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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.

Ali Esquembre Kucukalic
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1 Answers1

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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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  • 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
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    $|x\rangle$ 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
  • Hey, great explanation. Just one doubt which has lasted long. Wavefunction is the inner product of ket psi and basis vectors x. That is its an inner product so is a scalar. But function is vector, we know that. So can a scale of a larger space ( larger hibert space) be defined as a vector in another space( function space). – Shashaank Jul 10 '19 at 14:46
  • 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
  • On the position basis. So in that step you get scalars ( according to the inner product definition of your space in which ket psi and position kets are). After projection on to the position basis you are redefining inner product for functions which would be scalars in thatbspace but become vectors once you give them this construct and define and inner product for them..... That's what I have understood. Is that right – Shashaank Jul 10 '19 at 14:53
  • $\langle x\vert \psi\rangle$ for fixed $x$ is a component of the vector $\vert \psi \rangle$. Like all components, it is a number, i.e. a scalar. You can consider the set of all components of $\vert \psi \rangle$ as being a vector if you like (I do not), but a set is not the same thing as an element of the set. – mike stone Jul 10 '19 at 17:35
  • @mikestone ok. I understand. You mean that $\psi(x) $ at each x is a scalar but the complete set of these numbers is a vector which is the function $\psi(x) $. Right !But if that is so what is the difference between ket psi and wavefunction psi. Both are vectors. Are they in different spaces. And is my argument on the inner products above correct. For two kets you define the inner product as $\langle\phi\mid\psi\rangle$ without any reference to basis. But in you first equation itself $\delta(x - x') and \psi(x) $ are themselves position basis representations of ket x and ket psi. – Shashaank Jul 11 '19 at 17:16
  • You are then defining the innerproduct on these things( functions) through that integral. Isn't it. Is that right – Shashaank Jul 11 '19 at 17:17
  • if both ket psi and wavefunction psi are vectors in the hilbert space then what is the difference between them..... There should be a difference between wavefunction psi and ket psi... If both are vectors what is the difference. Shouldn't the difference be that the ket psi is an abstract vector in an abstract vector space. The wavefunction ( the set of all the function values at all x and not a single value of x which shall be scalar) is vector in another vector space, the function space.and you come a step down in that function space from that abstract vector space of ket psi in 2 operations – Shashaank Jul 11 '19 at 17:27
  • and you come a step down in that function space from that abstract vector space of ket psi in 2 operations the first being the projection on the x basis n then defining an inner product definition for the function ( the set of all values for x) – Shashaank Jul 11 '19 at 17:29