Chapter 3: Q13P (page 184)

Verify (14.7) and (14.8) Hints: Remember that a norm squared, like $$ , is a real and non-negative. Scalar, so its complex conjugate is just itself. But $$ is a complex scalar and $ = {< A | B >}^{}$ by (14.4). Show that $μ^{} = < A | B > = {}^{*}$ .

Short Answer

Expert verified

The given relation is (14.7) and (14.8) is verified.

Step by step solution

Given information

The given information is $f (x) = a_{0} + a_{1} x + a_{2} x^{2} + a_{3} x^{3}$

Definition of Complex Conjugate

The complex conjugate of a complex number is the number with an equal real part and an imaginary part equal in magnitude but opposite in sign.

Verify the given statement

The inner product space

$<A - μB | A - μB> = <A | A> + u^{*} - μ <A | B> + μ^{*} μ \geq 0$

And

$<A | A> - \frac{<A | B>}{} - \frac{}{} <A | B> + \frac{<A | B>}{} \frac{}{} = <A | A> - \frac{<A | B> {<A | B>}^{*}}{} = <A | A> - \frac{| A | B >^{2}}{} \geq 0$

Verify the following:

$<A - μ B | A - μ B> \geq 0$

Since,

This is the form of $A - μ B >$ , from (14.4b).

$<A - μ B | A - μ B> = <A | A - μ B> - μ^{*} = <A | A> - μ <A | B> - μ^{*} + μ^{*} μ $

Now,

$μ = \frac{}{} μ^{*} = \frac{}{}$
But,

${}^{*} = $

And

${}^{*} = <A | B>$

Therefore,

$μ^{*} = \frac{<A | B>}{}$

Then

$<A - μ B | A - μ B> = <A | A> - \frac{}{} <A | B> + \frac{<A | B>}{} \frac{}{} = <A | A> - \frac{<A | B> {<A | B>}^{*}}{} = <A | A> - \frac{| <A | B> R^{2}}{}$

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Verify (14.7) and (14.8) Hints: Remember that a norm squared, like $< B | B >$ , is a real and non-negative. Scalar, so its complex conjugate is just itself. But $< B | A >$ is a complex scalar and $< B | A > = {< A | B >}^{}$ by (14.4). Show that $μ^{} = < A | B > = {< B | B >}^{*}$ .

Short Answer

Step by step solution

Given information

Definition of Complex Conjugate

Verify the given statement

One App. One Place for Learning.

Most popular questions from this chapter

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Verify (14.7) and (14.8) Hints: Remember that a norm squared, like<B|B>, is a real and non-negative. Scalar, so its complex conjugate is just itself. But<B|A>is a complex scalar and<B|A>=<A|B>*by (14.4). Show thatμ*=<A|B>=<B|B>*.

Short Answer

Step by step solution

Given information

Definition of Complex Conjugate

Verify the given statement

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Thermodynamics

Particle Model of Matter

Physics of Motion

Translational Dynamics

Magnetism

Physical Quantities And Units

Study anywhere. Anytime. Across all devices.

Verify (14.7) and (14.8) Hints: Remember that a norm squared, like $< B | B >$ , is a real and non-negative. Scalar, so its complex conjugate is just itself. But $< B | A >$ is a complex scalar and $< B | A > = {< A | B >}^{}$ by (14.4). Show that $μ^{} = < A | B > = {< B | B >}^{*}$ .