# Field theory in non-relativistic QM

Hello fellow physicists!
I'm currently trying to learn some QFT and the reader gives an introduction by expressing the non-relativistic hamiltonian with integral and creation, destruction operators.

Later he writes:

|Psi, t > = \int d3x1...3xn Psi(x1, ..., xn; t) a+(x1) ... |0>

And i´m not really sure how to think about this. Isn't |Psi, t > the wavevector of Psi(x1, ..., xn; t) ?
Please give me some handwaving or mental words about why this is true..

A. Neumaier
|Psi, t > = \int d3x1...3xn Psi(x1, ..., xn; t) a+(x1) ... |0>

Isn't |Psi, t > the wave vector of Psi(x1, ..., xn; t) ?:

$$|x_1,...,x_n>:=a^*(x_1)...a^*(x_n)|0>$$ is a basis vector,
$$\Psi(x_1,...,x_n,t)$$ is the wave function, and
$$\Psi(t)=|\Psi,t>$$ is the wave vector.

In first quantized notation, the wave function is regarded as a wave vector in a function space. Then $$|x_1,...x_n>:=\delta_{x_1}\otimes...\delta_{x_n}$$, where $$\delta_z$$ is the delta function with support at z. Then your formula reduces to a trivial identity.

This is a functional version of the representation of a vector v in 3-space as
$$v=(v_1,v_2,v_3)^T = v_1|1>+v_2|2>+v_3|3>,$$
where |k> is the k-th unit vector.