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 $\hbar$s 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 $\hbar$s 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!