Thursday 25 June 2015

A particle has the wavefunction `Psi(x)=A(1-x^2)` for `|x|

The probability that a particle with the wavefunction `Psi(x)` will be found in some region `R` is  `int_R |Psi(x)|^2 dx.` In our problem it is


`int_(-oo)^0 |Psi(x)|^2 dx =int_(-1)^0 |Psi(x)|^2 dx =int_(-1)^0 (A(1-x^2))^2 dx.`


We can compute this integral in terms of `A` and can even find `A` from the normalization condition  `int_(-oo)^oo |Psi(x)|^2 dx = 1,` but it is not necessary for this particular problem.


Because this wavefunction is even, its square is also...

The probability that a particle with the wavefunction `Psi(x)` will be found in some region `R` is  `int_R |Psi(x)|^2 dx.` In our problem it is


`int_(-oo)^0 |Psi(x)|^2 dx =int_(-1)^0 |Psi(x)|^2 dx =int_(-1)^0 (A(1-x^2))^2 dx.`


We can compute this integral in terms of `A` and can even find `A` from the normalization condition  `int_(-oo)^oo |Psi(x)|^2 dx = 1,` but it is not necessary for this particular problem.


Because this wavefunction is even, its square is also even and the integral over the left semiaxis is the same as over the right semiaxis. Together these equal integrals give `1,` thus each of them is equal to `1/2.` This is the answer: the probability the particle will be found in the region `xlt0`  is `1/2.`


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