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I am wondering about the meaning of the scalar product and its relation with the wave function. In the Hilbert space, the scalar product is defined as

$$\langle \phi \rvert \psi \rangle = \int \phi^*\psi dx.$$

This defines the $\rvert \psi \rangle$ as a vector from the Hilbert vector space. Now, the wave function is defined from the scalar product

$$\psi (x,t) = \langle x \rvert \psi \rangle = \int x^*\psi dx.$$

First question: Is the last equality true? If so, which function $\psi$ has to be integrated? Isn't it a kind of recursive definition?

Let's now assume a Fock space. Let's expand the wave-vector in this basis, i.e., $\rvert \psi \rangle = \sum_m a_m\rvert m \rangle$. Now, the probability to find the system in a state $\rvert k \rangle$ is given by $P(k|\psi) = |a_k|^2 = |\langle k \rvert \psi \rangle|^2$. This makes sense since $\langle k \rvert \psi \rangle$ is a wave function.

Question 2: How is this wave function? Could I write it as $\psi(m,t) = \langle m \rvert \psi \rangle = \int m^*\psi dm$? I guess that somehow this integral should actually be a sum.

Thank you very much.

Aesku
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I think your formula is confused. The wavefunction is $$ \psi(x) = \langle x\vert \psi\rangle = \int \delta(x-x') \psi(x')\,dx' $$ where $\delta(x-x')= \langle x'\vert x\rangle$ is the wavefuction of the position eigenfunction $\vert x\rangle$ in the position eigenfunction basis. This not what you have written with the "$x$" operator. For the ocillator basis we have $$ \langle m\vert \psi\rangle= \int \varphi^*_m(x) \psi(x)dx $$ where $\varphi_m(x)$ is the oscillator wavefunction.

mike stone
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