Question

why do we multiply wave functions of different particles in a system rather than adding ? And is this question correct in the fist place?

Answer 1

I'm not sure in what context you are asking this question. The Schroedinger equation is a linear equation and therefore the general solution is a sum of solutions. ie\[\Psi (x)= \sum_{i=1}^{n} \alpha _{_{i}}\psi _{i} (x)\]. As an application suppose you had two particles in different states and these particle interacted via some force. As I recall to solve this problem ie find the final state due to this interaction you would assume the final state is a linear combination of the two wave functions\[\Psi _{f} = \alpha \psi _{1} + \beta \psi _{2}\] . and apply the Schroedinger equation for their interaction to find the mixing ( ie alpha and beta ) of the two states that would occur. then if you want the probability distribution you find \[\Psi(x) \Psi ^{*}(x)\] . Hope I didn't get myself into hot water.

Answer 2

I have a system with two particles 1 and 2 and two states a and b.
then if the two particles are distinguishable we can write two wave functions one as psi(1)sub a multiplied by psi(2) sub b and psi(2) sub a multiplied by psi(1) sub b. my question is why did we multiply?

Answer 3

In this case is the potential can be written as\[ V _{T}(x _{1}...z _{_{2}})=V(x _{1},..)+V(..,z _{2}) \] That is the potential for each particle depends only on its own coordinates ie they are non interacting, then the wave function for this system can be written as\[\Psi =\psi _{1} (x _{a},...)\psi _{2} (..,z _{b}) \] is a solution of the Schroedinger equation. Hope my notation is clear .

Answer 4

How did you just add the potentials ?

Answer 5

I'm not sure I understand three posts up, but I think this question is just related to the joint probability of two independent events. This represents the probability amplitude that particle 1 is in state a1.
\[\psi_{1} (a) |a_1 > \]
Here are the wave functions for the individual particles.
\[|\psi_{1}> = \psi_{1} (a) |a_1 > +\psi_{1} (b) |b_1 > \]
\[|\psi_{2}> = \psi_{2} (a) |a_2 > +\psi_{2} (b) |b_2 > \]
\[|\psi_{1}>|\psi_{2}> = ( \psi_{1} (a) |a_1 > +\psi_{1} (b) |b_1 > )(\psi_{2} (a) |a_2 > +\psi_{2} (b) |b_2 >) \]

\[= \psi_{1} (a)\psi_{2} (a)|a_1 >|a_2 > + \psi_{1} (a)\psi_{2} (b)|a_1 >|b_2 > + ...\]

So the probability amplitude that both particles are in their respective "a" state is
\[\psi_{1} (a)\psi_{2} (a)\]
And the probability for both particles in the "a" state is thus
\[\psi_{1}^* (a)\psi_{1} (a)\psi_{2}^* (a)\psi_{2} (a) = P(a_1)P(a_2)\]
So by multiplying the individual particle wave functions, we get that the probability for both particles to be in state "a" is the same as the product of the probabilities that the individual particles would be in the "a" state. This is the rule for independent events.

Answer 6

Regarding the issue of the added potential I meant if the potential can be written as the sum of two potentials each only affecting one of the particles then the solution to the Schroedinger equation can be written as the product of the wavefunctions of each particle.