Consider a particle whose wavefunction at some fixed time t is represented by

-A if -2≤x≤0
ψ(x)= A if 0≤x≤3
0 otherwise
(a) What is the normalization constant A?

(b) What is the probability of finding the particle at a position x≤0?

(c) What is the expected value of position x?

(d) What is the expected value of momentum p?

|psi(x)|^2 integrated from minus to plus infinity is 5 |A|^2, this has to be equal to 1, therefore you can choose A as 1/sqrt(5) (you are free to multiply this by any phase factor

exp(i theta)).
The probability of the aprticle being at negative x is the integral from minus infinity to zero of |psi(x)|^2 dx, which is 2/5.

Expacted position: Integrate

x |psi(x)|^2 dx from minus to plus infinity, you get <x> = 1/2.

And, of course <p> = 0, because the fact that psi(x) is (or can be chosen to be a) real function of x, implies that the modulus squared of the momentum space wavefunction

psi(p) is an even function of p. To see this, note that:

psi(x) = 1/sqrt(2 pi hbar) Integral from minus to plus infinity of
dp psi(p) exp(i p x/hbar)

Take the complex conjugate of this:

psi*(x) = 1/sqrt(2 pi hbar) Integral from minus to plus infinity of
dp psi*(p) exp(-i p x/hbar) =

1/sqrt(2 pi hbar) Integral from minus to plus infinity of
dp psi*(-p) exp(i p x/hbar)

Since psi*(x) = psi(x), this means that

psi*(-p) = psi(p)

therefore

|psi(-p)|^2 = psi*(-p)psi(-p) =

psi(p)psi*(p) = |psi(p)|^2

Answers to q6 and 8 ?

Please q6 and q8 !

prob 3 C

3C is 3/2

6A is pi/2 and pi/4
6B is 1/2 and 1/2
Looking for 6C and 6D
8A is 1/2, 1/2, 0, 0
8B is 0, 0, 1/2, 1/2
8C is 1, 1, 0, 0
Looking for 8D and 8E
8F is No

Thank you.

And waiting for the rest answers

5 plz....

7a 1/2 7b 1/2

Problem 9 plzz