A Eigenstates of the Klein Gordon Field Operator

[Jonathan Apps]

So I've got the field operator for a real, spin-0 field in quantum field theory. Usual expansion in terms of ladder operators and plane wave solutions you'll get in any QFT textbook. What are its eigenstates (ie states of definite field)? It would be similar to the states of definite A-field for quantized EM radiation but without the unnecessary complexity of having vector fields with polarization etc. The states of definite particle number are obviously NOT eigenstates - the expectation value of the field for those is zero. Additionally, what are the eigenvalues? Cheers :-)

 

[JimWhoKnew]

So I've got the field operator for a real, spin-0 field in quantum field theory. Usual expansion in terms of ladder operators and plane wave solutions you'll get in any QFT textbook. What are its eigenstates (ie states of definite field)? It would be similar to the states of definite A-field for quantized EM radiation but without the unnecessary complexity of having vector fields with polarization etc. The states of definite particle number are obviously NOT eigenstates - the expectation value of the field for those is zero. Additionally, what are the eigenvalues? Cheers :-)

As a start, you can try working out a simpler case:
For a 1D harmonic oscillator, what are the eigenstates of $a+a^\dagger$ ?

 

[Jonathan Apps]

Got the eigenstates of $a$ - poisson distributed in particle number, so-called "coherent state". It's easy enough to prove $a^{\dagger}$ doesn't have any eigenstates on the space of positive integer particle number. Not sure about $a + a ^{\dagger}$.

 

[JimWhoKnew]

Got the eigenstates of $a$ - poisson distributed in particle number, so-called "coherent state". It's easy enough to prove $a^{\dagger}$ doesn't have any eigenstates on the space of positive integer particle number. Not sure about $a + a ^{\dagger}$.

Welcome to PF (I should have written it in #2).

$a+a^\dagger$ is proportional to $\hat X$ , so the eigenstates are just $|x\rangle$ . We have$$|x\rangle=\sum |n\rangle\langle n|x\rangle=\sum\phi^*_n(x)|n\rangle=\sum\phi_n(x)|n\rangle$$where $\phi_n(x)$ are the wavefunctions of the energy eigenstates.
Consider now the real scalar field in the Schrödinger picture:$$\hat \phi(\vec{x})=\int \frac{dk^3}{(2\pi)^3}\frac{e^{i\vec{k}\vec{x}}a_{\vec{k}}+e^{-i\vec{k}\vec{x}}a^\dagger _{\vec{k}}}{\sqrt{2\omega_{\vec{k}}}} \quad .$$The $\vec{k}=0$ component is just like $a+a^\dagger$ of the 1D HO, but this time the amplitude of the normal mode 0 (the spatially constant component) takes the role of $x$ above. For the other modes, the reality constraint couples the modes $\vec{k}$ and $-\vec{k}$ , so it is less obvious. Assume that $\phi(\vec{x})$ is an arbitrary real Fourier-transformable function. We want to find the corresponding eigenstate $|\phi(\vec{x})\rangle$ such that$$\hat \phi(\vec{x})|\phi(\vec{x})\rangle=\phi(\vec{x})|\phi(\vec{x})\rangle \quad .$$A nice derivation of the solution can be found at stackexchange.

 

[JimWhoKnew]

A nice derivation of the solution can be found at stackexchange.

The referenced derivation (stackexchange) uses the continuum expansion, but relies on the discrete commutation relations $[a_k,a_{k'}^\dagger]=\delta_{k,k'}$ rather than $[a_k,a_{k'}^\dagger]=\delta(k-k')$ . This should be interpreted as some box-like (or lattice) regularization.

 

[Jonathan Apps]

Thanks! My Ph.D was on the effective action for various fields on finite spaces - I like the discrete stuff :-) Pretty sure the universe has a brick wall at the end of it somewhere anyway.

 

[JimWhoKnew]

For the other modes, the reality constraint couples the modes $\vec{k}$ and $-\vec{k}$ , so it is less obvious.

We can gain some more insight into the relation between the free field and the 1D harmonic oscillator:
For simplicity, I'll consider the field in one spatial dimension.
The field is real, so we can expand it with $\cos kx$ and $\sin kx$ . Define:$$b_k:=\frac{a_k+a_{-k}}{\sqrt{2}} \quad, \quad c_k:=\frac{a_k-a_{-k}}{\sqrt{2}} \quad.$$In this expansion the coupling between $k$ and $(-k)$ is removed and each "new" mode is truly normal (independent) to the others. This means that the amplitude of each mode (which is a real quantity) behaves like a 1D HO. Sounds abstract, but that is just what we would have expected from a guitar's string if it had only one polarization. We get$$[b_k,b_k^\dagger]=[c_k,c_k^\dagger]=1$$(see remark in #5) and$$[b_k,c_k]=[b_k,c_k^\dagger]=0 \quad .$$Define further$$\frac{b_k+b_k^\dagger}{\sqrt{2}} :=\alpha_k \hat X_{\cos,k} \quad, \quad \frac{c_k-c_k^\dagger}{i\sqrt{2}}:=\beta_k \hat P_{\sin,k}$$where $\alpha_k$ , $\beta_k$ are real proportionality factors. Absorbing $\omega_k$ and normalization factors into these proportionality factors, we have:$$\hat\phi(x)=\sum_{k\ge0} \left( \alpha_k \hat X_{\cos,k} \cos kx+\beta_k \hat P_{\sin,k} \sin kx \right)\quad.$$For a real function $f(x)$ expanded by$$f(x)=\sum_{k\ge0} \left(f_{\cos,k}\cos kx+f_{\sin.k}\sin kx \right) \quad,$$we have$$\hat\phi(x)|f\rangle=f(x)|f\rangle$$where$$|f\rangle=\prod_{k\ge0}|x_{\cos,k}=f_{\cos,k}/\alpha_k\rangle|p_{\sin,k}=f_{\sin,k}/\beta_k\rangle$$(which should be the same as the one discussed in post #4 when $f(x)=\phi(x)$ ). The 1D harmonic oscillator states $|x\rangle$ and $|p\rangle$ correspond to the "generalized" wavefunctions $\delta(x-x')$ and $\exp[ix(p-p')]$ respectively, so $|f\rangle$ should also be regarded as a "generalized" state ( $\langle f|f'\rangle=\delta^\infty (f-f')$ ).

 

[JimWhoKnew]

Pretty sure the universe has a brick wall at the end of it somewhere anyway.

In the down direction the universe is known to have stacked turtles.

 

[Jonathan Apps]

Are those turtles Neumann or Dirichlet?