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Why can鈥檛 you do integration-by-parts directly on the middle expression in Equation -1.29 pull the time derivative over onto x, note thatx/t=0 , and conclude thatd<x>/dt=0 ?

Short Answer

Expert verified

Because the derivative and integrals are taken with respect to different variables t and x,so the integration-by-parts is not possible.

Step by step solution

01

The given information

The equation 1.29 is,

dxdt=xt2dx=ih2mxt*t-*tdx

According to the given question, X/T=0. Integration-by-parts directly cannot be done on the middle expression dxdt=xt2dx=ih2mxt*t-*tdx.

02

The expected equation and the variables.

The expected value is the average of the results of a large number of measurements taken on separate systems.

The following is the expression for X's expected value:

X=*x,txx,tdxX=xx,t2dx

The wave function is x,t, and the average value or expectation value of the position operator x is x.

03

The differentiate of the equation and put the value of  x

The above equation have to be differentiated on both sides:

dxdt=xt2dx

The time derivative will be put on to xas the condition is given. So, the equation is:

dxdt=xtx2dx=t2dx+xt2dx=0+xt2dx=xt2dx

04

The limit of the equation

Taking the above equation in between aand b,the equation will be:

dxdt=batx2dx=x2dxab

The derivative is with regard to time, while the integration is with respect toxin the above integration.

As a result, part-by-part integration is not possible.

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Most popular questions from this chapter

Consider the wave function
(x,t)=Ae-|x|e-it

whereA, , and are positive real constants. (We鈥檒l see in Chapter for what potential (V) this wave function satisfies the Schr枚dingerequation.)

(a) Normalize .

(b) Determine the expectation values ofx and x2.

(c) Find the standard deviation of . Sketch the graph of2 , as a function ofx, and mark the points (x+)and (x-), to illustrate the sense in which蟽 represents the 鈥渟pread鈥 inx. What is the probability that the particle would be found outside this range?

Calculate d銆坧銆/dt. Answer:

dpdx=-Vx

This is an instance of Ehrenfest鈥檚 theorem, which asserts that expectation values obey the classical laws

Suppose you add a constantV0 to the potential energy (by 鈥渃onstant鈥 I mean independent ofxas well as t). In classical mechanics this doesn鈥檛 change anything, but what about quantum mechanics? Show that the wave function picks up a time-dependent phase factor:exp(-iV0t/h). What effect does this have on the expectation value of a dynamical variable?

We consider the same device as the previous problem, but this time we are interested in thex-coordinate of the needle point-that is, the "shadow," or "projection," of the needle on the horizontal line.

(a) What is the probability density (x)? Graph data-custom-editor="chemistry" (x) as a function of x, from -2rto +2r , where ris the length of the needle. Make sure the total probability is . Hint: data-custom-editor="chemistry" (x)dx is the probability that the projection lies between data-custom-editor="chemistry" xand data-custom-editor="chemistry" (x+dx). You know (from Problem 1.11) the probability that data-custom-editor="chemistry" is in a given range; the question is, what interval data-custom-editor="chemistry" dxcorresponds to the interval data-custom-editor="chemistry" 诲胃?

(b) Compute data-custom-editor="chemistry" <x>, data-custom-editor="chemistry" <x2>, and data-custom-editor="chemistry" , for this distribution. Explain how you could have obtained these results from part (c) of Problem 1.11.

For the distribution of ages in the example in Section 1.3.1:

(a) Computej2 andj2 .

(b) Determine 鈭j for each j, and use Equation 1.11 to compute the standard deviation.

(c) Use your results in (a) and (b) to check Equation 1.12.

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