1 Vector analysis -...

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1 Vector Analysis for quantum mechanics Masatsugu Sei Suzuki Department of Physics, SUNY at Binghamton (Date: February 03, 2014) Here we show Mathematica programs for the calculations of Laplacian, gradient, divergence, and rotation in the spherical co-ordinates and the cylindrical co-ordinates, as well as the Cartesian co-ordinates. These programs are very useful for the discussions on a variety of topics, such as the central field problem on the electron motion near a hydrogen atom. ________________________________________________________________________ 1. Gradient (; scalar) Nabra, gradient, del The gradient is defined as ) , , ( ˆ ˆ ˆ z y x z z y y x x ; gradient of the scalar ((Example)) f = f(r) with 2 2 2 z y x r . dr df dr df r z z y y x x dr df r z r dr df z y r dr df y x r dr df x z f z y f y x f x r f r r ˆ ) ˆ ˆ ˆ ( 1 ˆ ˆ ˆ ˆ ˆ ˆ ) ( where r x x r , r y y r , r z z r . 2. Divergence Now we define the divergence of the vector as

Transcript of 1 Vector analysis -...

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Vector Analysis for quantum mechanics Masatsugu Sei Suzuki

Department of Physics, SUNY at Binghamton (Date: February 03, 2014)

Here we show Mathematica programs for the calculations of Laplacian, gradient,

divergence, and rotation in the spherical co-ordinates and the cylindrical co-ordinates, as well as the Cartesian co-ordinates. These programs are very useful for the discussions on a variety of topics, such as the central field problem on the electron motion near a hydrogen atom. ________________________________________________________________________ 1. Gradient

(; scalar) Nabra, gradient, del The gradient is defined as

),,(ˆˆˆzyxz

zy

yx

x

; gradient of the scalar

((Example))

f = f(r)

with 222 zyxr .

dr

df

dr

df

r

zzyyxxdr

df

r

z

r

dr

dfz

y

r

dr

dfy

x

r

dr

dfx

z

fz

y

fy

x

fxrf

rr

ˆ

)ˆˆˆ(1

ˆˆˆ

ˆˆˆ)(

where

r

x

x

r

, r

y

y

r

, r

z

z

r

.

2. Divergence Now we define the divergence of the vector as

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z

F

y

F

x

F zyx

F .

(A) F is a real scalar. (B) Definition of solenoid

0 B . B is said to be soloenoid. _______________________________________________________________________ 3. F

We define the rotation of the vector as

zyx FFFzyx

zyx

ˆˆˆ

FW .

________________________________________________________________________ 4. Successive application of (i) )( This is defined by a Laplacian,

2

2

2

2

2

22

zyx

.

The equation 02 is called as the Laplace equation. (ii) is irrotational.

0 , since

0

ˆˆˆ

)(

zyx

zyx

zyx

.

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Thus is irrotational. (iii) )( F is solenoid.

0)(

zyx FFFzyx

xxx

F .

(iv) Formula

FFF 2)()( . ((Proof))

We use the formula given by

CBACABCBA )()()( . with A , B , and C = F. Then we find

FFF 2)()( . ((Example)) Calculations

If )2,2,( 2 yzxzyx A , find A , )( A , and ))(( A . We use the Mathematica.

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Clear"Gobal`";

Needs"VectorAnalysis`"SetCoordinatesCartesianx, y, z;

A1 x2 y, 2 x z , 2 y z;

CurlA12 x 2 z, 0, x2 2 z

CurlCurlA10, 2 2 x, 0

CurlCurlCurlA10, 0, 2

__________________________________________________________________ 5. Line and surface integral

We consider about the line integral

PQ

dI rA ,

where dsd r and the tangential component is assumed to be AS'

PQ

sdsAI .

If the contour is closed, we can write down as

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rA d .

In general the line integral depends on the choice of path. If F ( ; scalar)

P

Q

)()( PQddIPQPQ

rrF .

This value does not depend on the path of integral. 6. Surface Integral n normal vector to the surface

dad na . (da; area element) Then the surface integral is defined by

SS

dad nFaF .

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Fig. Right-hand rule for the positive normal. If F corresponds to the magnetic field; F = B,

S

daB is a magnetic flux through the area element S.

B

a

7. Gauss's theorem Here we define the volume integral as

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V

d ,

where is a scalar. (i) Gauss's theorem

SV

dd aFF .

First we consider the physical interpretation of F . Suppose that F = J (current density). The current coming out through ABCD is

dydzdxx

JJdydzJ x

xxdxxx )|(| 0

.

The current coming in through EFGH is equal to

dydzJ xx 0| .

Thus the net current through EFGH is equal to

dxdydzx

J x

.

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Thus the net current along the x direction through this small region is

dxdydzx

J x

.

Similarly for the y and z components, we have the net current along the y-direction and z-direction through the small region as

dxdydzy

J y

, dxdydzz

J z

,

respectively. Therefore the net current coming out through the volume element

dxdydzd can be expressed by

dd

surfaceSix

)( JaJ .

Summing over all parallel-pipes, we find that aJ d terms cancel out for all interior faces. Only the contributions of the exterior surface survive.

volume

surfaceexterior

dd )( JaJ ,

or

VA

dd JaJ . (Gauss’ theorem)

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(ii) Gauss' theorem Let F be a continuous and differentiable vector throughtout a region V of the space. Then

VSS

ddad FnFaF ,

where the surface integral is taken over the entire surface that encloses V. ((Example-1)) In the maxwell's equation, we have

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0

E ,

where is the charge density. From the Gauss's law, we have

SVV

ddd aEE

0

.

We assume that the volume V is formed of sphere with radius r. From the symmetry, E is perpendicular to the sphere surfaces,

rrE eE . Thus we have

S

rr

V

dEd ae

0

.

Since V

dQ , we get

204 r

QEr

. (Coulomb's law)

((Example-2))

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0 E ,

if = 0. From the Gauss's law, we have

SVV

ddd aEE 00

.

E1n and E2n are the normal components of E1 and E2. Then we have

0)( 21 aEE nn .

Therefore we have the boundary condition for E as

nn EE 21 .

8. Green's theorem

SV

dd a)()( 22 .

((Proof)) In the Gauss's theorem, we put

A . Then we have

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SVV

dddI aA )()(1 .

Noting that

2)( , we have

SV

ddI a)()( 21 .

By replacing , we also have

SV

ddI a)()( 21 .

Thus we find the Green's theorem

SV

ddII a)()( 2221 .

9. Stokes' theorem

x

y

A x0, y0 B

CD

1

dx

2

3

4 dy

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2

00

1

00

4321

1234

)},(),({)},(),({

)(

dyyxFydxxFdxdyyxFyxF

dyFdxFdyFdxF

ncirculatiod

yyxx

yxyx

lF

Note that

dxx

FyxFydxxF

dyy

FyxFdyyxF

yx

yyy

yx

xxx

00

00

,

00

,

00

),(),(

),(),(

Then we have

dxdydxdyy

F

x

Fncirculatio z

xy )()()( 1234 F

.

We can write down this as

1)()( aFeFlF ddxdyd z

sidesFour

,

where

dxdyd zea 1 . Imagine that paths 1 and 2 are expanded out until they coalesce with path C (or path 3). Since the line integrals of F along the potions that 1 and 2 have in common will cancel each other,

2,1

21

21

)()()( aFaFaF

lFlF

lF

ddd

dd

dC

.

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1 2

C Now let the surface S be divided up into a large number N of elements.

The above idea is extended to arrive at

SC

dd aFlF )( .

((Stoke's theorem)) Let S be a surface of any shape bounded by a closed curve C. If F is a vector, then

SSC

dadd nFaFlF )()( .

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________________________________________________________________________ 10 Curvilinear co-ordinates 10.1 General definition

We consider that new co-ordinate (q1, q2, q3) are related to (x, y, z) through

),,(

),,(

),,(

321

321

321

qqqzz

qqqyy

qqqxx

or

),,(

),,(

),,(

33

22

11

zyxqq

zyxqq

zyxqq

Since

j

jj

dqq

dqq

dqq

dqq

drrrr

r 33

22

11

,

we have

ji

jiijji

jiji

dqdqgdqdqqq

ddds,,

2 )(rr

rr ,

where

jijijijiij q

z

q

z

q

y

q

y

q

x

q

x

qqg

rr

(second rank tensor).

We now consider the general coordinate system. The relation between the constants h1, h2, and h3 and the tensor gij will be discussed later.

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e1

e2

e3

h1dq1e1

h2dq2e2

h3dq3e3

dr

iiiidqh

dqhdqhdqh

dsdsdsd

e

eee

eeer

333222111

332211

ji

ijiji idqdqhhddds

,

2 )( eerr ,

or we have

)(iijiij hhg ee ,

or

2iii hg .

Then

)(ii

jjii

ij

gg

gee .

Now we limit ourselves to orthogonal co-ordinate system.

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ijg for i ≠ j.

In order to simplify the notation, we use 2iii hg , so that

i

iidqhds 22

iiiidqh

dqhdqhdqh

dsdsdsd

e

eee

eeer

333222111

332211

Where e1, e2, and e3 are unit vectors which are perpendicular to each other.

3333

2222

1111

1

1

1

sqh

sqh

sqh

rre

rre

rre

where

jiiii qq

gh

rr2 ,

or

222

iiiii

iii q

z

q

y

q

x

qqgh

rr. (second rank tensor).

The volume element for an orthogonal curvilinear coordinate system is given by

321321333222111 )}(){( dqdqdqhhhdqhdqhdqhdV eee .

10.2 Spherical coordinate (i) Unit vectors

The position of a point P with Cartesian coordinates x, y, and z may be expressed in terms of r, , and of the spherical coordinates;

x rsin cos , y rsin sin , z rcos ,

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or

zy rrr eeer x cossinsincossin ,

jjj

jjj

j dqhdsd

3

1

3

1

eer .

x

y

z

q

dq

f df

r

drr cosq

r sinq

r sinq df

rdq

er

ef

eq

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sin

1

222

222

222

rzyx

h

rzyx

h

r

z

r

y

r

xhr

or

drdrdrdhdhdrhd rrrr eeeeeer sin ,

zyreee

re xr cossinsincossin

,

zyreee

re x

sinsincoscoscos1

,

yree

re x

cossinsin

1

.

This can be described using a matrix A as

z

y

x

z

y

xr

e

e

e

e

e

e

A

e

e

e

0cossin

sinsincoscoscos

cossinsincossin

.

or by using the inverse matrix A-1 as

e

e

e

e

e

e

A

e

e

e

A

e

e

e rrT

r

z

y

x

0sincos

cossincossinsin

sincoscoscossin1 .

or

eeee sincoscoscossin rx ,

eeee cossincossinsin ry ,

eee sincos rz .

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in the spherical coordinate systems.

(ii)

From the definition of , we have

3

1

3

1 j jjj

j jj qhs

ee ,

or,

sin

11

rrrr eee ,

where is a scalar function of r, , and . (iii) A

When a vector A is defined by

eeeA AAA rr .

The divergence is given by

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)]()sin()sin([sin

1

)]()()([1

22

rAArArrr

AhhAhhAhhrhhh

r

rrrr

A

or

A

rA

rAr

rr r

sin

1)(sin

sin

1)(

1 22A .

(iv) A

A is given by

ArrAA

r

rr

rAhAhAh

r

hhh

hhhr

r

rr

rr

r

sin

sin

sin

112

eeeeee

A

(v) Laplacian

]sin

1)(sin)([sin

sin

1

)]sin

1()(sin)sin([

sin

1

)]()()([1

2

22

2

22

2

rr

rr

rr

rr

h

hh

h

hh

rh

hh

rhhhrr

rr

or

2

2

2222

2

2

2222

22

sin

1)(sin

sin

1)(

1

sin

1)(sin

sin

1)(

1

rrr

rr

rrrr

rr

We can rewrite the first term of the right hand side as

)(1

)(1

22

2

rrrr

rrr

,

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which can be useful in shortening calculations. Note that we also use the expression for the operator

}sin

1)(sin

sin

1{

1)(

1

sin

1)(sin

sin

1)(

1

2

2

2222

2

2

2

2222

22

rrr

rr

rrrr

rr

11. Mathematica-1

We derive the above formula using the Mathematica. We use the Spherical co-rdinate. We need a Vector Analysis Package. We also need SetCordinatinates.In this system the vector is expressed in terms of (Ar, Aq, Af)

Clear"Gobal`"; Gra : Grad, r, , , "Spherical" &;

Diva : Div, r, , , "Spherical" &;

Curla : Curl, r, , , "Spherical" &;

Lap : Laplacian, r, , , "Spherical" &;

eq1 Lapr, , Simplify

1

r2 Csc2 0,0,2r, , Cot 0,1,0r, , 0,2,0r, , 2 r 1,0,0r, , r2 2,0,0r, ,

eq2 Grar, , 1,0,0r, , ,

0,1,0r, , r

,Csc 0,0,1r, ,

r

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A Arr, , , Ar, , , Ar, , ;

eq3 CurlaA Simplify

1rAr, , Cot

Csc A0,0,1r, , A0,1,0r, , ,

1rAr, , Csc Ar0,0,1r, ,

r A1,0,0r, , ,

1rAr, , Ar0,1,0r, , r A1,0,0r, ,

eq3 DivaA Simplify

1r2 Arr, ,

Ar, , Cot Csc A0,0,1r, , A0,1,0r, , r Ar1,0,0r, ,

_______________________________________________________________________ 12. Quantum mechanical orbital angular momentum The orbital angular momentum in the quantum mechanics is defined by

)( riprL , using the expression

sin

11

rrrr eee ,

in the spherical coordinate. Then we have

)sin

1(

)sin

11()(

ee

eeeerL

i

rrrrii rr

.

The angular momentum Lx, Ly, and Lz (Cartesian components) can be described by

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]sin

1)sinsincoscos(cos)cossin([

zyyi eeeeeL xx.

or

)coscot(sin

iLx ,

)sincotcos(

iLy ,

iLz .

We define L+ and L- as

)cot(

ieiiLLL iyx ,

and

)cot(

ieiiLLL i

yx .

We note that the operator can be expressed using the operator L as

2r

i

rr

Lre

.

The proof of this equation is given as follows.

)sin

1()

sin

1(

)(

eeeeeLr

rri r

,

or

rrrri rr

eeeLr

sin

11)(2

,

or

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2

)(

r

i

rr

Lre

.

From 2222zyx LLL L , we have

)](sinsin

1

sin

1[ 2

2

222

L ,

where the proof is given by Mathematica. Using

)( 2222

2

rr

rr

L,

we can also prove that

L

i

rrr )1(2 .

((Note))

)(11

)(11

)(11

2

22

22

22

222

22

222

2

rrrr

rr

rrr

rr

rrr

L

L

L

13. Mathematica ((Arfken))

_______________________________________________________________________ ((Arfken))

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________________________________________________________________________ ((Arfken))

________________________________________________________________________ ((Arfken))

________________________________________________________________________ ((Arfken))

((Arfken))

________________________________________________________________________ ((Mathematica))

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Vector analysisAngular momentum in the spherical coordinatesHere we use the angular momentum operatior in the unit of Ñ=1

Clear"Global`"; ux Sin Cos, Cos Cos, Sin;

uy Sin Sin, Cos Sin, Cos;

uz Cos, Sin, 0; ur 1, 0, 0;

Lap : Laplacian, r, , , "Spherical" &;

Gra : Grad, r, , , "Spherical" &;

Diva : Div, r, , , "Spherical" &;

Curla : Curl, r, , , "Spherical" &;

L : Crossur r, Gra & Simplify;

Lx : ux.L & Simplify; Ly : uy.L & Simplify;

Lz : uz.L & Simplify;

Arfken 2.5 .16 a

Lr, , Simplify

0, Csc 0,0,1r, , , 0,1,0r, ,

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Arfken 2.5.14, 2.5.16(b)

Lz

Lxr, , FullSimplify

Cos Cot 0,0,1r, , Sin 0,1,0r, , Lxr, , Lyr, , Simplify

Cos Sin Cot 0,0,1r, , 0,1,0r, , Lxr, , Lyr, , TrigToExp FullSimplify

Cos Sin Cot 0,0,1r, , 0,1,0r, , Lzr, , Simplify

0,0,1r, , Arfkem 2.5 .15

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LxLyr, , LyLxr, , Lzr, , Simplify

0

Arfken 2.5.16(c)

eq1 LxLxr, , LyLyr, , LzLzr, , Expand FullSimplify

Csc2 0,0,2r, , Cot 0,1,0r, , 0,2,0r, , eq2 r2 Lapr, , Dr2 Dr, , , r, r Simplify

Csc2 0,0,2r, , Cot 0,1,0r, , 0,2,0r, , eq1 eq2 Simplify

0

Arfken 2.5.17(a)

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eq3 ur Dr, , , r Cross1

rur, Lr, , Simplify

1,0,0r, , ,0,1,0r, ,

r,

Csc 0,0,1r, , r

eq4 Grar, ,

1,0,0r, , ,0,1,0r, ,

r,

Csc 0,0,1r, , r

eq3 eq4 Simplify

0, 0, 0

Arfken 2.5.17 (b)

eq5 ur r Lapr, , Simplify

1rCsc2 0,0,2r, , Cot 0,1,0r, , 0,2,0r, , 2 r 1,0,0r, , r2 2,0,0r, , , 0, 0

eq6 Grar, , r Dr, , , r Simplify

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2 1,0,0r, , r 2,0,0r, , ,

0,1,0r, , r

1,1,0r, , ,

Csc 0,0,1r, , r 1,0,1r, , r

eq7 CurlaLr, , Simplify

1r

Csc2 0,0,2r, , Cot 0,1,0r, , 0,2,0r, , ,

0,1,0r, , r 1,1,0r, ,

r,

Csc 0,0,1r, , r 1,0,1r, ,

r

seq1 eq5 eq6 eq7 Simplify

0, 0, 0The form of “2y(r) can be expressed by

Arfken 2.5.18

eq8 Lapr FullSimplify

2 rr

r

eq9 1

r2Dr2 Dr, r, r FullSimplify

2 rr

r

eq10 1

rDr r, r, 2 FullSimplify

2 rr

r

14. Radial momentum operator in the quantum mechanics (a) In classical mechanics, the radial momentum of the radius r is defined by

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)(1

pr r

prc .

(b) In quantum mechanics, this definition becomes ambiguous since the component

of p and r do not commute. Since pr should be Hermitian operator, we need to define as the radial momentum of the radius r is defined by

)(2

1

rrprq

rpp

r .

This symmetric expression is indeed the canonical conjugate of r.

irprp rqrq

.

Note that

rrr

irr

iprq

1

)()1

)(( .

((Mathematica))

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Clear"Global`"; ur 1, 0, 0;

Gra : Grad, r, , , "Spherical" &;

Diva : Div, r, , , "Spherical" &;

prc :—

ur.Gra & ;

prcr, , — 1,0,0r, ,

prq : —

2ur .Gra —

2Diva ur &;

prqr, , Simplify

— r, , r 1,0,0r, ,

r

Commutation relation

prqr r, , r prqr, , Simplify

— r, , 15. Expression of the Laplacian in the spherical coordinate Here we redefine prq simplily as

)(2

1

rrpr

rpp

r .

We show that

2

2222

rpr

L ,

by using the Mathematica. The total Hamiltonian H can be expressed by

)(22

1)(

2)(

2

12

222

22 rV

mrp

mrV

mrV

mH r

Lp

.

The effective potential Veff(r) is defined as

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)1(2

)()(2

2

llmr

rVrVeff

.

((Mathematica)) Spherical coordinate Clear"Global`"; Clearrux Sin Cos, Cos Cos, Sin;

uy Sin Sin, Cos Sin, Cos;

uz Cos, Sin, 0; ur 1, 0, 0;

L :

— Crossur r, Grad, r, , ,

"Spherical" & Simplify;

Lx : ux.L &; Ly : uy.L &;

Lz : uz.L &;

Lap : Laplacian, r, , , "Spherical" &;

Gra : Grad, r, , , "Spherical" &;

Diva : Div, r, , , "Spherical" &;

Cur : Curl, r, , , "Spherical" &;

prq : —

21, 0, 0.Gra

2Diva 1, 0, 0 & ;

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prqr, , Simplify

— r, , r 1,0,0r, ,

r

eq1 Nestprq, r, , , 2 Simplify;

eq2 1

r2LxLxr, , LyLyr, ,

LzLzr, , FullSimplify;

eq12 eq1 eq2 Simplify

1

r2 —2 Csc2 0,0,2r, , Cot 0,1,0r, , 0,2,0r, , 2 r 1,0,0r, , r2 2,0,0r, ,

eq3 —2 Lapr, , Simplify

1

r2 —2 Csc2 0,0,2r, , Cot 0,1,0r, , 0,2,0r, , 2 r 1,0,0r, , r2 2,0,0r, ,

eq12 eq3 FullSimplify

0 ________________________________________________________________________ 16 Cylindrical coordinates

The position of a point in space P having Cartesian coordinates x, y, and z may be expressed in terms of cylindrical co-ordinates

cosx , siny , z = z. The position vector r is written as

zyx zeeer sincos ,

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dzdddqhd zj

jjj eeeer

3

1

,

where

1

1

3

2

1

zhh

hh

hh

The unit vectors are written as

zz

z

yx

yx

zzh

h

h

err

e

eerr

e

eerr

e

1

cossin11

sincos1

The above expression can be described using a matrix A as

z

y

x

z

y

x

z e

e

e

e

e

e

A

e

e

e

100

0cossin

0sincos

.

or by using the inverse matrix A-1 as

zz

T

zz

y

x

e

e

e

e

e

e

A

e

e

e

A

e

e

e

100

0cossin

0sincos1 .

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17. Differential operations in the cylindrical coordinate

The differential operations involving are as follows.

zz

eee

1,

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zVz

VV

1

)(1

V ,

z

z

VVVz

eee

V1

,

2

2

2

2

22 1

)(1

z

.

where V is a vector and is a scalar. 18 Mathematica for the cylindrical coordinate Vector analysisAngular momentum in the cylindrical coordinatesHere we use the angular momentum operatior in the unit of Ñ=1

Clear"Global`"; r1 z2 2 ; ux Cos, Sin, 0;

uy Sin, Cos, 0; uz 0, 0, 1; r , 0, z;

ur 1

r1, 0, z; Gra : Grad, , , z, "Cylindrical" &;

Lap : Laplacian, , , z, "Cylindrical" &;

Curla : Curl, , , z, "Cylindrical" &;

Diva : Div, , , z, "Cylindrical" &;

Vector analysis in the cylindrical coordinate

eq1 Lap, , z Simplify

0,0,2, , z 0,2,0, , z2

1,0,0, , z

2,0,0, , z

eq2 Gra, , z Simplify

1,0,0, , z,0,1,0, , z

, 0,0,1, , z

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B B, , z, B, , z, Bz, , z;

eq3 CurlaB Simplify

B0,0,1, , z Bz0,1,0, , z

,

B0,0,1, , z Bz1,0,0, , z,

B, , z B0,1,0, , z B1,0,0, , z

eq4 DivaB Simplify

Bz0,0,1, , z B, , z B0,1,0, , z

B1,0,0, , z

Angular momentum in the cylindrical coordinate

L : Crossr, Gra &; Lx : ux.L & Simplify;

Ly : uy.L & Simplify;

Lz : uz.L & Simplify;

L, , z Simplify

z 0,1,0, , z

,

0,0,1, , z z 1,0,0, , z, 0,1,0, , z

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Lx, , z FullSimplify

Sin 0,0,1, , z z Cos 0,1,0, , z

z Sin 1,0,0, , z

Ly, , z FullSimplify

Cos 0,0,1, , z z Sin 0,1,0, , z

z Cos 1,0,0, , z

Lx, , z Ly, , z FullSimplify

2 0,0,1, , z z 0,1,0, , z z 1,0,0, , z

Lx, , z Ly, , z FullSimplify

2 0,0,1, , z z 0,1,0, , z z 1,0,0, , z

Lz, , z Simplify

0,1,0, , z

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The commutation of the angular momentum in the cylindrical coordinate

eq5 LxLy, , z LyLx, , z Lz, , z Simplify

0

L2 in the cylindrical coordinate

eq6 LxLx, , z LyLy, , z Lz Lz, , z FullSimplify

2 z 0,0,1, , z 1

2 4 0,0,2, , z z2 2 0,2,0, , z z2 2 1,0,0, , z

z 2 1,0,1, , z z 2,0,0, , z

Schrodionger equation with a speific wavefunction in the cylindrical coorrdinate

H1 Exp kz z Exp m ;

1

2 LapH1 V H1 Simplify

kz zm m2 kz2 2 2 2 V 2 2

19. Expression of the Laplacian in the cylindrical coordinate Here we redefine prq simply as

)(2

1

rrpr

rpp

r ,

with the commutation relation

ipr r ],[ , where

22 zr , zzeer .

We show that

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2

2222

rpr

L ,

by using the Mathematica. ((Mathematica))

Clear"Global`"; r1 z2 2 ; ux Cos, Sin, 0;

uy Sin, Cos, 0; uz 0, 0, 1; r , 0, z; ur 1

r1, 0, z;

Gra : Grad, , , z, "Cylindrical" &;

Lap : Laplacian, , , z, "Cylindrical" &;

Curla : Curl, , , z, "Cylindrical" &;

Diva : Div, , , z, "Cylindrical" &;

Angular momentum in the cylindrical coordinate

L : — Crossr, Gra &; Lx : ux.L & Simplify;

Ly : uy.L & Simplify;

Lz : uz.L & Simplify;

The linear momentum in the cylindrical coordinate

prq : —

2ur.Gra —

2Diva ur & ;

prq, , z Simplify

— , , z z 0,0,1, , z 1,0,0, , z

z2 2

Commutation relation in the cylindrical coordinate

r1 prq, , z prqr1 , , z Simplify

— , , zProof of

-Ñ2“2= pr 2 + L2

r2

eq1 Nestprq, , , z, 2 Simplify

1

z2 2 —2 2 z 0,0,1, , z z2 0,0,2, , z 2 1,0,0, , z 2 z 1,0,1, , z 2,0,0, , z

eq20 LxLx, , z LyLy, , z LzLz, , z FullSimplify;

eq2 eq20

r12 Simplify

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1

2 z2 2 —2 2 z 2 0,0,1, , z z2 0,2,0, , z

3 0,0,2, , z z2 1,0,0, , z 0,2,0, , z 1,0,0, , z 2 z 1,0,1, , z z2 2,0,0, , z

eq12 eq1 eq2 FullSimplify

1

2 —2 0,2,0, , z 1,0,0, , z 0,0,2, , z 2,0,0, , z

L2 in the cylindrical coordinate

eq3 —2 Lap, , z Expand FullSimplify

1

2 —2 0,2,0, , z 1,0,0, , z 0,0,2, , z 2,0,0, , z

eq12 eq3 Simplify

0 19 Jacobian

321321321321 ),,(

),,(dqdqdqhhhdqdqdq

qqq

zyxdxdydzdV

.

Jacobian determinant is defined as;

321

321

321

321 ),,(

),,(

q

z

q

z

q

zq

y

q

y

q

yq

x

q

x

q

x

qqq

zyx

.

(a) Spherical coordinate

ddrdrddrdhhhdqdqdqhhh r sin2321321 .

(b) Cylindrical co-ordinate

dzdddzddhhhdqdqdqhhh z 321321 .

((Mathematica))

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This is the program to determine the Jacobian determinant.

JacobianDeterminant[pt, coordsys]:

to give the determinant of the Jacobian matrix of the transformation from the coordinate system coordinate system to the Cartesian coordinate system at the point pt.

Clear"Gobal`"Jacobian determinant for transformation from cylindrical to Cartesian coordinates:

jdet JacobianDeterminant, , z, Cylindrical

Jacobian determinant for transformation from cylindrical to Spherical coordinates:

jdet JacobianDeterminantr, , , Sphericalr2 Sin

________________________________________________________________________ APPENDIX Formula

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