Canonical quantization and the parallels between QFT and QM [closed]

Question

Questions similar to this have been asked multiple times on stack exchange about quantum field theory and the role of operators and the similarity of the theory to regular quantum mechanics. However, most of them do not answer the questions here or sometimes just claim that QFT is a theory to determine scattering amplitudes without explaining QFT's theoretical justification or motivation. I have not studied QFT and its applications in detail yet and so I am unfamiliar with a lot of concepts. So, my question is mostly about the theoretical framework of canonical quantization. In order to make the question readable, I will divide the question into 3 parts.

The role of operators in canonical quantization:

I understand that a classical field can be quantized by promoting the field and momentum to operators and imposing commutation relations. This is analogy with quantum mechanics. In quantum mechanics, we assume that the value of a measurement can only be one of the eigenvalues of the operator associated to the measurement and the eigenvectors are the possible states. In QFT, the hilbert space of quantum mechanics is upgraded to the fock space $ F = H^1 \bigoplus H^2 \bigoplus H^3 ...$ . Now, in the case of the klein gordon field, the field operator is hermitian. If the principles of quantum mechanics carry over to QFT, what observable would the field operator correspond to? This is even more confusing because the field operator assigns a different operator to each point of space. As far as I understand, we don't expand the states into the eigen basis of the operators because the operator formalism is more useful and easy to work with. But in theory, is it possible to expand a state such as $a^\dagger _{p_n}...a^\dagger_{p_1} |0 \rangle$ in the eigenbasis of the field operator (at a certain point) and what does it mean?. Are there any other observables (hermitian operators) in QFT?

Probabilistic interpretation of QFT:

Assuming again that the reasoning from quantum mechanics can be directly applied to quantum field theory, how is the probabilistic nature of quantum mechanics realised in QFT? I am not knowledgeable enough in QFT to ask specific questions on this topic but is it possible to have a superposition of states such as $| \psi \rangle = c_1 a^\dagger _{p_1} |0 \rangle +c_2a^\dagger_{p_2}|0 \rangle$ and does it carry the ususal probabilistic interpretation of $c_1^2$ and $c_2^2$ being the probabilities of the state collapsing to the corresponding states? On another note, why is it that we don't introduce QFT with the measurement problem as we do for QM?

Fundamental definition of states:

When I look at an expression like $a^\dagger _p |0 \rangle$, I think of exciting a single harmonic oscillator in momentum space. However, what field is the particle fundamentally on oscillation of? I understand how the creation operator works mathematically but contrary to the claim "Particles are excitations of fields", it seems to me that when a particle of momentum $p$ is created as in the state above,it is neither an oscillation of the field operator or the momentum operator. So isn't the statement incorrect? In addition, there are no position or momentum bases in QFT. In that case, is a state an abstract mathematical object with no spatial (like $\langle x| \psi \rangle$ in quantum mechanics) or momentum space representation? In other words, is a particle not an excitation in a physical or mathematical field?

Answer 1

I think you are confused by trying to think of QFT as a many-body extension of QM. Quantum Field theory is not quantizing a many-particle theory. It is quantising a field. The field may be $\hat z(x,y)$ the quantum operator correspondingt to the vertical displacement of a drum skin at position $(x,y)$. It may be the electromagnetic field -- or something else. If it is the drum skin then there are position eigenstates $|y\rangle$ where $y$ is shorthand for the function $y(x,y)$ and such that $\hat y(x,y) |y\rangle = y(x,y) |y\rangle$. The "position" is the vertical position of the drumskin at $(x,y)$, not the position of a "particle". In QED the "position" eigenstates will be eigenstates of the $A_\mu$ (mod gauge transforms) field, or in QED "momentum" space represntation we have eigenstates of the $E=-\dot A$ field.

In some special cases involving translation invariance there may be particle-like energy eigenstates with energy $E=E( k)$ where $k$ labels a translation normal mode of the classical system, but these are special cases (although imporant ones). In the case of a drumskin the energy eigenstates are called "phonons" and they behave a bit like quantized lumps of sound, but they are occupation numbers of normal modes of the drumskin dymnamical system an not "particles'. These particle-like states emerge only after solving the quantum mechanics of the drumskin, they are not inputs.

In the general case, for a expample electromagnetism in a curved space, there are no particle-like eigenstates and Fock space is not much use. What does count is how some external and (x,y,z) localized detector respons to the local quantum field operator.

Answer 2

The underlying guiding principle that I want to emphasize in this answer is that a quantum field theory is a quantum mechanical theory, i.e., a theory built within the framework of quantum mechanics. Or, in other words, the set of all quantum field theories is a subset of all quantum mechanical theories. So, all the (basic) principles of quantum mechanics necessarily hold true for a quantum field theory. To further belabor the point, a quantum field theory is simply a quantum mechanical theory of a field. Just like a classical field theory is a classical mechanical theory of a field. In particle physics, we are interested in only a specific class of quantum field theories, namely, the quantum field theories that respect the Poincare symmetry. However, again, this specification does not mean that such a quantum field theory ceases to be a theory built within the framework of quantum mechanics. OK, so, let's get to your specific questions which I will formulate in my own words (and order).

What observable does the field operator in a quantum field theory correspond to?

• Well, first of all, a field operator in a quantum field theory does not need to be Hermitian, for example, you can have a quantum field theory of a complex scalar field. In which case, the field operator would not be a Hermitian operator. As you'd expect from the rules of quantum mechanics, such a field operator would not correspond to an observable.
• Furthermore, even when a field operator is Hermitian, it need not correspond to an observable for the simple reason that it might not be gauge invariant. This is nothing peculiar to a quantum field theory. A gauge-dependent quantity is simply not observable -- either classically or quantum-mechanically. You can't measure the classical field that is the four-potential $A^\mu$ in the classical field theory of electromagnetism (and you can't measure it in the quantum field theory of electromagnetism). This simply reflects the fact that in formulating a gauge theory, we artificially introduce fictitious degrees of freedom in our theory that are not there in the physical system that we are describing. Of course, such an introduction of fictitious degrees of freedom is done in a systematic manner so that we can keep track of this fictitiousness and ultimately get rid of it in order to construct such quantities that are gauge-invariant (i.e., independent of the fictitious degrees of freedom) and are thus observables. This, again, is true in both classical and quantum field theories. So, gauge-invariant quantities constructed out of $A^\mu$ are observables, e.g., the field-strength tensor or the Wilson loops.
• Finally, when a field operator is Hermitian and is gauge-invariant, it is an observable.$^\dagger$ The observable that it corresponds to is simply the value or the strength of the said field.$^\ddagger$ You can measure it via measuring its effect on a probe -- just like you'd measure the electric field with a pith ball (keep in mind that the quantum field theory of electromagnetism is formulated in terms of $A^\mu$, not $F^{\mu\nu}$. I am just giving an example of how you'd measure the strength of a field).

Can you expand the state of the system in the eigenbasis of the field operator(s)?

Yes, absolutely. Of course, the eigensubspaces of the field operator at a given point in space are degenerate over the full Hilbert space of the quantum field theory. So you need to consider the shared eigenstates of all the field operators (one at each point in space). The non-degenerate eigenstate(s) that you'd correspondingly get is the eigenstate(s) with an eigenvalue(s) that represents the field configuration over the whole of space. For simplicity, I will stick to the scalar field theory. In such a case, the field-configuration over the whole of space would be represented by the function $\phi:\mathbb{R}^3\to\mathbb{R}$. Correspondingly, you can write the eigenvalue equation(s) as

$$\hat{\phi}(x)\vert\psi_{\phi}\rangle = \phi(x)\vert\psi_{\phi}\rangle,\forall x\in\mathbb{R}^3, \phi : \mathbb{R}^3\to\mathbb{R}$$

Here, the $\hat{\phi}(x)$ is the field operator at $x\in\mathbb{R}^3$. You should really treat the $x$ here as a label and not as an input to a function. The state $\vert\psi_\phi\rangle$ is the eigenstate of the field such that its eigenvalue is a specific field-configuration $\phi:\mathbb{R}^3\to\mathbb{R}$. Now, you can write a generic state as

$$\vert \Psi\rangle = \int \delta \phi \vert \psi_{\phi}\rangle\langle \psi_{\phi}\vert \Psi\rangle $$

The wavefunction(al) of such a state in the field-basis is $\Psi[\phi]=\langle \psi_\phi\vert \Psi\rangle$ where the square-brackets are to emphasize that $\phi:\mathbb{R}^3\to\mathbb{R}$ is a function over the whole of space and not a particular value. You can see it as the uncountably infinite version of a wavefunction such as $\psi(n,l,m)$ that we are used to in quantum mechanics. In other words, you can see it as the probability amplitude corresponding to a given configuration of the field over the whole of space, this is exactly what a given $\phi:\mathbb{R}^3\to \mathbb{R}$ specifies (i.e., it specifies as to what value of the field do we attach to a given point in space. If you change the value of the field at any given point in space, you are talking about a new field configuration corresponding to a different mapping $\phi$). When mathematicians are not around, you can imagine it as $\Psi(\phi(x_1), \phi(x_2),...)$ where the "$\dots$" run over all the uncountably many infinite points in $\mathbb{R}^3$.

Position/Momentum Basis in QFT

It is indeed possible to write down the position/momentum basis for the Hilbert space of a quantum field theory. Of course, there are internal degrees of freedom (such as the spin or isospin) that are not captured in the probability amplitudes over the positions/momenta -- both in quantum mechanics and in quantum field theory. Having said that, to the extent that it is possible to write down the position/momentum basis in quantum mechanics, it is possible to write down the same in quantum field theory. In particular, such a basis would be a complete basis only for a spinless scalar field. Of course, you can construct such a basis for other quantum field theories in the same manner but it won't be a complete basis (or, if you prefer, it would be degenerate w.r.t. the spin degrees of freedom, etc.). OK, so how to construct a position/momentum basis?

As you already mentioned, the Hilbert space of a scalar field theory is the same as the Fock space.$^{\ast}$ For each $n-$particle sector of the Fock space, we already know what the position basis for each such sector is. It is $S_n\equiv\{\vert x_i\rangle\vert x_i\in\mathbb{R}^3, i=1,\dots,n\}$. Now, since the Fock space is the vector sum of all such sectors, the position-basis for the Fock-space is simply $\cup_nS_n$. Similarly, for the momentum basis. As you can imagine, this is an incredibly cumbersome way of keeping track of things -- especially when you have multiple fields where the Hilbert space of the QFT would be the tensor-product of the Fock spaces of each of the fields. Furthermore, there is a more fundamental reason as to why this is not a good way of doing things, see my footnote with the $\ast$ mark.

What field is $a^\dagger _p\vert 0\rangle$ an excitation of?

I am not sure if this answer will satisfy you but it is the excitation of the field whose creation operator $a^\dagger_p$ is.

¯\(ツ)/¯

In particular, if you have multiple fields in a quantum field theory then you'll have multiple different creation operators. You can imagine them as being labeled by the different field-names, e.g., $a^\dagger_p{^{\phi_1}}$, $a^\dagger_p{^{\phi_2}}$, etc. If it helps, you can treat the ladder operators as being constructed out of the field operators so that the fact that a given creation/annihilation operator is tied to a particular field is more explicit.

What happens to the probabilistic interpretation in QFT?

I think that the answer so far already makes it clear that nothing happens to the probabilistic interpretation in quantum field theory. A quantum field theory is just as probabilistic as any other theory in quantum mechanics. A superposition of two eigenstates of the field operator does mean that if you measure the field, it has the Born rule probabilities of being found in either of the two eigenstates. As to why we don't talk about the measurement problem in quantum field theory, it is simply because the framework of quantum field theory doesn't change anything about the measurement problem -- so people usually talk about it in the simplest possible settings of quantum mechanics.

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Footnotes

$^\dagger$ This is not exactly true but it's good enough to address the question as it is asked. In particular, as to why it is not exactly true has to do with the fact that it's not true that every Hermitian operator needs to be an obserable -- either in quantum mechanics or in quantum field theory. In the Hilbert space formulation of quantum mechanics, one implicitly assumes that all Hermitian operators ought to correspond to observables but there is no reason as to why it should be so (what is needed of necessity is only the converse, i.e., every observable needs to be a Hermitian operator). There are other ways to formulate quantum mechanics in which you posit upfront as to which operators are observables, e.g., the $C^\star-$algebraic formulation or the Von Neumann algebraic formulation. See this question for more details.

$^\ddagger$ Notice that what you practically end up measuring is not the strength of a field at a point but rather some average of fields in a given vicinity. For example, if you measure the electric field with pith balls then you'd be measuring $\int_V d^3 xf(x)\vec{E}(x)$ where $f(x)$ is some "smearing" function and the integral is over the volume $V$ of the pith ball.

$^\ast$ I should mention that it is not really true that the Hilbert space of a quantum field theory is the same as the Fock space (or the tensor product of Fock spaces). In particular, it is only true for free quantum field theories where the relation between fields and particles is rather direct. In an interacting field theory, the notion of a particle is much more subtle and you have to treat the field as the fundamental object and treat a quantum field theory in its own right as a quantum theory of a field -- not as just another way of talking about multi-particle quantum mechanics. This is what is emphasized in the answer by @mikestone. In particular, a particle-state in an interacting field theory is such states that are the in-coming or out-going states of a scattering experiment that are constructed via the LSZ procedure. The Hilbert space of such states is smaller than the Hilbert space of all states in the quantum field theory. This smaller subset of the full Hilbert space of the quantum field theory can still be seen as a Fock space (or the tensor product of Fock spaces) but you can't see the full Hilbert space of the quantum field theory (which does contribute to the scattering amplitudes) as such Fock space(s). See, this answer for more details on this point.