p}$ are simply complex numbers, and $\dagger$ is complex conjugation). Thus even the classical field has two types of excitations: Positive-frequency solutions (with coefficients $a_\mathbf{p}$) and negative-frequency solutions (with coefficients $b_\mathbf{p}^\dagger$). After quantizing, they correspond to particles and antiparticles.

[Physics] Derive canonical commutation relations from Schwinger’s principle

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

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

The Hamiltonian phase space path integral reads$^1$

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

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

If we change infinitesimally the action (1), we derive the Schwinger action principle:

$$\begin{align} \frac{\hbar}{i}\delta\langle q_f,t_f &| q_i, t_i \rangle \cr ~\stackrel{(2)}{=}~& \int_{q(t_i)=q_i}^{q(t_f)=q_f} \!{\cal D}q~{\cal D}p~ \delta_0 S_H(t_i,t_f)~\exp\left\{\frac{i}{\hbar} S_H(t_i,t_f)\right\}\cr ~\stackrel{(3)}{=}~&\langle q_f,t_f | \widehat{\delta_0 S}_H(t_i,t_f)|q_i, t_i \rangle . \end{align} \tag{4} $$

Similarly, we are interested in calculating the change to

$$\begin{align} \langle q^{\prime}_i,t_i |& \hat{F}(t_f)|q^{\prime\prime}_i, t_i \rangle \cr ~=~& \int \! dq^{\prime}_f~dq^{\prime\prime}_f~ \langle q^{\prime}_i,t_i |q^{\prime}_f,t_f \rangle ~ \langle q^{\prime}_f,t_f | \hat{F}(t_f)|q^{\prime\prime}_f, t_f \rangle~\langle q^{\prime\prime}_f,t_f |q^{\prime\prime}_i,t_i \rangle ,\end{align} \tag{5} $$

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

$$\begin{align} \langle q^{\prime}_i,t_i | &\delta\hat{F}(t_f)|q^{\prime\prime}_i,t_i\rangle\cr ~=~& \delta\langle q^{\prime}_i,t_i | \hat{F}(t_f)|q^{\prime\prime}_i,t_i\rangle\cr ~\stackrel{(5)}{=}~& \int \! dq^{\prime}_f~dq^{\prime\prime}_f~ \delta\langle q^{\prime}_i,t_i |q^{\prime}_f,t_f \rangle ~ \langle q^{\prime}_f,t_f | \hat{F}(t_f)|q^{\prime\prime}_f, t_f \rangle~\langle q^{\prime\prime}_f,t_f |q^{\prime\prime}_i,t_i \rangle\cr ~+~&\int \! dq^{\prime}_f~dq^{\prime\prime}_f~ \langle q^{\prime}_i,t_i |q^{\prime}_f,t_f \rangle ~ \langle q^{\prime}_f,t_f | \hat{F}(t_f)|q^{\prime\prime}_f, t_f \rangle~\delta\langle q^{\prime\prime}_f,t_f |q^{\prime\prime}_i,t_i \rangle\cr ~\stackrel{(4)}{=}~&\frac{i}{\hbar}\langle q^{\prime}_i,t_i | [\hat{F}(t_f),\widehat{\delta_0 S}_H(t_i,t_f)]|q^{\prime\prime}_i,t_i\rangle ,\end{align} \tag{6}$$

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.$^2$

$$\begin{align} \frac{\hbar}{i} \delta\hat{q}^k(t_f) ~\stackrel{(7)}{\simeq}~& [\hat{q}^k(t_f),\widehat{\delta_0 S}_H(t_i,t_f)]\cr ~\stackrel{(1)}{\simeq}~& \left[\hat{q}^k(t_f), \sum_{\ell=1}^n \hat{p}_{\ell}(t_f)~ \delta_0\{ \hat{q}^{\ell}(t_f) - \hat{q}^{\ell}(t_i)\} \right]\cr ~=~&\sum_{\ell=1}^n [\hat{q}^k(t_f), \hat{p}_{\ell}(t_f)]~\delta_0\hat{q}^{\ell}(t_f) . \end{align} \tag{8}$$

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

$$\begin{align} [\hat{q}^k(t), \hat{p}_{\ell}(t)]~=~&i\hbar \delta_{\ell}^k~{\bf 1} , \cr k,\ell~\in~&\{1,\ldots, n\},\end{align} \tag{10}$$

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

$$\begin{align} \hat{F}(t_f)~=~&\exp\left\{\frac{i}{\hbar} \hat{H} \Delta t \right\} \hat{F}(t_f)\exp\left\{-\frac{i}{\hbar} \hat{H} \Delta t \right\}, \cr \Delta t~:=~&t_f-t_i. \end{align} \tag{11}$$

Moreover, $| q, t \rangle $ are instantaneous position eigenstates in the Heisenberg picture,

$$\begin{align} \hat{q}^k(t)| q, t \rangle~=~&q^k| q, t \rangle, \cr k~\in~&\{1,\ldots, n\}.\end{align} \tag{12}$$

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 CCR

$$\begin{align} [\hat{z}^I(t), \hat{z}^K(t)] ~\stackrel{(16)}{=}~&i\hbar\omega^{IK}~{\bf 1} , \cr I,K~\in~&\{1,\ldots, 2n\}, \end{align} \tag{15}$$

is $i\hbar$ times the equal-time Peierls bracket

$$\begin{align} \{z^I(t), z^K(t^{\prime})\} ~\stackrel{(17)}{\simeq}~&\omega^{IK} , \cr I,K~\in~&\{1,\ldots, 2n\}. \end{align} \tag{16}$$

The Peierls bracket is defined as

$$\begin{align} \{ F,G \}~:=~&\iint_{[t_i,t_f]^2}\!dt~dt^{\prime}~\sum_{I,K=1}^{2n} \frac{\delta F }{\delta z^I(t)}~G^{IK}_{\rm ret}(t,t^{\prime})~\frac{\delta G }{\delta z^K(t^{\prime})} \cr &- (F\leftrightarrow G)\cr ~\stackrel{(22)}{\simeq}~&\iint_{[t_i,t_f]^2}\!dt~dt^{\prime}~\sum_{I,K=1}^{2n} \frac{\delta F }{\delta z^I(t)}~\omega^{IK}~\frac{\delta G }{\delta z^K(t^{\prime})}. \end{align} \tag{17}$$

Classically, if the Hamiltonian action (1) is recast in symplectic notation

$$\begin{align} S_H(t_i,t_f)~\sim~&\int_{[t_i,t_f]}\!dt\left(\frac{1}{2}\sum_{I,K=1}^{2n} z^I ~\omega_{IK}~ \dot{z}^K-H\right)\cr ~\sim~&\frac{1}{2} \iint_{[t_i,t_f]^2}\!dt~dt^{\prime}\sum_{I,K=1}^{2n} z^I(t)~\omega_{IK}~ \delta^{\prime}(t\!-\!t^{\prime}) ~z^K(t^{\prime})\cr &-\int_{[t_i,t_f]}\!dt~H, \end{align} \tag{18}$$

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

$$\begin{align} H_{IK}(t,t^{\prime}) ~:=~& \frac{\delta^2 S_H(t_i,t_f)}{\delta z^I(t) \delta z^K(t^{\prime})}\cr ~=~&\omega_{IK}~\delta^{\prime}(t\!-\!t^{\prime}) -\partial_I\partial_K H ~\delta(t\!-\!t^{\prime})\cr ~\simeq~&\omega_{IK}~\delta^{\prime}(t\!-\!t^{\prime}). \end{align} \tag{21}$$

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

$$\begin{align} \delta z^I(t_f) ~\stackrel{(20)+(23)}{=}&-\int_{[t_i,t_f]}\! dt^{\prime} \sum_{K=1}^{2n} G^{IK}_{\rm ret}(t_f,t^{\prime})~\frac{\delta ~\delta_0 S_H(t_i,t_f)}{\delta z^K(t^{\prime})}\cr ~\stackrel{(17)}{=}~~&\{\delta_0 S_H(t_i,t_f), z^I(t_f)\} \cr ~\simeq~~~&\left\{ \sum_{J=1}^{2n} z^J(t_f)~\omega_{JK} \delta_0z^K(t_f) , z^I(t_f)\right\}\cr ~=~~~&-\sum_{J=1}^{2n} \delta_0 z^J(t_f)~\omega_{JK} \{z^K(t_f) , z^I(t_f)\}\cr ~\stackrel{(16)}{\simeq}~~&-\delta_0z^I(t_f),\end{align} \tag{25}$$

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.

Best Answer

Related Solutions

Quantum Field Theory – Quantizing a Complex Klein-Gordon Field and Its Two Types of Excitations

[Physics] Derive canonical commutation relations from Schwinger’s principle

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