At time t = 0 a particle is represented by the wave function ψ (x,0) = A(x/a) , o ≤ x ≤ a, A(b-x)/(b-a) ,a ≤ x ≤ b, 0 , otherwise, where A, a, and b are (positive) constants. (a) Normalize ψ(that is, find A, in terms of a and b). (b) Sketch ψ(x,0) , as a function of x. (c) Where is the particle most

Question

At time t = 0 a particle is represented by the wave function

[latex] Ψ (x,0) =\begin{cases} A(x,0) , & 0\leq x\leq a, \ A(b-x)/(b-a) ,& a \leq x\leq b, \ 0,& otherwise,\end{cases} [/latex].

where A, a, and b are (positive) constants.

(a) Normalize Ψ(that is, find A, in terms of a and b).

(b) Sketch Ψ(x,0) , as a function of x.

(c) Where is the particle most likely to be found, at t = 0?.

(d) What is the probability of finding the particle to the left of a? Check your result in the limiting cases b = a and b= 2a.

(e) What is the expectation value of x?.

Accepted Answer

(a)

[latex] 1 = \frac{\left|A ight|^2 }{a^2} \int_{0}^{a}{x^2}dx + \frac{\left|A ight|^2 }{(b-a)^2} \int_{0}^{b} {(b-a)^2}dx = \left|A ight|^2 \left\{\frac{1}{a^2}\Bigl(\frac{x^3}{3}\Bigr)\mid ^a_0 + \frac{1}{(b-a)^2 } \biggl(-\frac{(b-x)^3}{3}\biggr)\mid ^b_a ight\}[/latex]

[latex] = \left|A ight|^2 \biggl[\frac{a}{3} + \frac{b-a}{3}\biggr] = \left|A ight|^2 \frac{b}{3}[/latex] ⇒ [latex] A = \sqrt{\frac{3}{b} }[/latex].

(b)

(c) At x = a.

(d)

[latex] P = \int_{0}^{a}{\left|Ψ ight|^2}dx = \frac{\left|A ight|^2 }{a^2} \int_{0}^{a}{x^2}dx = \left|A ight|^2 \frac{a}{3} = \frac{a}{b}[/latex], [latex] \begin{cases} P=1 & if b=a \P=1/2 & if b=2a \end{cases} [/latex].

(e)

[latex] \left\langle x ight angle = \int{x \left|\Psi ight|^2 } dx = \left|A ight| ^2 \left\{\frac{1}{a^2}\int_{0}^{a}{x^3}dx + \frac{1}{(b-a)^2} \int_{a}^{b}{x(b-x)^2} dx ight\} [/latex].

[latex] = \frac{3}{b} \left\{\frac{1}{a^2}\Bigl(\frac{x^4}{4}\Bigr) \mid ^a_0 + \frac{1}{(b-a)^2}\biggl(b^2\frac{x^2}{2} -2 b \frac{x^3}{3} + \frac{x^4}{4} \biggr) \mid ^b_a ight\} [/latex].

[latex] = \frac{3}{4b(b - a)^2} [a^2(b - a)^2 + 2b^4 - 8b^4/3 + b^4 - 2a^2b^2 + 8a^3b/3 - a^4] [/latex].

[latex] = \frac{3}{4b(b - a)^2} \biggl(\frac{b^4}{3} - a^2 b^2 + \frac{2}{3}a^3b \biggr) = \frac{1}{4(b - a)^2}(b^3 - 3a^2b + 2a^3) = \frac{2a + b}{4} [/latex] .

Question 1.4: At time t = 0 a particle is represented by the wave function...

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