Free relativistic particle (wave function)

[bjogae]

Homework Statement

The hamiltonian of a free relativistic particle moving along the x-axis is taken to be [tex]H=\sqrt{p^2c^2+m^2c^4}[/tex] where [tex]p[/tex] is the momentum operator. If the state of the wave function at time [tex]t=0[/tex] is described by the wave function [tex]\psi_0(x)[/tex] what is the wave function at time [tex]t>0[/tex] Hint: solve the time-dependent Schrödinger equation in momentum space. The answer can be left in the form of an integral.

Homework Equations

The Attempt at a Solution

In momentum space [tex] \psi(x)=\frac{1}{\sqrt{2\pi}} \int_k \phi(k) e^{i k x} [/tex]
does this mean that [tex] \psi_0(x)=\frac{1}{\sqrt{2\pi}} \int_k \phi_0(k) e^{i k x}[/tex]
and how do i know what [tex] \phi_0(k)[/tex] is?

Is the right answer something in form of [tex] \psi_0(x)=\frac{1}{\sqrt{2\pi}} \int_k \phi_0(k) e^{i k x}e^{i E t/\hbar}[/tex] where i just kind of write down the usual derivation of the time-dependent schrödinger equation?

 

[gabbagabbahey]

You can get [itex]\phi_0(k)[/itex] from [itex]\psi_0(x)[/itex] by taking its Fourier transform.

Your final answer will involve a double integral, and come from

[tex]\psi(x,t)=\frac{1}{\sqrt{2\pi}}\oint\phi(k,t)e^{ikx}dx[/tex]

You will need to determine [itex]\phi(k,t)[/itex] from [itex]\phi_0(k)[/itex] by solving the time-dependent Schroedinger equation.

 

[bjogae]

You can get [itex]\phi_0(k)[/itex] from [itex]\psi_0(x)[/itex] by taking its Fourier transform.

So i get [tex]\phi_0(p)=\frac{1}{\sqrt{2\pi}} \int_k \psi_0(x) e^{i p x}[/tex] ?

Your final answer will involve a double integral, and come from

[tex]\psi(x,t)=\frac{1}{\sqrt{2\pi}}\oint\phi(k,t)e^{ikx}dx[/tex]

You will need to determine [itex]\phi(k,t)[/itex] from [itex]\phi_0(k)[/itex] by solving the time-dependent Schroedinger equation.

Then i solve the time dependent as following:
[tex]H\phi_0 = E \phi_0 \,[/tex] where [tex]H=\sqrt{p^2c^2+m^2c^4} [/tex]Is it so, that:
[tex] \phi(p,\,t)= A(t) \phi_0(p) \ [/tex]
which leads to
[tex] \Phi(p,\,t) = \phi_0(p) e^{-i{E t/\hbar}} \,[/tex]
and then
[tex]\psi(x,t)=\frac{1}{\sqrt{2\pi}}\int dpe^{-i{E t/\hbar}}e^{ip x} \frac{1}{\sqrt{2\pi}} (\int \psi_0(x) e^{i p x} dx)[/tex]

 

[bjogae]

Is this correct? It seems to be ok to me.

 

[gabbagabbahey]

Then i solve the time dependent as following:
[tex]H\phi_0 = E \phi_0 \,[/tex] where [tex]H=\sqrt{p^2c^2+m^2c^4} [/tex]

No, that's the time-independent Schroedinger equation...

 

[bjogae]

No, that's the time-independent Schroedinger equation...

Of course. So the time dependent looks like
[tex]i \hbar{\partial \over \partial t} \phi(p,\,t) = \hat H \phi(p,\,t)[/tex]
and for
[tex] \hat H=\sqrt{p^2c^2+m^2c^4} [/tex]
it gives
[tex]\phi(p,\,t) = \phi(p,\,0)e^{-i\sqrt{p^2c^2+m^2c^4}t/\hbar}[/tex]
which would lead to (by the previous reasoning)
[tex]\psi(x,t)=\frac{1}{2\pi}\int dpe^{-i\sqrt{p^2c^2+m^2c^4}t/\hbar}e^{ip x} (\int \psi_0(x) e^{i p x} dx)[/tex]
is this the answer?

 

[gabbagabbahey]

Of course. So the time dependent looks like
[tex]i \hbar{\partial \over \partial t} \phi(p,\,t) = \hat H \phi(p,\,t)[/tex]
and for
[tex] \hat H=\sqrt{p^2c^2+m^2c^4} [/tex]
it gives
[tex]\phi(p,\,t) = \phi(p,\,0)e^{-i\sqrt{p^2c^2+m^2c^4}t/\hbar}[/tex]

Good

which would lead to (by the previous reasoning)
[tex]\psi(x,t)=\frac{1}{2\pi}\int dpe^{-i\sqrt{p^2c^2+m^2c^4}t/\hbar}e^{ip x} (\int \psi_0(x) e^{i p x} dx)[/tex]
is this the answer?

Be careful, [itex]k=\frac{p}{\hbar}[/itex]...so when you rewrite your Fourier transforms in terms of [itex]p[/itex] instead of [itex]k[/itex], you should get some [itex]\hbar[/itex]s in there somewhere.

 

[bjogae]

Good



Be careful, [itex]k=\frac{p}{\hbar}[/itex]...so when you rewrite your Fourier transforms in terms of [itex]p[/itex] instead of [itex]k[/itex], you should get some [itex]\hbar[/itex]s in there somewhere.

one more try

[tex]\psi(x,t)=\frac{1}{2\pi\hbar}\int dpe^{-i\sqrt{p^2c^2+m^2c^4}t/\\hbar}e^{ip x/\hbar} (\int \psi_0(x) e^{i p x/\hbar} dx)[/tex]
which shold be correct if
[tex] k=\frac{p}{\hbar}[/tex]

 

[gabbagabbahey]

Looks good to me!