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In regular quantum mechanics of particles, I have the Schrodinger evolution picture for a general state

$$ i\hbar \frac{d}{dt} \left|\psi(t)\right> = \hat H \left|\psi(t)\right> $$

then we take the inner product with respect to $ \left< x\right| $ to obtain the equation in the position representation.

$$ i\hbar \frac{d}{dt} \psi(x,t) = -\frac{\hbar^2}{2m}\nabla^2\psi(x,t) - V(x)\psi(x,t) $$

where $$ \psi(x,t) = \left< x |\psi(t)\right> $$ and $$ \left< x |\hat H|\psi(t)\right> = -\frac{\hbar^2}{2m}\nabla^2\psi(x,t) - V(x)\psi(x,t) $$

In the case of quantum fields, the position and momentum and demoted from operator status to simply paramaters and the field operator (here for KG real field) is given by

$$ \hat \psi(\vec x) = \int \frac{d^3p}{(2\pi)^3} \frac{1}{\sqrt{2\omega_p}} \Bigl[\hat a(\vec p) e^{i\vec k. \vec x} + \hat a^\dagger(\vec p) e^{-i\vec k. \vec x} \Bigr] $$ in the Schrodinger picture. The states are generated from vacuum by the operator

$$ \hat \psi(x) \left|0\right> = \int \frac{d^3p}{(2\pi)^3}\frac{1}{2E_p}e^{-i\vec p.\vec x} \left|\vec p\right>$$ Now what exactly are these states, are they eigenstates of some operator (seems like the field operator) or quite arbitrary ? In what representation is the field operator given here ?

EDIT :

I think I understood the first part of the question, these states are the eigen-states of Hamiltonian of the field. But the question about the choice of basis of the operator still remains.

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  • $\begingroup$ It's funny you ask this. See the following recent post and especially the comments physics.stackexchange.com/questions/94554/… $\endgroup$ – joshphysics Jan 21 '14 at 6:30
  • $\begingroup$ @joshphysics : I am not sure if I am being misunderstood, but let me put it this way. From linear algebra I know that an abstract linear operator is given in by $$ [F]_{\left|a\right>} = \left< a \right| \hat F\left|a'\right> $$ where $ \left|a\right> $ is a choice of representation. In that sense, in what representation is the field operator given here. It can't be in position or momentum since they are no more operators. $\endgroup$ – user35952 Jan 21 '14 at 9:04
  • $\begingroup$ Hmmm. Now I'm really confused. You should be thinking of $x$ as a label similar to the label $i$ on the position operator components $X_i$ in the quantum mechanics of a particle. For each $x$, the expression $\hat \phi(x)$ is not an operator being written in some specific basis. $\endgroup$ – joshphysics Jan 21 '14 at 9:16
  • $\begingroup$ @joshphysics : I understand that the field has infinite degrees of freedom (grossly speaking infinite continuous components). My question is not concerned with that, it is rather about the choice of basis for the field operator (like in the case of quantum mechanics). $\endgroup$ – user35952 Jan 21 '14 at 9:24
  • $\begingroup$ Are you asking if there is a "continuous basis" of field operator eigenstates that is analogous to the position basis in ordinary quantum of a single-particle? $\endgroup$ – joshphysics Jan 21 '14 at 9:29
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Some points :

i) Quantum mechanics "is" Quantum Field Theory (QFT) with zero spatial dimension (stricly speaking this concerns the relation between bosonic QFT and the harmonic oscillator in Quantum mechanics)

ii) "Heinsenberg" representation is (at $99.99\%$) used in Quantum Field theory

iii) "Schrodinger" representation is (at $99.99\%$) not used in Quantum Field theory

Consider, in quantum field theory, a scalar field theory for simplicity, the operators depend on parameters $x,t$, one write them $\Phi(x,t)$. The equivalent in Quantum mechanics, taking zero spatial dimension, is an operator $\Phi(t)$, so you recognize the Heisenberg representation, where the operators depend on time (for instance $X(t)$ or $P(t)$).

Now, to complete the comparison for the effects of this operators on the vaccuum, we may think to the quantum harmonic oscillator in Quantum mechanics. So, in QFT, you may think of the state $\Phi(x,t)|0\rangle$, in the same way, that, in the case of the harmonic oscillator, you may think about states $X(t)|0\rangle$ or $P(t)|0\rangle$

Now, the "Schrodinger" formalism, in QFT, is not used, but we may guess what it is, thanks to the i) prescription. A "wave-function" in QFT, is simply : $\psi(\phi;x,t)$ , and if you go to zero spatial dimension, you get the usual "wave-function" $\psi(\phi;t)$, where $\phi$ could be $x$ or $p$, for instance.

[EDIT 1]

So $\psi(\phi;x,t)$ must be understood as $<\phi|\psi(x,t)>$, as usual, taking zero spatial dimension, and we recover quantum mechanics

This kind of formalism ("wave-function" formalism) is practically not used in QFT.

Finally , $\Phi(x,t)$, is an operator which does not commute with the hamiltonian (here written without the zero energy) $H = \sum\limits_{\vec p} E_p ~a^\dagger(\vec p)a(\vec p)$, and the commutator is not proportionnal to $\Phi(x,t)$ (just calculate it), so $\Phi(x,t)|0\rangle$ is not an eigenstate for the hamiltonian.

[EDIT 2]

The Heisenberg representation is the most enlightening representation, because Heisenberg discovers Quantum Mechanics at Heligoland, in june $1925$, under the the name of Matrix Mechanics,without any use or concept of "wavefunction" (which was "invented" at the end of $1925$ by Schrodinger). The original paper of Heinsenberg (traduced in English) is here. A paper to help you understand the original paper is here. You have a blog article here. In the original paper of Heinsenberg, you will never see "wavefunctions", you will only see quantities (see formulae $7,8$ page $12$) like $\mathcal U (n,n-\alpha)$, which are "matrices".

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  • $\begingroup$ Thanks a lot. This clarifies a lot of things. But firstly, what are the advantages of Heisenberg picture in QFT ? Secondly, the from prescription (i) you have suggested the wavefunction is $ \psi(\phi(x),t) = \left< \Phi (x) |\psi (t)\right> $ ? But in that sense you treat field as state and not an operator. $\endgroup$ – user35952 Jan 21 '14 at 14:19
  • $\begingroup$ For the first part of your question, I edited the answer [EDIT 2], and you will see which are the origins of the Quantum theory. For the second part, I have deleted my sentence about the "more readable way" because it is disturbing, and I edited the answer too [EDIT 1]. $\endgroup$ – Trimok Jan 21 '14 at 17:59
  • $\begingroup$ See also this previous answer $\endgroup$ – Trimok Jan 21 '14 at 18:07

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