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In the canonical quantization of QFT we talk about:

  • states representing a field.
  • field operators.

The quantum Klein-Gordon equation is expressed in terms of the field φ. Is φ (in the equation) the state or the operator? And whichever the answer is, can you find a way to express the same equation in terms of the other possibility?

Qmechanic
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TrentKent6
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    $\varphi$ is an operator and the states of the free theory can be obtained by acting with this operator on the vacuum state. – Silas Oct 25 '23 at 13:46
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    Fields are not states and states are not fields. The KG equation governs the time-evolution of the field operators (in the Heisenberg picture). – Tobias Fünke Oct 25 '23 at 14:06
  • @Tobias Fünke I talked about the state representing a field. – TrentKent6 Oct 25 '23 at 14:20
  • @Silas Do we obtain proper states from $\phi |0\rangle$? For the scalar field (solution to the KG equation), $\phi = \int d^{3}\vec{p} a^{\dagger}_{p}e^{-ipx} + \text{h.c.}$ up to a normalisation - what of the exponential factor that just remains? – ShKol Oct 25 '23 at 14:31
  • @ShKol. "proper"? phase? what of it? – Cosmas Zachos Oct 25 '23 at 17:48
  • @CosmasZachos I am very new to QFT and I have just learnt that we define 1 particle states as $a_{p}^{\dagger}|0\rangle$ - since there's an infinity of such $a_{p}^{\dagger}$ along with the phase (which is also integrated upon) here, I was just asking what kind of states these represent. – ShKol Oct 25 '23 at 18:39
  • They are superpositions of one particle states, obviously. – Cosmas Zachos Oct 25 '23 at 18:46
  • @Silas you are wrong: states of the free theory can be obtained by acting with the creation operators on the vacuum state. the field operator can be composed of many creation and many annihilation operators, and most importantly, does not only act on the vacuum, but on any generic state of the Fock space – TrentKent6 Oct 25 '23 at 21:02
  • @TrentKent6 of coarse you can act with an operator on a different state. Acting with an annihilation operator on the vacuum annihilates it per definition so there is no difference between acting with $\varphi$ or a superposition of creation operators on the vacuum. – Silas Oct 26 '23 at 13:31

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The free fields are a repackaging of an infinity of de-coupled (normal mode) oscillator operators $a_{\vec p}$, $$ \phi(\vec x,t)= \int\!\!\frac{d^3p}{(2\pi)^3 \sqrt{2\sqrt{m^2+|\vec p|^2}}} ~ e^{it \sqrt{m^2+|\vec p|^2} -\vec p\cdot \vec x } ~a^\dagger _{\vec p} +\mathrm {h.c.} $$ in a Lorentz-invariant manner (which you are not asking about), with coefficients specifically chosen in the packaging so these fields obey the K-G equation, manifestly.

The states of the theory consist of infinite products of powers of such oscillators acting on the vacuum, $|0\rangle$, of any and all types (momenta, $ \vec p$): the Fock space. In general, they have no clue about the K-G equation, as they lack the variables it controls, anymore than oil paints have any clue about the Mona Lisa. The K-G equation relates and circumscribes the field operators.

Nevertheless, if you construct states by operating on the vacuum with such quantum fields which depend on t and $\vec x$s, those states will naturally reflect the above arrangement, Lorentz invariance, symmetries, and K-G time-evolution outlined above, in general in a complicated manner of your design. For example, the state $\phi |0\rangle$ will trivially satisfy the K-G equation. Why do you care?

Cosmas Zachos
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