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

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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