How Is the Wavefunction Ψ(x) Derived from Momentum Space?

  • Thread starter Thread starter Eole
  • Start date Start date
  • Tags Tags
    Wavefunction
Eole
Messages
2
Reaction score
0
Here's a derivation of wavefunction of State Ψ in representations of coordinates and momentum
Ψ (x)=<x|Ψ >=<x|∫dp|p><p|Ψ >=∫dp<x|p><p|Ψ>=∫dp exp{ipx/h}Ψ(p)
Ψ (p)=<p|Ψ >=∫dx exp{-ipx/h}Ψ(x)


Ψ (x)=<x|Ψ >=<x|∫dp|p><p|Ψ >=∫dp<x|p><p|Ψ>=∫dp exp{ipx/h}Ψ(p)
i don't understand how ∫dp<x|p><p|Ψ> become ∫dp exp{ipx/h}Ψ(p)
Could you please tell me the drivation of this formula?

and another question is why Ψ (x) could be denoted as <x|Ψ >?
 
Physics news on Phys.org
I probably won't be able to help, but I'm sure that if I'm struggling to make sense of that, then other people are too. Some things just can't be written without proper equations. Maybe try writing the equation on Microsoft word's equation tool and attach the document.
 
Eole said:
Here's a derivation of wavefunction of State Ψ in representations of coordinates and momentum
Ψ (x)=<x|Ψ >=<x|∫dp|p><p|Ψ >=∫dp<x|p><p|Ψ>=∫dp exp{ipx/h}Ψ(p)
Ψ (p)=<p|Ψ >=∫dx exp{-ipx/h}Ψ(x)
Ψ (x)=<x|Ψ >=<x|∫dp|p><p|Ψ >=∫dp<x|p><p|Ψ>=∫dp exp{ipx/h}Ψ(p)
i don't understand how ∫dp<x|p><p|Ψ> become ∫dp exp{ipx/h}Ψ(p)
Could you please tell me the drivation of this formula?
and another question is why Ψ (x) could be denoted as <x|Ψ >?

It seems to me that you are asking why
<x|p>=exp{ipx/h}
but surely you must have this derivation in your notes? BTW, the above equation isn't actually quite right (needs a factor in order to normalize it) and, in my notes anyway, the 'h' is actually an 'h-bar'.

As regards your other question, I can only assume that it is very sloppy notation because I have never seen a state use the same symbol as the wave function (although I am still an undergraduate so my experience is rather limited).
 
Eole said:
Ψ (x)=<x|Ψ >=<x|∫dp|p><p|Ψ >=∫dp<x|p><p|Ψ>=∫dp exp{ipx/h}Ψ(p)
i don't understand how ∫dp<x|p><p|Ψ> become ∫dp exp{ipx/h}Ψ(p)
Could you please tell me the drivation of this formula?
and another question is why Ψ (x) could be denoted as <x|Ψ >?

\psi(x) = \langle x | \psi \rangle by definition. It is the component of |x> in the expansion of | \psi \rangle in the position basis.

∫dp<x|p><p|Ψ> = ∫dp exp{ipx/h}Ψ(p), because \langle x|p \rangle =\exp{ipx/\hbar} (apart from a constant factor) and <p|Ψ>=Ψ(p), again by definition.
 
Thread 'Need help understanding this figure on energy levels'
This figure is from "Introduction to Quantum Mechanics" by Griffiths (3rd edition). It is available to download. It is from page 142. I am hoping the usual people on this site will give me a hand understanding what is going on in the figure. After the equation (4.50) it says "It is customary to introduce the principal quantum number, ##n##, which simply orders the allowed energies, starting with 1 for the ground state. (see the figure)" I still don't understand the figure :( Here is...
Thread 'Understanding how to "tack on" the time wiggle factor'
The last problem I posted on QM made it into advanced homework help, that is why I am putting it here. I am sorry for any hassle imposed on the moderators by myself. Part (a) is quite easy. We get $$\sigma_1 = 2\lambda, \mathbf{v}_1 = \begin{pmatrix} 0 \\ 0 \\ 1 \end{pmatrix} \sigma_2 = \lambda, \mathbf{v}_2 = \begin{pmatrix} 1/\sqrt{2} \\ 1/\sqrt{2} \\ 0 \end{pmatrix} \sigma_3 = -\lambda, \mathbf{v}_3 = \begin{pmatrix} 1/\sqrt{2} \\ -1/\sqrt{2} \\ 0 \end{pmatrix} $$ There are two ways...
Back
Top