Field theory in non-relativistic QM

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SUMMARY

The discussion focuses on the interpretation of the non-relativistic Hamiltonian in quantum field theory (QFT), specifically how the state vector |Psi, t> is expressed in terms of creation and annihilation operators. Participants clarify that |Psi, t> represents the wave vector of the wave function Psi(x1, ..., xn; t), and they emphasize the relationship between the wave function and basis vectors in function space. The conversation highlights the equivalence between the functional representation of vectors in QFT and classical vector representation in three-dimensional space.

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

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

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

In first quantized notation, the wave function is regarded as a wave vector in a function space. Then [tex]|x_1,...x_n>:=\delta_{x_1}\otimes...\delta_{x_n}[/tex], where [tex]\delta_z[/tex] 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
[tex]v=(v_1,v_2,v_3)^T = v_1|1>+v_2|2>+v_3|3>,[/tex]
where |k> is the k-th unit vector.
 

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