Question

through a sketch of the probability density, P(x). a) For a quantum particle which exhibits a wave function, as y(x)= A(x/L)'e twin, where the given parameter, L, has dimension of length, and the particle is only contained in the infinite positive domain, x = [0,-), determine the normalization coefficient, A, so that the wave function is properly normalized, . Then, write down the properly normalized wave function, y(x), and the probability density, P(x)=\w (x)}", which is a function of L. Also, briefly sketch the probability density, P(x), as a function of position (no details are needed), but to make this specific, use the parameter value, L=1. b) Determine the expectation value of position, or (*)= (w|xv) = dxy" (x)xv (x) = [dxxP(x), as a function of the general parameter, L. Also, determine the result using the parameter choice, L=1, and note if it is reasonable in comparison to the sketch in a). c) Determine the maximum position, xur where the probability density function, P(x), is maximum, as a function of L, and then use the parameter value, L = 1, in order to note if this result is close to the expectation value of X, or (x), found above. d) Determine the position uncertainty, or Ar = \(+?)-(x), where (x2)=(w/x?v) = dorys" (*)x?v"(x)=1dxx?P(x), as a function of L. Finally, using the parameter choice, L=1, provide the result, Ar, and note if this seems reasonable with respect to the sketch in a).

Answer 1

Normolisotion voor Ke let gye? x= 12.9%) xdx = 1² dt = 1? [%cf11dl) gadu = {²014 914 1 x 193% % di 1946)%$****** (A12x L3 247 4722 -1A124*3*+ Alormalised L, you - 14:13) 1992 pay = 141 92 x 10 222 Fox max or min demo da la do htumpe **** 12 tue 12x-4a R 19 17:30 AX Nole

4 12 3 1/2 to JX6 = +) Ho = + L Hence P(m) will be maa at n=fL , no (1) = 5 pln dx = LAR Lys - 2 de kepada p e lo lar de = 192 1 1 2 ) ² et 12 dt 27 LY Jo Ly .. 2 Jo ju = 2 4 x 2 2 = 3 1 2 4 le * (x) = 1 de 2 pm dx = (A12 L x 6 e sa de = (A²llues) e (244) (1 dn) - !" E9h et 14/12 4 six(+2)%*4.84-da , 222 n OLse e su= x - 2232 = 2/1.25 -100630 =0.4312 Nole your Please problem set ch yourself L=R a genral . I solved thank your
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