I Wave function using the time dependent Schrodinger equation

Click For Summary
To determine the wavefunction ψ(x, t) of a free particle at time t using the time-dependent Schrödinger equation, one can express it as ψ(x, t) = exp(-iHt/ħ)ψ(x, 0), where H is the Hamiltonian operator. For a free particle, the Hamiltonian is given by H = p²/(2m), leading to the propagator G(x, x', t) = ⟨x|exp(-iHt/ħ)|x'⟩. This propagator can be evaluated using momentum eigenstates, resulting in G(x, x', t) = ∫ dp ⟨x|exp[-ip²t/(2mħ)]|p⟩⟨p|x'⟩. Calculating the propagator directly can be complex, but it is essential for obtaining the wavefunction at time t.
ThiagoSantos
Messages
2
Reaction score
1
Given a wavefunction ψ(x, 0) of a free particle at initial time t=0, I need to write the general expression of the function at time t. I used a Fourier transform of ψ(x, t) in terms of ψ(p, t), but, i don't understand how to use green's functions and the time dependent schrodinger equation to get my answer. What's the relationtship between them?
 
Physics news on Phys.org
The time-dependent Schrödinger equation reads
$$\mathrm{i} \hbar \partial_t \psi(x,t)=\hat{H} \psi(x,t).$$
Let's assume for simplicity that
$$\hat{H}=\frac{\hat{p}^2}{2m} + V(\hat{x}),$$
i.e., that ##\hat{H}## is not explicitly time dependent. Then the formal solution of the equation above is
$$\psi(x,t)=\exp \left (-\frac{\mathrm{i} \hat{H} t}{\hbar} \right) \psi(x,0).$$
This you can write in the form
$$\psi(x,t)=\int_{\mathbb{R}} \mathrm{d} x' \left \langle x \left |\exp \left (-\frac{\mathrm{i} \hat{H} t}{\hbar} \right) \right| x' \right \rangle \psi(x',0)= \int_{\mathbb{R}} \mathrm{d} x' G(x,x',t) \psi(x',0),$$
i.e., the propator is
$$G(x,x',t)=\left \langle x \left |\exp \left (-\frac{\mathrm{i} \hat{H} t}{\hbar} \right) \right| x' \right \rangle.$$
Usually it's of course difficult to really calculate the propagator.

For a free particle, where ##\hat{H}=\hat{p}^2/(2m)## you can use the momentum eigenstates to evaluate it:
$$G(x,x',t)=\int_{\mathbb{R}} \mathrm{d} p \langle x |\exp[-\mathrm{i} \hat{p}^2 t/(2m \hbar)]|p \rangle \langle p|x' \rangle.$$
 
For the quantum state ##|l,m\rangle= |2,0\rangle## the z-component of angular momentum is zero and ##|L^2|=6 \hbar^2##. According to uncertainty it is impossible to determine the values of ##L_x, L_y, L_z## simultaneously. However, we know that ##L_x## and ## L_y##, like ##L_z##, get the values ##(-2,-1,0,1,2) \hbar##. In other words, for the state ##|2,0\rangle## we have ##\vec{L}=(L_x, L_y,0)## with ##L_x## and ## L_y## one of the values ##(-2,-1,0,1,2) \hbar##. But none of these...

Similar threads

  • · Replies 39 ·
2
Replies
39
Views
1K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 6 ·
Replies
6
Views
2K
  • · Replies 21 ·
Replies
21
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K
Replies
1
Views
1K
  • · Replies 17 ·
Replies
17
Views
3K
Replies
3
Views
2K