Chapter 1: Q6P (page 18)

Why can’t you do integration-by-parts directly on the middle expression in Equation -1.29 pull the time derivative over onto x, note that $\partial x / \partial t = 0$ , and conclude that $d < x > / dt = 0$ ?

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

The given information

The equation 1.29 is,

$\frac{d <x>}{d t} = \int x \frac{\partial}{\partial t} {|Ψ|}^{2} d x = \frac{i h}{2 m} \int x \frac{\partial}{\partial t} (Ψ^{*} \frac{\partial Ψ}{\partial t} - \frac{\partial Ψ^{*}}{\partial t} Ψ) d x$

According to the given question, $\partial X / \partial T = 0$ . Integration-by-parts directly cannot be done on the middle expression $\frac{d <x>}{d t} = \int x \frac{\partial}{\partial t} {|Ψ|}^{2} d x = \frac{i h}{2 m} \int x \frac{\partial}{\partial t} (Ψ^{*} \frac{\partial Ψ}{\partial t} - \frac{\partial Ψ^{*}}{\partial t} Ψ) d x$ .

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:

$<X> = \int Ψ^{*} (x, t) x Ψ (x, t) d x <X> = \int x {|Ψ (x, t)|}^{2} d x$

The wave function is $Ψ (x, t)$ , and the average value or expectation value of the position operator $x$ is $<x>$ .

The differentiate of the equation and put the value of x

The above equation have to be differentiated on both sides:

$\frac{d <x>}{d t} = \int x \frac{\partial}{\partial t} {|Ψ|}^{2} d x$

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

$\frac{d <x>}{d t} = \int (x \frac{\partial}{\partial t} x {|Ψ|}^{2}) d x = \int \frac{\partial}{\partial t} {|Ψ|}^{2} d x + \int x \frac{\partial}{\partial t} {|Ψ|}^{2} d x = 0 + \int x \frac{\partial}{\partial t} {|Ψ|}^{2} d x = \int x \frac{\partial}{\partial t} {|Ψ|}^{2} d x$

The limit of the equation

Taking the above equation in between $a$ and $b$ ,the equation will be:

$\frac{d <x>}{d t} = \int_{b}^{a} \frac{\partial}{\partial t} (x {|Ψ|}^{2}) d x = (x {|Ψ|}^{2}) d x_{a}^{b}$

The derivative is with regard to time, while the integration is with respect to $x$ in the above integration.

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

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Why can’t you do integration-by-parts directly on the middle expression in Equation -1.29 pull the time derivative over onto x, note that $\partial x / \partial t = 0$ , and conclude that $d < x > / dt = 0$ ?

Short Answer

Step by step solution

The given information

The expected equation and the variables.

The differentiate of the equation and put the value of x

The limit of the equation

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Why can’t you do integration-by-parts directly on the middle expression in Equation -1.29 pull the time derivative over onto x, note that∂x/∂t=0 , and conclude thatd<x>/dt=0 ?

Short Answer

Step by step solution

The given information

The expected equation and the variables.

The differentiate of the equation and put the value of x

The limit of the equation

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Radiation

Translational Dynamics

Modern Physics

Force

Turning Points in Physics

Physical Quantities And Units

Study anywhere. Anytime. Across all devices.

Why can’t you do integration-by-parts directly on the middle expression in Equation -1.29 pull the time derivative over onto x, note that $\partial x / \partial t = 0$ , and conclude that $d < x > / dt = 0$ ?