Calculation of probabilities in QM

[facenian]

I posted this question in the homework page but I got no answers, so will try here.
Given the wave function of a (spinless) particle I need to expres in terms of \psi(\vec{r}) the probability for simultaneous measurements of X y P_z to yield :

x_1 \leq x \leq x_2
p_z \geq 0

I got the result:
\int_{-\infty}^{\infty}dz\int_{-\infty}^{\infty}dy\int_{x_1}^{x_2}dx \int_{-\infty}^{\infty}dp_x\int_{-\infty}^{\infty}dp_y\int_0^{\infty}dp_z <\vec{p}|\vec{r}>\psi(\vec{r})<\psi|\vec{p}>

To get this I evaluated the expression <\psi|P_2P_1|\psi> where P_1 and P_2 are the proyectors:
P_1=\int_{-\infty}^{\infty}dz\int_{-\infty}^{\infty}dy\int_{x_1}^{x_2}dx|x,y,z><x,y,z|
P_2=\int_{-\infty}^{\infty}dp_x\int_{-\infty}^{\infty}dp_y\int_0^{\infty}dp_z|p_x,p_y,p_z><p_x,p_y,p_z|

I need to know two things: 1) is my result correct? 2) in case it is correct, is there any other more simple or concrete answer?

 

[Avodyne]

1) Yes.
2) Yes.

 

[facenian]

1) Yes.
2) Yes.

Thank you Avodyne, could you be so kind as to explain a little how to get an easier or more concrete answer?

 

[Avodyne]

Well, notice that if it wasn't for the pz question, you could do everything with the position-space wave function. So you really only need to Fourier transform on z. Your version has a lot of extra Fourier-transforming and untransforming in x and y.

 

[javierR]

You can go further by writing <p|r> and <Psi|p> in a function form like you did with <x|Psi>. Generally, you want to put things in a single representation, like the position space, so you might ask yourself how far you can go to do that.

If you have a free particle, e.g., then Psi(r)=Psi_x Psi_y Psi_z, and similarly for the momentum space wavefunction. Can you see how doing the above steps would make things much more simplified?

 

[facenian]

Thank you both. I think what javier says is simple to write explictly the expresions for <p|r> and <psi|p>
in space representation but that would only complicate even more the final expression(sorry if I misunderstood you)
So will try Avodyne's approach and I will very much appretiate any comments on this :
Since we need to express the final result in terms of \psi(\vec{r}) only, we will change momentarily from X,Y,Z to the C.S.C.O X,Y,P_z. The desired probabilty is &lt;\psi|P_n|\psi&gt;

where
P_n=\int_{-\infty}^\infty dy\int_{x_1}^{x_2}dx\int_0^\infty dp_z |x,y,p_z&gt;&lt;x,y,p_z|
&lt;\psi|P_n|\psi&gt;=\int_{-\infty}^\infty dy \int_{x_1}^{x_2}dx \int_0^\infty dp_z |&lt;x,y,p_z|\psi&gt;|^2
where
&lt;x,y,p_z|\psi&gt;=\int&lt;x,y,p_z|\vec{r&#039;}&gt;&lt;\vec{r&#039;}|\psi&gt;d^3r&#039;=\int&lt;x,y,p_z|x&#039;,y&#039;,z&#039;&gt;&lt;x&#039;,y&#039;,z&#039;|\psi&gt;d^3r&#039;
=\int\delta(x-x&#039;)\delta(y-y&#039;)&lt;p_z|z&#039;&gt;\psi(\vec{r&#039;})d^3r&#039;=\frac{1}{\sqrt{2\pi\hbar}}\int_{-\infty}^\infty dz&#039; \psi(x,y,z&#039;)e^{-ip_z z&#039;/\hbar}
is this what you meant Avodyne?

 

[Avodyne]

Yep!

 

[facenian]

Thank you Avodyne! I really appreate your help because a have no other means to deal with this matters.