Hello!
The probability density is the square of the absolute value of a wave function, in our case it is `P_d(x)=|Psi(x)|^2=c^2 e^(-2x/L)` for positive x's and `c^2 e^(2x/L)` for negative x's.
The integral over the real axis of a probability density must be `1` (the total probability). From this statement we can find `c:`
`int_(RR) p_d(x) dx = 2 int_0^(+oo) c^2 e^(-2x/L) dx = -2c^2 L/2e^(-2x/L) |_(x=0)^(+oo) = c^2 L=1.`
So really `c = 1/sqrt(L) approx...
Hello!
The probability density is the square of the absolute value of a wave function, in our case it is `P_d(x)=|Psi(x)|^2=c^2 e^(-2x/L)` for positive x's and `c^2 e^(2x/L)` for negative x's.
The integral over the real axis of a probability density must be `1` (the total probability). From this statement we can find `c:`
`int_(RR) p_d(x) dx = 2 int_0^(+oo) c^2 e^(-2x/L) dx = -2c^2 L/2e^(-2x/L) |_(x=0)^(+oo) = c^2 L=1.`
So really `c = 1/sqrt(L) approx 0.707.` (b)
(c) the probability of finding the particle within 1 mm of the origin is the integral of the probability density from `-1` to `1.` It is evidently
`2 int_0^1 1/L e^(-2x/L) dx = -2/L L/2e^(-2x/L) |_(x=0)^1 = 1-e^(-2/L) = 1-e^(-1) approx 0.632,`
which is really 63.2%.
(a), (d) look at the graph at https://www.desmos.com/calculator/mmija615t1
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