Solving a differential equation generally goes something like this:

  • One is given a set of initial conditions. There is a large freedom of choice in doing so. For example, in Quantum Mechanics any normalizable $\psi(x,0)$ will do
  • The differential equation then determines how the state evolves, giving $\psi(x,t)$ for all times $t$.

Consider now QFT, in particular the free real scalar field as a simple example. The (Klein-Gordon) equation of motion for fields in the Heisenberg picture is solved by


For the real scalar field, the explicit solution is written $$\phi(\vec{x},t)=\int (a(\vec{p})e^{-ip\cdot x}+a^{\dagger}(\vec{p})e^{ip\cdot x}) \frac{d^3\vec{p}}{\sqrt{(2\pi)^3 2E_0}}$$

In QFT lecture, it is often stated that this is the most general $\phi(x,t)$ which solves the equation of motion. Let's investigate that claim. In particular, plugging in $t=0$, we have assumed the initial condition

$$\phi(\vec{x},0)=\int (a(\vec{p})e^{i\vec{p}\cdot \vec{x}}+a^{\dagger}(\vec{p})e^{-i\vec{p}\cdot \vec{x}}) \frac{d^3\vec{p}}{\sqrt{(2\pi)^3 2E_0}}$$

But as far as I can tell, the only constraint on $\phi(x,0)$ introduced in lecture is that it should obey the commutation relations. $\phi(x,0)$ is a very important quantity, because in the Schrödinger picture it is the field operator, just as analogously $x$ and $p=i\nabla$ were the position and momentum operators in Quantum Mechanics.

Therefore I'd like to ask for one of two possible solutions:

  • Can someone provide a proof that we must have this particular initial condition for the fields? My feeling is that the commutation relations alone are not enough to give this specific form, as operators have a huge amount of degrees of freedom.
  • Just as interesting would be to provide a counterexample: an initial condition $\phi(\vec{x})$ which satisfies the commutation relations but is not equal to $\phi(\vec{x},0)$ above.

Either of these answers would be very elucidating for me. If not a proof, then a reference to a proof would of course also be just as helpful.

EDIT: After more thought, the answer to this question is sort of a mixture of knzhou and ~Cosmos Zachos' answers. As knzhou remarked, the form $\phi(x,0)$ is not uniquely defined by the commutation relations or KG equation, because you could just redefine $\phi(x,0)$ to be the field with $t=1$ and all equations would still hold. This would be equivalent to redefining the operators $a(\vec{p})$ by adding a phase.

However, there is a strong restriction on the $a(\vec{p})$ operators, namely* they can only be redefined up to a momentum-dependent phase $a(\vec{p}) \to e^{i \alpha(\vec{p})}a(\vec{p})$. The proof is along the lines of what Cosmos Zachos' was mentioning and the treatment is done in many QFT textbooks. The essential reason this is possible is that redefining $a(\vec{p}) \to e^{i \alpha(\vec{p})}a(\vec{p})$ does not change the commutation relations.

*The momentum-dependence is my own realization and not taken from a book, but seeing as the commutation relations remain unchanged under the redefinition, it seems clear that they cannot constrain the phase of $a(\vec{p})$. In addition, redefining $\phi(\vec{x},0) \to \phi(\vec{x},1)$ is a momentum-dependent change in the $a$ operators: $a(\vec{p}) \to e^{i\sqrt{\vec{p}^2+m^2} t} a(\vec{p})$

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    $\begingroup$ Possibly related: physics.stackexchange.com/questions/196460/… $\endgroup$ – GodotMisogi Oct 25 '18 at 13:31
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    $\begingroup$ A free scalar field is no more and no less than a fancy repackaging of quantum oscillators. Convince yourself you appreciate the analog expressions for just one oscillator, first. $\endgroup$ – Cosmas Zachos Oct 25 '18 at 14:05
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    $\begingroup$ As @CosmasZachos notes, that's the key preliminary step. Quantum field theory is mapping dynamical variables to an infinite number of harmonic oscillators via a Fourier transform, so the physical requirements will pop out when you compute the commutators after this exercise. $\endgroup$ – GodotMisogi Oct 25 '18 at 15:14
  • $\begingroup$ When you say quantum oscillators, I assume you mean the general solution to the free particle, not the harmonic oscillator, is that true? And, if the solution is an analogy to the free particle solution in ordinary QM, that means to me that it is an assumption in QFT, and not a direct consequence of the commutators or equation of motion (option 2). Would you agree? $\endgroup$ – doublefelix Oct 25 '18 at 15:45
  • $\begingroup$ No offense meant, but what I am looking for is a straightforward yes or no answer to the question: Can the usual form of $\phi(x,0)$ for real free field be derived as the only possible form consistent with the KG equation and commutation relations, or not? I am glad to review any topics in order to answer it myself, but having just thoroughly reviewed my lecture notes for second quantization, I don't see any straightforward answer, so I don't know where to continue. $\endgroup$ – doublefelix Oct 26 '18 at 15:17

First off, we rarely want to solve explicitly for the field operators as a function of time in QFT. We're usually far more interested in things like correlation functions or $S$-matrix elements.

Second off, this is usually not the way we think about Heisenberg picture. In Heisenberg picture, $$\langle A \rangle(t) = \langle \psi | A(t) | \psi \rangle.$$ So to evaluate an expectation value $\langle A \rangle(t)$, we need two pieces of information: the initial state $|\psi \rangle$ and the initial operator $A(0)$. Usually, we think of $|\psi \rangle$ as specifying the initial condition, and $A(0)$ as specifying what operator we're talking about. So there is no freedom in $A(0)$, if you change it you're talking about a physically different operator.

As a simple example, you can show that in Heisenberg picture, the operators $x(t)$ and $p(t)$ rotate into each other, because classically the simple harmonic oscillator rotates in phase space. If you decided to define a new position operator $x'(t) = x(t + \pi /2)$ it would satisfy all the equations, but it won't mean the position anymore, it would in fact be measuring the momentum.

As another example, you could define a new field operator $\phi'(\mathbf{x}, t) = \phi(\mathbf{x}, t+1)$. It will satisfy all the required relations, but it just won't mean the same thing: rather than give the field value at time $t$ it will give what the field value will be at time $t+1$.

  • $\begingroup$ Whoa! Now it's so obvious, thank you! phi(x,0) is not uniquely determined, of course, because you could just have chosen phi(x,1) as your initial condition - and it is consistent with commutators. I do understand that (as in QM with x) phi(0) is a static quantity in the theory, that is why I was under the impression that it is especially important. $\endgroup$ – doublefelix Oct 27 '18 at 19:59

An additional note: It is easy to get confused at the meaning of the field operator. The field operator in QFT is not the value of the field. It is the hermitian operator that corresponds to the fact that you can measure the field. Particular arrangements of the field are encoded in states, not the value of the operator.

The analogy is in regular quantum mechanics. You have states corresponding to every possible position, $|x=a\rangle$, and you have a position operator that measures the position, $\hat{X}|a\rangle = a|a\rangle$. There's a different state for every possible position, but there's only one position operator.

In a simple QFT, there is a state for every possible arrangement of the field $|\varphi\rangle$, and a field operator that measures the field in a particular state, $\phi(x)|\varphi\rangle = \varphi(x)|\varphi\rangle$. There's an infinite number of states, but only one field operator.

With this in mind it should not be so surprising that the initial value of the field operator is uniquely determined. Saying that $\phi(p,t) = a_p e^{-iE_pt} +a_p^†e^{iE_p t}$ is the analogue of saying that $\hat{X}(t) = ae^{-i\omega t} +a^† e^{i\omega t}$ for a regular QM harmonic oscillator. Both statements are true by definition of the operators $a,a^†$ and the canonical commutation relations. And both statements don't tell you anything about any specific position or field arrangement or state.

  • $\begingroup$ In other words, the quantum field solution (the operator $\varphi$) does not determine and is not determined with the initial state occupation numbers. $\endgroup$ – Vladimir Kalitvianski Oct 27 '18 at 15:56
  • $\begingroup$ Right, I'm aware that just as x was in QM, phi(x,t=0) is a static operator in the theory (I had written this in my question). And based on knzhou's answer, it is actually not true that the initial field is uniquely determined! However the analogy to the form of the x operator in the harmonic oscillator was helpful to notice, thank you! $\endgroup$ – doublefelix Oct 27 '18 at 20:03

Work in momentum space, where the infinite oscillator modes decouple, so $$ [a^\dagger (\vec k), a(\vec p)]=\delta^3 (\vec p -\vec k), \\ [a^\dagger (\vec k), a^\dagger (\vec p)]=0=[a (\vec k), a(\vec p)], $$ and so on, where $a(\vec p)\equiv \int d^3x \exp(-i \vec p\cdot \vec x ) \bigl (E(p)\phi(\vec x,0)+i \partial_t \phi(\vec x,0)\bigr )/\sqrt{(2\pi)^3 2E(p)}$, etc.

In momentum space, the K-G equation is an ODE, $$ \bigl (\partial_t^2 +\vec p ^2 +m^2\bigr ) {\tilde \phi}(\vec p ,t)=0. $$ The most general classical solution is then, $$ {\tilde \varphi}(\vec p ,t) = c(\vec p) e^{-iE(p)t} +c^* (\vec p) e^{iE(p)t}, $$ for $E=\sqrt{\vec p^2 +m^2}$, and arbitrary complex functions c. Setting t =0 provides the arbitrary IC.

The corresponding general quantum solution is, likewise, $$ {\tilde \phi}(\vec p ,t) = \hat c(\vec p) e^{-iE(p)t} +\hat c^\dagger (\vec p) e^{iE(p)t}. $$ Setting t=0 provides a hitherto arbitrary IC.

Fourier transform suitably adapting normalizations in the definition, $$\phi(\vec{x},t)=\int \bigl (\hat c(\vec{p})e^{-ip\cdot x}+\hat c ^{\dagger}(\vec{p})e^{ip\cdot x}\bigr ) \frac{d^3\vec{p}}{\sqrt{(2\pi)^3 2E(p)}},$$ and apply the canonical commutators $$ i[\partial_t \phi(\vec x), \phi (\vec y)]=\delta^3(\vec x-\vec y), ... $$ ...etc, to peg down those for the $\hat c , \hat c ^\dagger$, which are the very ones obeyed above by the $a, a^\dagger$. (For cringing detail, see chapter 12 of Bjorken & Drell, vII.)

That is to say, in drab monotony, your "feeling" is unwarranted, and the extraordinarily powerful field commutations do peg down $$ \tilde{\phi}(\vec p,0)= \frac{a(\vec p)+ a^\dagger (\vec p) }{\sqrt{(2\pi)^3 2E(p)}} ~, $$ uniquely, up to the phase of a, which is labile in the commutation relations: Each oscillator merely rotates rigidly in phase space ($a, a^\dagger$), just like a classical point, so, then, semi classically, which underlies canonical quantization and makes the classical-quantum correspondence so natural.


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