How to interpret wave function as a matrix

Black Integra
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As we all know, we can write schrodinger equation in Linear algebraic form.
Also, Dirac had introduced his matrix mechanics.
And we can write any linear operator as matrix.
and so on...

How can we write wave function as matrix?
What is the dimension of this matrix?
 
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Black Integra said:
As we all know, we can write schrodinger equation in Linear algebraic form.
Also, Dirac had introduced his matrix mechanics.
And we can write any linear operator as matrix.
and so on...

How can we write wave function as matrix?
What is the dimension of this matrix?
\infty\times 1. You can write \psi=\sum_{k=1}^\infty a_k e_k, where the e_k are basis vectors. The "matrix" of components of \psi relative to the ordered basis \langle e_i\rangle_{k=1}^\infty is \begin{pmatrix}a_1\\ a_2\\ \vdots\end{pmatrix}
 
Thank you for your reply.
but if the dimension is inf how can I apply this to, for example, http://en.wikipedia.org/wiki/Dirac_equation" , where the dimension of its operator is 4x4.
 
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Black Integra said:
Thank you for your reply.
but if the dimension is inf how can I apply this to, for example, http://en.wikipedia.org/wiki/Dirac_equation" , where the dimension of its operator is 4x4.
You wouldn't. If you're talking about a theory in which a solution to the classical Dirac equation is considered an (\mathbb R^4-valued) wavefunction, then you could write \psi=\sum_{\mu=0}^3\sum_{k=1}^\infty a^\mu_k e_\mu u_k where the e_\mu are the standard basis vectors for the space of 4×1 matrices, and the u_k are members of an orthonormal basis for the space of square-integrable functions. I suppose you could arrange the a^\mu_k into another infinite column matrix if you want to, but I think that would be a rather pointless thing to do.

Anyway, the theory that interprets a classical Dirac field as a wavefunction is obsolete. The Dirac equation is still used in quantum field theory, but now the Dirac field (the solution to the equation) isn't a wavefunction. It's an operator.
 
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Wow, that's very interesting.
Thanks for those information!
 
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