Quantum Mechanics in Simple Matrix Form: chapter 3 of (2-28)

October 19, 2009 by theproblemyard

Once again I have added letters to differentiate sub-exercises.

3-1. Find the matrices for

A> \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}+\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}=\begin{pmatrix} 1 & -i \\ i & -1 \end{pmatrix}

B> \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}+\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}=\begin{pmatrix} 0 & 1-i \\ 1+i & 0 \end{pmatrix}

C> \frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}+ \frac{1}{2}\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}=\frac{1}{2}\begin{pmatrix} 2 & 0 \\ 0 & 0 \end{pmatrix} = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}

D> \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}+i\begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}+\begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 2 \\ 0 & 0 \end{pmatrix}

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3-2. Compute the matrix products

A> \begin{pmatrix} 1 & -1 & 1 \\ 0 & 1 & 0 \\ 2 & 0 & 3 \end{pmatrix} \begin{pmatrix} 3 & 3 & -1 \\ 0 & 1 & 0 \\ -2 & -2 & 1 \end{pmatrix}=

\begin{pmatrix} 3+0-2 & 3-1-2 & -1+0+1 \\ 0+0+0 & 0+1+0 & 0+0+0 \\ 6+0-6 & 6+0-6 & -2+0+3 \end{pmatrix}=

\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}

B> \begin{pmatrix} 3 & 3 & -1 \\ 0 & 1 & 0 \\ -2 & -2 & 1 \end{pmatrix}\begin{pmatrix} 1 & -1 & -1 \\ 0 & 1 & 0 \\ 2 & 0 & 3 \end{pmatrix}=

\begin{pmatrix} 3+0-2 & -3+3+0 & 3+0-3 \\ 0+0+0 & 0+1+0 & 0+0+0 \\ -2+0+2 & 2-2+0 & -2+0+3 \end{pmatrix}=

\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}

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3-3. Find the matrices for these products:

A> \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}=

\frac{1}{4}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}=

\frac{1}{4}\begin{pmatrix} 1+1 & 1+1\\ 1+1 & 1+1 \end{pmatrix}=

\frac{1}{4}\begin{pmatrix} 2 & 2 \\ 2 & 2 \end{pmatrix}=

\begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\\frac{1}{2} &\frac{1}{2} \end{pmatrix}

B> \frac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}\frac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}=

\frac{1}{4}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}=

\frac{1}{4}\begin{pmatrix} 1+1 & -1-1 \\ -1-1 & 1+1 \end{pmatrix}=

\begin{pmatrix} \frac{1}{2} & -\frac{1}{2} \\ -\frac{1}{2} &\frac{1}{2} \end{pmatrix}

C> \frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}\frac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}=

\frac{1}{4}\begin{pmatrix} 1-1 & -1+1 \\ 1-1 & -1+1 \end{pmatrix}=

\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}

D> \frac{1}{2}\begin{pmatrix} 1 & -1 \\ -1 & 1 \end{pmatrix}\frac{1}{2}\begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}=

\frac{1}{4}\begin{pmatrix} 1-1 & 1-1 \\ -1+1 & -1+1 \end{pmatrix}=

\begin{pmatrix} 0 & 0 \\ 0 & 0 \end{pmatrix}

Again the answers should be matrices you recognize. Notice that the product to two matrices can be be zero when neither one is zero.

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3-4. Show that the inverse of the matrix

M=\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{pmatrix}

is

M^{-1}=\begin{pmatrix} 1 & -3 & 2 \\ -3 & 3 & -1 \\ 2 & -1 & 0 \end{pmatrix}

A> M times M inverse> \begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{pmatrix} \begin{pmatrix} 1 & -3 & 2 \\-3 & 3 & -1 \\ 2 & -1 & 0 \end{pmatrix}=

\begin{pmatrix} 1-6+6 & -3+6-3 & 2-2+0 \\ 2-12+10 & -6+12-5 & -4-4+0 \\ 3-15+12 & -9+15-6 & 6-5+0 \end{pmatrix}=

\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}

M inverse times M> \begin{pmatrix} 1 & -3 & 2 \\-3 & 3 & -1 \\ 2 & -1 & 0 \end{pmatrix}\begin{pmatrix} 1 & 2 & 3 \\ 2 & 4 & 5 \\ 3 & 5 & 6 \end{pmatrix}=

\begin{pmatrix} 1-6+6 & 2-12+10 & 3-15+12 \\ -3+6-3 & -6+12-5 & -9+15-6 \\ 2-2+10 & 4-4+0 & 6-5+0 \end{pmatrix}=

\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}

Since we can see the products are both equal to the diagonal matrix then the M inverse we were given must be the true M inverse.

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3-5. Suppose M has an inverse. Show that if A commutes with M inverse, then A commutes with M and, conversely, if A commutes with M, then A commutes with M inverse. This means there is an unambiguous matrix AM^{-1}=M^{-1}A for A divides by M if and only if there is an unambiguous matrix AM=MA for the product of A and M.

Show that if AM^{-1}=M^{-1}A then A commutes with M>

MA=MA

MAM^{-1}=MAM^{-1}

MM^{-1}A=MAM^{-1}

A=MAM^{-1}

AM=MAM^{-1}M

AM=MA

Show that if AM=MA then A commutes with M inverse>

AM^{-1}=AM^{-1}

MAM^{-1}=MAM^{-1}

MAM^{-1}=AMM^{-1}

MAM^{-1}=A

M^{-1}MAM^{-1}=M^{-1}A

AM^{-1}=M^{-1}A

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3-6. Let A, B, and M be matrices the same size and z a complex number. Suppose A and M commute and B and M commute. Show that A + B and M commute, that AB and M commute, and that zB and M commute.

given: AM=MA and BM=MB

Show that A+B and M commute> (A+B)M = (A+B)M

AM+BM = MA+MB

(A+B)M = M(A+B)

Show that AB and M commute> ABM = ABM

ABM = AMB

ABM = MAB

Show that zB and M commute>

z=x+yi

zBM= zBM

xBM+yiBM= xBM+yiBM

xBM+yiBM= xMB+yiMB

xBM+yiBM= MxB+MyiB (x and yi are scalers hence they give the same result whether placed before or after a matrix)

zBM= MzB

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3-7. Show that the inverse of the matrix

M=\begin{pmatrix} 2 & 2-i \\ 2+i & -2 \end{pmatrix}

is M^{-1}=\frac{1}{9}M.

M times (1/9)M>  \begin{pmatrix} 2 & 2-i \\ 2+i & -2 \end{pmatrix} \begin{pmatrix}\frac{2}{9} & \frac{2-i}{9} \\ \frac{2+i}{9} & \frac{-2}{9} \end{pmatrix}=

\begin{pmatrix} 2\cdot\frac{2}{9}+(2-i)\cdot\frac{2+i}{9} & 2\cdot\frac{2-i}{9}-(2-i)\cdot\frac{2}{9}\\(2+i)\cdot\frac{2}{9}-2\cdot\frac{2+i}{9}& (2+i)\cdot\frac{2-i}{9}+2\cdot\frac{2}{9} \end{pmatrix}=

\begin{pmatrix} \frac{4}{9}+ \frac{2^2+i^2}{9} & \frac{4-2i}{9}-\frac{4-2i}{9}\\ \frac{4+2i}{9}-\frac{4+2i}{9} & \frac{2^{2}-i^{2}}{9}+\frac{4}{9}\end{pmatrix}=

\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

(1/9)M times M> \begin{pmatrix} \frac{2}{9} & \frac{2-i}{9} \\ \frac{2+i}{9} & \frac{-2}{9}\end{pmatrix}\begin{pmatrix} 2 & 2-i \\ 2+i & -2 \end{pmatrix}=

\begin{pmatrix} \frac{2}{9}\cdot2+\frac{2-i}{9}\cdot(2+i) & \frac{2}{9}\cdot(2-i)-\frac{(2-i)}{9}\cdot 2 \\ \frac{2+i}{9}\cdot 2-\frac{2}{9}\cdot(2+i)& \frac{2+i}{9}\cdot(2-i)+\frac{2}{9}\cdot 2 \end{pmatrix}=

\begin{pmatrix} \frac{4}{9}+ \frac{2^2+i^2}{9} & \frac{4-2i}{9}-\frac{4-2i}{9}\\ \frac{4+2i}{9}-\frac{4+2i}{9} & \frac{2^{2}-i^{2}}{9}+\frac{4}{9}\end{pmatrix}=

\begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}

For any inverse matrix A^{-1} is defined as satisifying AA^{-1}=1=A^{-1}A thefore (1/9)M must be the inverse of M.

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Quantum Mechanics in Simple Matrix Form: chapter 2 of (2-28)

October 5, 2009 by theproblemyard

Note, I added letters to help distinguish between sub-exercises. Also see my friend’s post of the same exercises.

Write all answers in the form x + iy with x and y real. (note: I didn’t write the imaginary parts like they said because I think it looks weird, instead I put the number before the i)
2-1. Work out the following sums and products:

A> (3-i) + (2 + 4i) = 5 + 3i

B> (1 + 3i) + 2 = 3 + 3i

C> (-5 + 2i) - (2 + 2i)= -7 + 0i

D> (-2 + i) + (2 + 2i)= 0 + 3i

E> (3 - i)(2 + 4i) = 6 + 12i -2i + 4 = 10 + 10i

F> (1 + 3i)2 = 2 + 6i

G> i(1 + 3i)= i - 3 = -3 + i

H> (-5 + 2i)(2 + 3i) = -10 - 15i + 4i - 6 = -16 -11i

I> (2 + 3i)(-2 + 3i) = -4 + 6i - 6i - 9 = -13 + 0i

J> (2 + 3i)(3 + 2i) = 6 + 4i + 9i - 6 = 0 + 15i

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2-2. Use the formula for z^{-1} to find 1/i and 1/-i and check that your answers give i(1/i)=1 and -i(1/-i)=1.

z^{-1}=\frac{1}{x^2+y^2}(x-iy)

A> \frac{1}{i}=i^{-1}=\frac{1}{0^2+1^2}(0-i)

\frac{1}{i}=(0-i)=-i

check i(\frac{1}{i})=1 for correctness: i\cdot i(\frac{1}{i})=i\cdot(-i)=1

B> \frac{1}{-i}=(-i)^{-1}=\frac{1}{0^2+(-1)^2}(0+i)

\frac{1}{-i}=\frac{1}{1}(0+i)

\frac{1}{-i}=(0+i)=i

Check -i(\frac{1}{-i})=1 for correctness: -i(\frac{1}{-i})=-i\cdot i=1

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2-3. ….. blah blah

A> i^*i = (-i)\cdot i=1

B> (-i)^*(-i)=i \cdot -i=1

C> |i|^2=(\sqrt{0^2+1^2})^2=1

D> |-i|^2=(\sqrt{0^2+(-1)^2})^2=1

E> |i|=\sqrt{0^2+1^2}=1

F> |-i|=\sqrt{0^2+(-1)^2}=1

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2-4 Let z=3+4i………..

A> z^*=3-4i

B> z^*z=(3-4i)\cdot (3+4i)= 9+12i-12i+16=9+16=25

C> |z|=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5

D> z^2 = (3+4i)\cdot (3+4i)=9+12i+12i-16=-7+24i

E> \frac{1}{z}=\frac {1}{3+4i} = \frac {1}{3+4i}\cdot\frac{3-4i}{3-4i}=\frac{3-4i}{25}

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2-5. Show that |w+z|<|w|+|z| for w=3 and z=4i.

|w+z|<|w|+|z|

|3+4i|<|3+0i|+|0+4i|

\sqrt{3^2+4^2}<\sqrt{3^2+0} + \sqrt{0^2+(4i)^2}

\sqrt{9-16}<\sqrt{9} + \sqrt{-16}

\sqrt{-7}<3 + 4i

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2-6. User the formula for \sqrt{z} to find \sqrt{i} and \sqrt{-i} and then check that your answers give (\sqrt{i})^2=i and (\sqrt{-i})^2=-i.

\sqrt{z}=\sqrt{ \frac{\sqrt{x^2+y^2}}{2} }(\sqrt{1 + \frac{x}{\sqrt{x^2+y^2}} }+i \sqrt{1-\frac{x}{x^2+y^2}} )

For y\geq 0

\sqrt{z}=\sqrt{ \frac{\sqrt{x^2+y^2}}{2} }(-\sqrt{1 + \frac{x}{\sqrt{x^2+y^2}} }+i \sqrt{1-\frac{x}{x^2+y^2}} )

For y\leq 0

Therefore:

A> \sqrt{i} = \sqrt{0+1i}=\sqrt{ \frac{\sqrt{0^2+1^2}}{2} }(\sqrt{1 + \frac{0}{\sqrt{0^2+1^2}} }+i \sqrt{1-\frac{0}{0^2+1^2}} )

\sqrt{i} =\sqrt{ \frac{1}{2} }(1+i)

\sqrt{i} =\sqrt{ \frac{1}{2} }+i\sqrt{ \frac{1}{2} }

Check (\sqrt{i})^2=i for correctness> \sqrt{i}^2 =(\sqrt{ \frac{1}{2} }+i\sqrt{ \frac{1}{2} })^2

\sqrt{i}^2 =\frac{1}{2} + \frac{1}{2}i+ \frac{1}{2}i-\frac{1}{2}

\sqrt{i}^2 =i

B>\sqrt{-i} = \sqrt{0-1i}=

\sqrt{ \frac{\sqrt{0^2+(-1)^2}}{2} }(-\sqrt{1 + \frac{0}{\sqrt{0^2+(-1)^2}} }+i \sqrt{1-\frac{0}{0^2+(-1)^2}} )

\sqrt{-i} =\sqrt{ \frac{1}{2} }(-1+i)

\sqrt{-i} =-\sqrt{\frac{1}{2} }+i\sqrt{ \frac{1}{2} }

Check (\sqrt{-i})^2=-i for correctness> \sqrt{-i}^2 =(-\sqrt{ \frac{1}{2} }+i\sqrt{ \frac{1}{2} })^2

\sqrt{-i}^2 =\frac{1}{2} - \frac{1}{2}i- \frac{1}{2}i-\frac{1}{2}

\sqrt{-i}^2 =-i

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2-7 Show that the following are solutions to the quadratic equation az^2+bz+c=0 :

1st eq: z=-(\frac{b}{2a})+\sqrt{(\frac{b}{2a})^2-\frac{c}{a}}

2nd eq: z=-(\frac{b}{2a})-\sqrt{(\frac{b}{2a})^2-\frac{c}{a}}

I will first simplify z, and find z squared:

z=-(\frac{b}{2a})+\sqrt{\frac{b^2}{4a^2}-\frac{4ac}{4a^2}}

z=\frac{-b}{2a}+\frac{1}{2a}\sqrt{b^2-4ac}

z=\frac{1}{2a}(-b+\sqrt{b^2-4ac})

Therefore for z squared:

z^2=\frac{1}{4a^2}(-b+\sqrt{b^2-4ac})^2

z^2=\frac{1}{4a^2}(b^2-2b\sqrt{b^2-4ac}+b^2-4ac)

z^2=\frac{1}{4a^2}(2b^2-2b\sqrt{b^2-4ac}-4ac)

z^2=\frac{1}{2a^2}(b^2-b\sqrt{b^2-4ac}-2ac)

Hence we see that:

1st eq simplified: z=\frac{1}{2a}(-b+\sqrt{b^2-4ac})

2nd eq simplified: z=\frac{1}{2a}(-b-\sqrt{b^2-4ac})

1st eq squared and simplified: z^2=\frac{1}{2a^2}(b^2-b\sqrt{b^2-4ac}-2ac)

2nd eq squared and simplified: z^2=\frac{1}{2a^2}(b^2+b\sqrt{b^2-4ac}-2ac)

A (plugging 1st eq into az^2+bz+c=0)>

a\cdot \frac{1}{2a}(b^2-b\sqrt{b^2-4ac}-2ac)+b\cdot\frac{1}{2a}(-b+\sqrt{b^2-4ac})+c=0

\frac{1}{2a}(b^2-b\sqrt{b^2-4ac}-2ac-b^2+b \sqrt{b^2-4ac})+c=0

\frac{1}{2a}(b^2-2ac-b^2)+c=0

\frac{1}{2a}(-2ac)+c=0

-c+c=0

B (plugging 2nd eq into az^2+bz+c=0)>

a\cdot \frac{1}{2a}(b^2+b\sqrt{b^2-4ac}-2ac)+b\cdot\frac{1}{2a}(-b-\sqrt{b^2-4ac})+c=0

\frac{1}{2a}(b^2+b\sqrt{b^2-4ac}-2ac-b^2-b\sqrt{b^2-4ac})+c=0

\frac{1}{2a}(b\sqrt{b^2-4ac}-2ac-b\sqrt{b^2-4ac})+c=0

\frac{1}{2a}(-2ac)+c=0

-c+c=0

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Quantum Mechanics in Simple Matrix Form

October 5, 2009 by theproblemyard

This post starts my series of problems from this book:

Quantum Mechanics Book

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