THE PROBLEMYARD

Quantum Mechanics in Simple Matrix Form: chapter 5 of (2-28)

February 27, 2011

: Vectors

I use “x” for vectors that are pointing farthest to the left (that is $\pi$ radians) and I use “y” for the other vector, or rather, a vector not pointing farthest to the left

5-1.

Find x_1 y_1 + x_2 y_2 + x_3 y_3

(a)

Given:

$x_1 = 0 \newline x_2 =4 \cdot cos( \frac{\pi}{4} ) \newline x_3 = 4 \cdot sin( \frac{\pi}{4})$

$y_1 = 0 \newline y_2 = 5 \newline y_3 = 0$

> $x_1 y_1 + x_2 y_2 + x_3 y_3=$

$0 \cdot 0+ 4 \cdot \cos( \frac{ \pi}{4}) \cdot 5 + 4 \cdot \sin(\frac{\pi}{4}) \cdot 0 =$

$4 \cdot \cos( \frac{ \pi}{4}) \cdot 5 = 14.1421356$

(b)

Given:

$x_1 = 0 \newline x_2 = 0 \newline x_3 = 4$

$y_1 = 0 \newline y_2 = 5 \newline y_3 = 0$

>


_

(c)

Given:




>


_
(d)

Given:




>

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Chicken chicken chicken? …..

September 6, 2010

$Chicken chicken chicken =$

$\sqrt{\frac{1}{2}}(1+ichicken)chickens\sqrt{\frac{1}{2}}(1-ichicken)=$

$\frac{1}{2}(1+ichicken)chickens(1-ichicken)=$

$\frac{1}{2}(1+ichicken)(chickens-ichickenschicken)=$

$\frac{1}{2}(1+ichicken)[chickens-i(chickenschicken)]=$

$\frac{1}{2}(1+ichicken)[chickens-i(ichickening)]=$

$\frac{1}{2}(1+ichicken)(chickens+chickening)=$

$\frac{1}{2}[(1+ichicken)chickens+(1+ichicken)chickening)]=$

$\frac{1}{2}(chickens+ichickenchickens+chickening+ichickenchickening)=$

$\frac{1}{2}[chickens+i(chickenchickens)+chickening+i(chickenchickening)]=$

$\frac{1}{2}[chickens+i(-ichickening)+chickening+i(ichickens)]=$

$\frac{1}{2}(chickens+chickening+chickening-chickens)=$

$\frac{1}{2}(2chickening)=$

$chickening$

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Quantum Mechanics in Simple Matrix Form: chapter 4 of (2-28)

December 24, 2009

Common matrices:

$\Sigma_1 = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$

$\Sigma_2 = \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix}$

$\Sigma_3 = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

Properties:

$\Sigma_1^{\hspace{1.5 mm} 2} = 1$

$\Sigma_2^{\hspace{1.5 mm} 2}= 1$

$\Sigma_3^{\hspace{1.5 mm} 2} = 1$

$\Sigma_1\Sigma_2 = i\Sigma_3$

$\Sigma_2\Sigma_1 = -i\Sigma_3$

$\Sigma_2\Sigma_3 = i\Sigma_1$

$\Sigma_3\Sigma_2 = -i\Sigma_1$

$\Sigma_3\Sigma_1 = i\Sigma_2$

$\Sigma_1\Sigma_3 = -i\Sigma_2$

Formula for finding inverse of matrix:

$M^{-1} = \frac{1}{z_0^{\hspace{1.5 mm}2}-z_1^{\hspace{1.5 mm}2}-z_2^{\hspace{1.5 mm}2}-z_3^{\hspace{1.5 mm}2} }(z_01-z_1\Sigma_1-z_2\Sigma_2-z_3\Sigma_3)$

Formulas for z values in $z_0 1+ z_1 \Sigma_1 + z_2 \Sigma_2 + z_3 \Sigma_3$

$z_0 = \frac{1}{2}(a_{12} + a_{22}) \hspace{1 cm} z_1 = \frac{1}{2}(a_{21} + a_{12})$

$z_2 =-i\frac{1}{2}(a_{21} - a_{12})\hspace{1 cm} z_3 = \frac{1}{2}(a_{11} - a_{22})$

4-1. Use the formulas for the squares and products of the Pauli matrices to calculate

A> $[\frac{1}{2}(1+\Sigma_1)]^2=$

$\frac{1}{4}(1+\Sigma_1)^2=$

$\frac{1}{4}(1^2+2\Sigma_1+(\Sigma_1)^2)=$

$\frac{1}{4}+\frac{1}{2}\Sigma_1+\frac{1}{4}(\Sigma_1)^2=$

$\frac{1}{4}+\frac{1}{2}\Sigma_1+\frac{1}{4}=$

$\frac{1}{2}+\frac{1}{2}\Sigma_1$

B> $[\frac{1}{2}(1-\Sigma_1)]^2=$

$\frac{1}{4}(1-\Sigma_1)^2=$

$\frac{1}{4}(1^2-2\Sigma_1+(\Sigma_1)^2)=$

$\frac{1}{4}-\frac{1}{2}\Sigma_1+\frac{1}{4}(\Sigma_1)^2=$

$\frac{1}{4}-\frac{1}{2}\Sigma_1+\frac{1}{4}=$

$\frac{1}{2}-\frac{1}{2}\Sigma_1$

C> $\frac{1}{2}(1+\Sigma_1)\frac{1}{2}(1-\Sigma_1)=$

$\frac{1}{4}(1+\Sigma_1)(1-\Sigma_1)=$

$\frac{1}{4}(1^2-\Sigma_1^{\hspace{1.5 mm} 2})=$

$\frac{1}{4}(1-1)=$

$0$

D> $(i\Sigma_3)\Sigma_1(-i\Sigma_3)=$

$i(-i)\Sigma_3\Sigma_1\Sigma_3=$

$i(-i)(\Sigma_3\Sigma_1)\Sigma_3=$

$i(-i)(i\Sigma_2)\Sigma_3=$

$i(-i)i(\Sigma_2\Sigma_3)=$

$i(-i)i(i\Sigma_1)=$

$-iiii\Sigma_1=$

$ii\Sigma_1=$

$-\Sigma_1$

E> $(i\Sigma_3)\Sigma_2(-i\Sigma_3)=$

$i(-i)\Sigma_3\Sigma_2\Sigma_3=$

$i(-i)(\Sigma_3\Sigma_2)\Sigma_3=$

$i(-i)(-i\Sigma_1)\Sigma_3=$

$i(-i)(-i)(\Sigma_1\Sigma_3)=$

$i(-i)(-i)(-i\Sigma_2)=$

$(-i)(-i\Sigma_2)=$

$-\Sigma_2$

F> $\sqrt{\frac{1}{2}}(1+i\Sigma_3)\Sigma_1\sqrt{\frac{1}{2}}(1+i\Sigma_3)=$

$\frac{1}{2}(1+i\Sigma_3)\Sigma_1(1-i\Sigma_3)=$

$\frac{1}{2}(1+i\Sigma_3)[\Sigma_1-i(\Sigma_1\Sigma_3)]=$

$\frac{1}{2}(1+i\Sigma_3)[\Sigma_1-i(-i\Sigma_2)]=$

$\frac{1}{2}(1+i\Sigma_3)(\Sigma_1-\Sigma_2)=$

$\frac{1}{2}[\Sigma_1(1+i\Sigma_3)-\Sigma_2(1+i\Sigma_3)]=$

$\frac{1}{2}[\Sigma_1+i\Sigma_1\Sigma_3-\Sigma_2-i\Sigma_2\Sigma_3]=$

$\frac{1}{2}[\Sigma_1+i(\Sigma_1\Sigma_3)-\Sigma_2-i(\Sigma_2\Sigma_3)]=$

$\frac{1}{2}[\Sigma_1+i(-i\Sigma_2)-\Sigma_2-i(i\Sigma_1)]=$

$\frac{1}{2}(\Sigma_1+\Sigma_2-\Sigma_2+\Sigma_1)=$

$\frac{1}{2}(\Sigma_1+\Sigma_1)=$

$\frac{1}{2}(2\Sigma_1)=$

$\Sigma_1$

G> $\sqrt{\frac{1}{2}}(1+i\Sigma_3)\Sigma_2\sqrt{\frac{1}{2}}(1-i\Sigma_3)=$

$\frac{1}{2}(1+i\Sigma_3)\Sigma_2(1-i\Sigma_3)=$

$\frac{1}{2}(1+i\Sigma_3)(\Sigma_2-i\Sigma_2\Sigma_3)=$

$\frac{1}{2}(1+i\Sigma_3)[\Sigma_2-i(\Sigma_2\Sigma_3)]=$

$\frac{1}{2}(1+i\Sigma_3)[\Sigma_2-i(i\Sigma_1)]=$

$\frac{1}{2}(1+i\Sigma_3)(\Sigma_2+\Sigma_1)=$

$\frac{1}{2}[(1+i\Sigma_3)\Sigma_2+(1+i\Sigma_3)\Sigma_1)]=$

$\frac{1}{2}(\Sigma_2+i\Sigma_3\Sigma_2+\Sigma_1+i\Sigma_3\Sigma_1)=$

$\frac{1}{2}[\Sigma_2+i(\Sigma_3\Sigma_2)+\Sigma_1+i(\Sigma_3\Sigma_1)]=$

$\frac{1}{2}[\Sigma_2+i(-i\Sigma_1)+\Sigma_1+i(i\Sigma_2)]=$

$\frac{1}{2}(\Sigma_2+\Sigma_1+\Sigma_1-\Sigma_2)=$

$\frac{1}{2}(2\Sigma_1)=$

$\Sigma_1$

Write your answers in the form $z_0 1+ z_1 \Sigma_1 + z_2 \Sigma_2 + z_3 \Sigma_3$

4-2. Use the formulas to find the numbers z0, z1, z2, z3 for the matrix

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

Formulas:

$z_0 = \frac{1}{2}(a_{12} + a_{22}) \hspace{1 cm} z_1 = \frac{1}{2}(a_{21} + a_{12})$

$z_2 =-i\frac{1}{2}(a_{21} - a_{12})\hspace{1 cm} z_3 = \frac{1}{2}(a_{11} - a_{22})$

Plugging the “a” values in we get:

$z_0 = \frac{1}{2}(0) =0$

$z_1 = \frac{1}{2}(1) =\frac{1}{2}$

$z_2 = -i\frac{1}{2}(-1) = i \frac{1}{2}$

$z_3 = \frac{1}{2}(0) = 0$

> $\frac{1}{2} \cdot \Sigma_1 + i\frac{1}{2} \cdot \Sigma_2$

Check:

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

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

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

Test for inverse:

If $z_0^{\hspace{1.5 mm} 2} - z_1^{\hspace{1.5 mm} 2} - z_2^{\hspace{1.5 mm} 2} - z_3^{\hspace{1.5 mm} 2} \ne 0$

then there is an inverse. So we plug the z’s in:

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

$-(\frac{1}{2})^{2} - (i \frac{1}{2})^{2}=$

$-\frac{1}{4} + \frac{1}{4}=$

$0$

So sorry no inverse.

4-3 Write out the 2 by 2 matrix for

$x_1\Sigma_1 + x_2\Sigma_2 + x_3\Sigma_3$

> $x_1 \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix} + x_2 \begin{pmatrix} 0 & -i \\ i & 0 \end{pmatrix} + x_3 \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$

multiply the matrix by itself, and thus verify that

$(x_1\Sigma_1+x_2\Sigma_2+x_3\Sigma_3)^2=(x_1^{\hspace{1.5 mm} 2} + x_2^{\hspace{1.5 mm} 2} + x_3^{\hspace{1.5 mm} 2})1$

by a second method that is different from the one already used.

They mean using matrix algebra:

> $(x_1\Sigma_1+x_2\Sigma_2+x_3\Sigma_3)(x_1\Sigma_1+x_2\Sigma_2+x_3\Sigma_3)=$

$x_1^2\Sigma_1^2+x_1x_2 \Sigma_1\Sigma_2+x_1 x_3 \Sigma_1 \Sigma_3 +$

$x_2 x_1 \Sigma_2 \Sigma_1+x_2^2 \Sigma_2^2+x_2 x_3 \Sigma_2 \Sigma_3 +$

$x_3 x_1 \Sigma_3 \Sigma_1+x_3 x_2 \Sigma_3 \Sigma_2+ x_3^2 \Sigma_3^2=$

$x_1^2 + x_1 x_2 i \Sigma_3 - x_1 x_3 i \Sigma_2$

$- x_2 x_1 i \Sigma_3 + x_2^2 + x_2 x_3 i \Sigma_1$

$+ x_3 x_1 i \Sigma_2 - x_3 x_2 i \Sigma_1 + x_3 ^2 =$

$x_1^{\hspace{1.5 mm} 2}+x_2^{\hspace{1.5 mm} 2}+x_3^{\hspace{1.5 mm} 2}$

The method already used was:

> $\begin{pmatrix} x_3 & (x_1-ix_2) \\ (x_1+x_2i) & -x_2 \end{pmatrix} \cdot \begin{pmatrix} x_3 & (x_1-ix_2) \\ (x_1+x_2i) & -x_2 \end{pmatrix}=$

$\begin{pmatrix} x_3^{\hspace{1.5 mm} 2}+(x_1^2-x_2^2i)(x_1^2+x_2^2i) & x_3(x_1-x_2i)-(x_1+x_2i)x_3 \\ (x_1-x_2i)x_3-x_3(x_1+x_3) & (x_1+x_2i)(x_1-x_2i)+x_3x_3 \end{pmatrix}=$

$\begin{pmatrix} x_3^{\hspace{1.5 mm} 2}+(x_1^{\hspace{1.5 mm} 2}+x_2^{\hspace{1.5 mm} 2}) & x_3x_1+x_3x_2i-x_3x_1-x_3x_2i \\x_1x_3+x_2x_3i-x_3x_1-x_3x_2i & x_1^{\hspace{1.5 mm} 2}-x_1x_2i+x_2x_1i+ x_2^{\hspace{1.5 mm} 2} + x_3^{\hspace {1.5 mm} 2} \end{pmatrix}=$

$\begin{pmatrix} x_3^{\hspace{1.5 mm} 2}+x_1^{\hspace{1.5 mm} 2}+x_2^{\hspace{1.5 mm} 2} & 0 \\ 0 & x_1^{\hspace{1.5 mm} 2}+ x_2^{\hspace{1.5 mm} 2} + x_3^{\hspace {1.5 mm} 2} \end{pmatrix}$

4-4. Use the formula to find the inverse $M^{-1}$ for

$M = \sqrt{\frac{1}{2}}(1-i\Sigma_3)$

and then check that your answer gives

$M^{-1}M=1=MM^{-1}$

From simplifying

> $M = \sqrt{\frac{1}{2}}(1-i\Sigma_3)=$

$\sqrt{\frac{1}{2}}-\sqrt{\frac{1}{2}}i\Sigma_3$

we see what goes into the formula

$M^{-1} = \frac{1}{z_0^{\hspace{1.5 mm}2}-z_1^{\hspace{1.5 mm}2}-z_2^{\hspace{1.5 mm}2}-z_3^{\hspace{1.5 mm}2} }(z_01-z_1\Sigma_1-z_2\Sigma_2-z_3\Sigma_3)$

So plugging in we get:

> $\frac{1}{(\sqrt{\frac{1}{2}})^2-(-\sqrt{\frac{1}{2}i})^2}=$

$\frac{1}{1}(\sqrt{\frac{1}{2}}+\sqrt{\frac{1}{2}}i\Sigma_3)$

$\sqrt{\frac{1}{2}}+\sqrt{\frac{1}{2}}i\Sigma_3$

Check:

> $(\sqrt{\frac{1}{2}}+\sqrt{\frac{1}{2}}i\Sigma_3)(\sqrt{\frac{1}{2}}-\sqrt{\frac{1}{2}}i\Sigma_3)=$

$\frac{1}{2}+\frac{1}{2}\Sigma_3^{\hspace{1.5 mm}2 }=$

$\frac{1}{2}+\frac{1}{2}=$

$1$

4-5. Find all the matrices $z_0 1+ z_1 \Sigma_1 + z_2 \Sigma_2 + z_3 \Sigma_3$

that commute with $\Sigma_3$

> $\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \begin{pmatrix}1 & 0 \\ 0& -1 \end{pmatrix}=$

$\begin{pmatrix}a_{11} & -a_{21} \\ a_{12} & -a_{22} \end{pmatrix}$

> $\begin{pmatrix}1 & 0 \\ 0& -1 \end{pmatrix} \begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}=$

$\begin{pmatrix}a_{11} & a_{21} \\ -a_{12} & -a_{22} \end{pmatrix}$

We see from these two matrices that:

$a_{12}=-a_{12}$

$a_{21}=-a_{21}$

Therefore $a_{12}=0$ and $a_{21}=0$

So the set of matrices that commute would be:

> $z_01+z_3\Sigma_3 =$

$\begin{pmatrix} z_0+z_3 & 0 \\ 0 & z_0+z_3 \end{pmatrix}$

4-6 Use the formula to find the inverse for $M^{-1}$ for

$M=2\Sigma_1+\Sigma_2+2\Sigma_3$

and then check that your answer gives

$M^{-1}M=1=MM^{-1}$

So we plug into the formula we saw in 4-4 for getting the inverse:

> $\frac{1}{-2^2-1^2-2^2}(-2\Sigma_1-\Sigma_2-2\Sigma_2)=$

$\frac{1}{-9}(-2\Sigma_1-\Sigma_2-2\Sigma_2)=$

$\frac{2}{9}\Sigma_1+\frac{1}{9}\Sigma_2+ \frac{2}{9}\Sigma_2$

Check:

> $M^{-1}M=$

$(\frac{2}{9}\Sigma_1+\frac{1}{9}\Sigma_2+\frac{2}{9}\Sigma_3)(2\Sigma_1+\Sigma_2+2\Sigma_3)=$

$\frac{2}{9}\Sigma_1(2\Sigma_1+\Sigma_2+2\Sigma_3)+$

$\frac{1}{9}\Sigma_2(2\Sigma_1+\Sigma_2+2\Sigma_3)+$

$\frac{2}{9}\Sigma_3(2\Sigma_1+\Sigma_2+2\Sigma_3)=$

$\frac{4}{9}\Sigma_1\Sigma_1+\frac{2}{9}\Sigma_1\Sigma_2+\frac{4}{9}\Sigma_1\Sigma_3+$

$\frac{2}{9}\Sigma_2\Sigma_1+\frac{1}{9}\Sigma_2\Sigma_2+\frac{2}{9}\Sigma_2\Sigma_3+$

$\frac{4}{9}\Sigma_3\Sigma_1+\frac{2}{9}\Sigma_3\Sigma_2+\frac{4}{9}\Sigma_3\Sigma_3=$

$\frac{4}{9}+\frac{2}{9}i\Sigma_3-\frac{2}{9}i\Sigma_2-\frac{4}{9}\Sigma_2+\frac{1}{9}+\frac{2}{9}i\Sigma_1+\frac{4}{9}i\Sigma_2-\frac{2}{9}\Sigma_1+\frac{4}{9}=$

$\frac{4}{9}+\frac{1}{9}+\frac{4}{9}=$

$1$

> $MM^{-1}=$

$(2\Sigma_1+\Sigma_2+2\Sigma_3)(\frac{2}{9}\Sigma_1+\frac{1}{9}\Sigma_2+\frac{2}{9}\Sigma_3)=$

$2\Sigma_1(\frac{2}{9}\Sigma_1+\frac{1}{9}\Sigma_2+\frac{2}{9}\Sigma_3)+$

$\Sigma_2(\frac{2}{9}\Sigma_1+\frac{1}{9}\Sigma_2+\frac{2}{9}\Sigma_3)+$

$2\Sigma_3(\frac{2}{9}\Sigma_1+\frac{1}{9}\Sigma_2+\frac{2}{9}\Sigma_3)=$

$(\frac{4}{9}\Sigma_1\Sigma_1+\frac{2}{9}\Sigma_1\Sigma_2+\frac{4}{9}\Sigma_1\Sigma_3)+$

$(\frac{2}{9}\Sigma_2\Sigma_1+\frac{1}{9}\Sigma_2\Sigma_2+\frac{2}{9}\Sigma_2\Sigma_3)+$

$\frac{4}{9}\Sigma_3\Sigma_1+\frac{2}{9}\Sigma_3\Sigma_2+\frac{4}{9}\Sigma_3\Sigma_3)=$

$\frac{4}{9}-\frac{2}{9}i\Sigma_3-\frac{4}{9}i\Sigma_2-\frac{2}{9}i\Sigma_3+\frac{1}{9}+\frac{2}{9}i\Sigma_1+\frac{4}{9}i\Sigma_2-\frac{2}{9}\Sigma_1+\frac{4}{9}=$

$\frac{4}{9}+\frac{1}{9}+\frac{4}{9}$

$1$

Compare this with problem 3-7.

Tags: Pauli Matrices, Quantum Mechanics
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Quantum Mechanics in Simple Matrix Form: chapter 3 of (2-28)

October 19, 2009

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

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

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

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

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.

3-4. Show that the inverse of the matrix

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

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

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.

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>

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

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

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

$MAM^{-1}=A$

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

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

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)

$zBM= MzB$

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.

Tags: Imaginary Numbers, Matrices, Quantum Mechanics
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Quantum Mechanics in Simple Matrix Form: chapter 2 of (2-28)

October 5, 2009

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

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: $-i(\frac{1}{-i})=-i\cdot i=1$

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

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

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

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

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

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

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

$\sqrt{-7}<3 + 4i$

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

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

$\sqrt{-i}^2 =-i$

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 $az^2+bz+c=0$ )>

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

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

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

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

$-c+c=0$

Tags: Imaginary Numbers, Quantum Mechanics
Posted in Quantum Mechanics | Leave a Comment »

Quantum Mechanics in Simple Matrix Form

October 5, 2009

This post starts my series of problems from this book:

Tags: Quantum Mechanics
Posted in Quantum Mechanics | Leave a Comment »

THE PROBLEMYARD

Quantum Mechanics in Simple Matrix Form: chapter 5 of (2-28)

Chicken chicken chicken? …..

$Chicken chicken chicken =$

Quantum Mechanics in Simple Matrix Form: chapter 4 of (2-28)

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

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

Quantum Mechanics in Simple Matrix Form

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