2 particles in a 1-dimensional box

[nemisis42]

If there are 2 particles of mass m1 and m2 in a one dimensional box of width a, I'm trying to find 1)what values will be found if the energy is measured, and with what probability these values will take occur. and 2) what is the probability of finding particle 1 with mass m1 in the interval (0,a/2) (all of this is at time t=0) (the particles are not symmetric)The wave equation is:


$$\Psi$$ (X1,X2,0)=(3$$\Phi$$₇(X1)*$$\Phi$$₆(X2)+8$$\Phi$$₃(X1)*$$\Phi$$₂(X2))/(sqrt(73))

I ended up with energy E=(((n7)^2/(m1))+((n6)^2/(m2)))*$$\hbar$$^2*$$\pi$$^2/(2*a^2)+(((n3)^2/(m1))+((n2)^2/(m2)))*$$\hbar$$^2*$$\pi$$^2/(2*a^2))

With (9/73) chance for E_7,6 and (64/73) chance for E_3,2

Would anybody be able to tell me if what I have looks correct(and point me in the right direction if its not) and tell me where to start with the probability of finding particle 1 in the interval (0,a/2). I did change the values from the original equation. I'm really just interested in the principal behind this.

 

[azntoon]

Yes, your answer looks correct. The probability of finding particle 1 in the interval (0,a/2) is given by the following formula: P(X1 in (0,a/2)) = |\int_0^{a/2}\Psi (X1,X2,0)|^2dX1 where \Psi (X1,X2,0) is the wave equation you have written above.