Free relativistic particle (wave function)

[bjogae]

Homework Statement

The hamiltonian of a free relativistic particle moving along the x-axis is taken to be H=\sqrt{p^2c^2+m^2c^4} where p is the momentum operator. If the state of the wave function at time t=0 is described by the wave function \psi_0(x) what is the wave function at time t>0 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 \psi(x)=\frac{1}{\sqrt{2\pi}} \int_k \phi(k) e^{i k x}
does this mean that \psi_0(x)=\frac{1}{\sqrt{2\pi}} \int_k \phi_0(k) e^{i k x}
and how do i know what \phi_0(k) is?

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

 

[gabbagabbahey]

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

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

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

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

 

[bjogae]

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

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

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

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

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

Then i solve the time dependent as following:
H\phi_0 = E \phi_0 \, where H=\sqrt{p^2c^2+m^2c^4}Is it so, that:
\phi(p,\,t)= A(t) \phi_0(p) \
which leads to
\Phi(p,\,t) = \phi_0(p) e^{-i{E t/\hbar}} \,
and then
\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)

 

[bjogae]

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

 

[gabbagabbahey]

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

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
i \hbar{\partial \over \partial t} \phi(p,\,t) = \hat H \phi(p,\,t)
and for
\hat H=\sqrt{p^2c^2+m^2c^4}
it gives
\phi(p,\,t) = \phi(p,\,0)e^{-i\sqrt{p^2c^2+m^2c^4}t/\hbar}
which would lead to (by the previous reasoning)
\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)
is this the answer?

 

[gabbagabbahey]

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

Good

which would lead to (by the previous reasoning)
\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)
is this the answer?

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

 

[bjogae]

Good



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

one more try

\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)
which shold be correct if
k=\frac{p}{\hbar}

 

[gabbagabbahey]

Looks good to me!