Quantum Field Configurations and Wavefunctions

[rocdoc]

Could anyone explain what a quantum field configuration is, and any relation this concept may have to the idea of a wavefunction?

Perhaps for a scalar, quantum field?

 

[hilbert2]

A field configuration is a function $A(x)$ that gives the strength of the field $A$ at a point $x$. It's the same thing in classical and quantum mechanics. A "wave function" for a field $A$ would be a functional $\psi$ which produces a complex number $\psi\left[A(x)\right]$ when it acts on a field configuration $A(x)$. It's the equivalent to the wave function $\psi(x_1 ,x_2 ,x_3 ,\dots)$ of a particle system with degrees of freedom $x_i$. But usually quantum fields are not described with this kind of functionals, it's done with abstract state vectors and particle creation and annihilation operators.

 

[Peter Morgan]

Another way to talk about a "configuration of a quantum field" would be to talk about a "quantum state". We can write the vacuum state, the most common starting point for constructing other quantum states, as a complex-valued function of any quantum field operator. For a self-adjoint operator $\hat A$, for example, $\langle 0|\hat A|0\rangle$ tells us the expected value associated with that operator if we measure it. Higher powers, such as $\langle 0|\hat A^n|0\rangle$, tell us the higher moments of the probability distribution associated with $\hat A$ in the vacuum state.
If we have a quantum field operator $\hat B$, then we can construct a different state, $\frac{\langle 0|\hat B^\dagger\hat A\hat B|0\rangle}{\langle 0|\hat B^\dagger\hat B|0\rangle}$; for this to exist, $\langle 0|\hat B^\dagger\hat B|0\rangle$ has to exist.
All that's exactly as you'd expect for ordinary quantum mechanics. In quantum mechanics we can construct a wave function by introducing a basis of position vectors, $|x\rangle$, so that for any vector $|U\rangle, |\psi\rangle$, ... we can construct a wave function $U(x)=\langle x|U\rangle, \psi(x)=\langle x|\psi\rangle$, ... . A position basis in quantum mechanics is usually said to be "improper", because $\langle x|x\rangle$ is not finite, so that sometimes we have to be careful; for quantum field theory, for any $n$ we can construct $B(x_1, ..., x_n)=\langle 0|\hat\psi(x_n)^\dagger\cdots\hat\psi(x_1)^\dagger\hat B|0\rangle$, using the lowest-level operator-valued distribution $\hat\psi(x)$ as a building block, but we have to be even more careful when we use this construction than in the ordinary QM case (for interacting fields, very much so, indeed we don't know how to be careful enough in any simple way).
Although one can discuss the quantities $B(x_1, ..., x_n)$, there is a significant difference from the QM case, where $|x\rangle$ and $|y\rangle$ are orthogonal when $x$ and $y$ are space-like separated: $\langle 0|\hat\psi(x)^\dagger\hat\psi(y)|0\rangle$ is always non-zero, so $\hat\psi(x)|0\rangle$ and $\hat\psi(y)|0\rangle$ are not orthogonal.

 

[rocdoc]

I was thinking of asking ' How is a field strength function $\mathbf {A(x)}$ related to the field operator?' .

Material on pg. 39 of Bailin and Love, see reference 1, seems relevant to this. The relevant material I now quote,

'We may define eigenstates of $\hat \phi(t,\mathbf x)$ denoted by $| \phi(\mathbf x), t\rangle$ such that
$$ \hat \phi(t,\mathbf x) | \phi(\mathbf x), t\rangle= \phi(\mathbf x) | \phi(\mathbf x), t\rangle~~~~~~~~~~~~~~~~~~~~~(4.1)$$
'.Here $\hat \phi(t,\mathbf x)$ is a scalar quantum field.

$\phi(\mathbf x)$ seems to be a field strength function in the sense of hilbert2's post, see post1 of this thread.
Note how similar EQ(4.1) is to eigenvalue equations in non-relativistic quantum mechanics
$$ \hat O | \Psi(\mathbf x, t)\rangle= \lambda |\Psi (\mathbf x, t)\rangle$$
In non-relativistic quantum mechanics one would tend to think of $\lambda$ as a property of the eigenstate ( I would ), so it appears that, the strength of the quantum field should be thought of as associated with whichever of it's states, it is in.

References

1) D. Bailin and A.Love , Introduction to Gauge Field Theory, IOP Publishing Ltd, 1986.

 

[Peter Morgan]

In non-relativistic quantum mechanics one would tend to think of $\lambda$ as a property of the eigenstate ( I would ), so it appears that, the strength of the quantum field should be thought of as associated with whichever of it's states, it is in.

"the strength of the quantum field", though you haven't defined what you mean, is just one observable that one might measure in a given state. A state, say $\rho$, gives an expected value for $\hat\phi(t,\mathbf{x})$, $\rho\bigl(\hat\phi(t,\mathbf{x})\bigr)$ (with the vacuum state being $\rho_0\bigl(\hat\phi(t,\mathbf{x})\bigr)=\langle 0|\hat\phi(t,\mathbf{x})|0\rangle$.) One has to be careful, however, because the second moment, $\rho\bigl(\hat\phi(t,\mathbf{x})^2\bigr)$, is infinite in all states in the vacuum sector.
It's important to note that to determine what the quantum state is, it's not enough to measure just the local strength of the field, it's necessary to measure all the n-point correlation functions (at least that's necessary for a Wightman field).