0
$\begingroup$

I understand that each mode has its own wavefunction. I would like to know, how do I calculate the wavefunction of a quantum field? Should I multiply or sum the wavefunctions of the individual modes?

$\endgroup$
2
$\begingroup$

Not sure what you mean by "wavefunction of a quantum field". Quantum fields involve an infinity of oscillator operators acting on the Fock vacuum. If you (over-)simplified the field as a 2 mode system, with subscripts 1 and 2, $$\phi = c_1 a^\dagger _1 + c_1^* a _1 + c_2 a^\dagger _2 + c_2^* a _2, $$ then the implied tensor-product structures would generate product Fock states $$ a_1^{\dagger k} a_2^{\dagger j} |0,0\rangle, $$ where I denote the empty vacuum with the vanishing occupation number of each mode, so a symmetrize tensor product of 1-2-states. If you absolutely had to consider space wave functions, you'd have a conventional product of wavefunctions, $$ \langle x_1| \langle x_2|n_1,n_2\rangle = \psi (x_1) \psi(x_2), $$ (real) Hermite functions whose arguments $x_1$ and $x_2$ have absolutely nothing to do with our spacetime, and merely keep track of the decoupled mode indices in notional spaces.

I fear you are actually asking a messier question, namely eigenfunctions of such field operators, as, e.g. this and this and this and this ... questions. The site is full of them. Again, the most direct answers revolve around states, not eigenfunctions.

However, having solved that problem, you may mimic a functional Schroedinger equation.

All of these structures involve products of wavefunctions.

$\endgroup$
2
$\begingroup$

Yes, you should multiply the wavefunctions of individual oscillators. If you have several independent systems, the wavefunction of the composite system is a product of the wavefunctions of individual systems (like a probability distribution of a system consisting of several independent variables is a product of individual probability distributions).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.