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(a) From the definition (Equation 4.46), construct n1(x)andn2(x).

(b) Expand the sines and cosines to obtain approximate formulas forn1(x)androle="math" localid="1656329588644" n2(x), valid whenx1.. Confirm that they blow up at the origin.

Short Answer

Expert verified

a)

The functions are,

n1(x)=-cosxx2-sinxx

n2(x)=-(3x3-1x)cosx-3x2sinx

b)

The new functions are,

n1(x)-1x2n2(x)-3x3

Step by step solution

01

Concept used

The general expression for the function ni(x) is given by,

ni(x)-(-x)i1xddxicosxx

02

Construct  n1(x) and  n2(x)

We need to construct n1xand n2x, as:

n1x=-(-x)1xddxcosxx=-cosxx2-sinxxn1x=-cosxx2-sinxx

Similarly solving for n2(x),

n2x=-(-x)21xddx2cosxx=-x21xddx1xddxcosxx=-xddx1x·-xsinx-cosxx2=xddxsinxx2+cosxx3

Further solving above equation,

n2(x)=xx2cosx-2xsinx+-x3sinx-3x2cosxx4n2(x)=cosxx-2x2x2-sinxx2-3cosxx3n2(x)=-3x3-1xcosx-3x2sinxn2(x)=-3x3-1xcosx-3x2sinx

Thus, the function n1(x), and n2(x) are n1(x)=-cosxx2-sinxxand n2(x)=-3x3-1xcosx-3x2sinxrespectively.

03

Calculate approximate values of the sine and cosine function

We can approximate the sine and cosine functions as sinxxandcosx1,so:

n1(x)-1x2+1x

But since is x very small then 1/x2is much larger than 1, thus we can neglect 1, so get:

n1(x)-1x2

And for we have:

n2(x)-3x3-1x-3x2xn2(x)-3x3

Thus, approximate formulas for n1(x), and n2(x) are n1(x)-1x2and n2(x)-3x2respectively.

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Most popular questions from this chapter

Work out the radial wave functions R30,R31,andR32using the recursion formula. Don’t bother to normalize them.

(a) For a functionf(ϕ)that can be expanded in a Taylor series, show that f(ϕ+φ)=eiLzφ/f(ϕ) (where is an arbitrary angle). For this reason, Lz/ is called the generator of rotations about the Z-axis. Hint: Use Equation 4.129 , and refer Problem 3.39.More generally, L·n^/ is the generator of rotations about the direction n^, in the sense that exp(iL·n^φ/)effects a rotation through angleφ (in the right-hand sense) about the axis n^ . In the case of spin, the generator of rotations is S·n^/. In particular, for spin 1/2 χ'=ei(σ·n^)φ/2χtells us how spinors rotate.

(b) Construct the (2×2)matrix representing rotation by 180about the X-axis, and show that it converts "spin up" χ+into "spin down"χ- , as you would expect.

(c) Construct the matrix representing rotation by 90about the Y-axis, and check what it does to

χ+

(d) Construct the matrix representing rotation by 360about the -Zaxis, If the answer is not quite what you expected, discuss its implications.

(e) Show thatei(σ·n^)φ/2=cos(φ/2)+i(n^·σ)sin(φ/2)

(a) If you measured the component of spin angular momentum along the x direction, at time t, what is the probability that you would get +h/2?

(b) Same question, but for the ycomponent.

(c) Same, for the z component.

If the electron were a classical solid sphere, with radius

rc=e24πO0˙mc2

(the so-called classical electron radius, obtained by assuming the electron's mass is attributable to energy stored in its electric field, via the Einstein formula E=mc2), and its angular momentum is (1/2)h then how fast (in m/sm/s) would a point on the "equator" be moving? Does this model make sense? (Actually, the radius of the electron is known experimentally to be much less than5.156×1010m/src, but this only makes matters worse).

Deduce the condition for minimum uncertainty inSx andSy(that is, equality in the expression role="math" localid="1658378301742" σSxσSy(ħ/2)|<Sz>|, for a particle of spin 1/2 in the generic state (Equation 4.139). Answer: With no loss of generality we can pick to be real; then the condition for minimum uncertainty is that bis either pure real or else pure imaginary.

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