Understanding the Role of k in the Wavefunction of Quantum Systems

[Arman777]

I am reading a textbook quantum physics by stephen gasiorowicz. And he defines a wavefunction in this form,

$$Ψ(x,t)=\int_{-∞}^{∞}A(k)e^{i:(kx-ωt)}dk$$

I did not understand why its a function of $k$ or why even we are taking integral with respect to $k$ ? Is $k$ actually means momentum somehow ? Or what's the importance of $k$ since we the wavefunction is actually a function of $x$ and $t$.

 

[PeroK]

I am reading a textbook quantum physics by stephen gasiorowicz. And he defines a wavefunction in this form,

$$Ψ(x,t)=\int_{-∞}^{∞}A(k)e^{i:(kx-ωt)}dk$$

I did not understand why its a function of $k$ or why even we are taking integral with respect to $k$ ? Is $k$ actually means momentum somehow ? Or what's the importance of $k$ since we the wavefunction is actually a function of $x$ and $t$.

It's not a function of $k$. $k$ is a dummy variable over which the integration is done.

The set of functions $e^{i(kx - wt)}$, $k \in \mathbb{R}$, is an uncountable (or continuous) basis for the space of wavefunctions.

Mathematically you integrate over a "density" function $A(k)$, which essentially gives you the weighing for each eigenfunction in your expansion.

You should compare what you have with the discrete case:
$$\Psi(x, t) = \sum_{n = 1}^{\infty}A_n f_n(x, t)$$
Your equation is the continuous version of this.

The comparison is: sum over $n$ (discrete case) against integral over $k$ (continuous case).

 

[Arman777]

$k$ is not a wavenumber right ? I understand what you said I guess but its still confusing for me somehow

 

[PeroK]

$k$ is not a wavenumber right ? I understand what you said I guess but its still confusing for me somehow

$k$ represents the momentum for that particular eigenfunction. Whether you also want to call that a wavenumber? I would go with the terminology in your book.

I think it isn't that easy to grasp. The important thing is that physically you always have a range of $k$, hence a wave-packet.

 

[Arman777]

I agree that it is not easy. I am watching some MIT videos to understand it. And later on I ll get back to you

 

[Arman777]

I guess I understand it. Its the adding the all possible wavefunctions in momentum space and getting some new wavefunction.

 

[Nugatory]

I guess I understand it. Its the adding the all possible wavefunctions in momentum space and getting some new wavefunction.

Yep, although it might be better to think of $\Psi(x,t)$ as the wavefunction of some physical system (not a "new" wavefunction) and the integral is a way of rewriting it in a more computationally tractable form, a sum of momentum eigenfunctions.