# Is the scalar product also a wave function?

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.

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.
• This is a good answer, but just to nitpick, isn't $\delta(x-x')$ not a wavefunction (it isn't a function)? – Will May 11 '19 at 17:49
• $|x\rangle$ is the ket of definite $x$, which is not $x$. That is, if a particle is at $x=3$, its ket is $|3\rangle$, which corresponds to $\delta(x-3)$. – JEB May 11 '19 at 19:24
• Great answer! However, I am still wondering about something. Let's take now a time dependent state, so that $\psi(x,t) = \langle x,t \rvert \psi \rangle$. So, now we've got the vector $\rvert x,t \rangle$ that is related with $\rvert x \rangle$ by $\rvert x,t \rangle = e^{-iHt}\rvert x \rangle$. This leads obviously to $\psi(x,t) = e^{iHt}\psi(x)$. Now the question is, how can I compute a scalar product like $\langle x',t' \rvert x,t \rangle$? I mean, how can I write this scalar product as an integral, according to the definition of the scalar product? – Ali Esquembre Kucukalic May 12 '19 at 10:30
• In the integral $\psi(x')$ is the wavefunction. The ket has already been projected onto the the position basis and then put in the integral. The inner product construct of the space in which your ket psi is is different from this space. In that space it's just an inner product between 2 abstract vectors without reference to any basis. In you explanation you have given an extra construct to your wacefunction. First you project the ket psi and the position kets on the position basis. Then you get the wavefunction and the dirac delta functions respectively after you have projected them – Shashaank Jul 10 '19 at 14:50