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The following is problem number 7 of section 2 in the book

I have no problems proving the first two statements.

To prove the first statement, we do the following:

[tex]\alpha \beta - \beta \alpha = 1[/tex]

[tex]\Rightarrow ( \alpha \beta - \beta \alpha ) \beta = 1 (\beta)[/tex]

[tex]\Rightarrow \alpha \beta^2 - \beta \alpha \beta = \beta[/tex]

[tex]\Rightarrow \alpha \beta^2 - \beta ( 1 + \beta \alpha ) = \beta[/tex]

[tex]\Rightarrow \alpha \beta^2 - \beta - \beta^2 \alpha = \beta[/tex]

[tex]\Rightarrow \alpha \beta^2 - \beta^2 \alpha = 2 \beta [/tex]

[tex]\square[/tex]

To prove the second statement, we start with the conclusion of the first statement:

[tex] \alpha \beta^2 - \beta^2 \alpha = 2 \beta [/tex]

[tex]\Rightarrow ( \alpha \beta^2 - \beta^2 \alpha ) \beta = 2 \beta ( \beta )[/tex]

[tex]\Rightarrow \alpha \beta^3 - \beta^2 \alpha \beta = 2 \beta^2[/tex]

[tex]\Rightarrow \alpha \beta^3 - \beta^2 ( 1 + \beta \alpha ) = 2 \beta^2[/tex]

[tex]\Rightarrow \alpha \beta^3 - \beta^2 - \beta^3 \alpha = 2 \beta^2[/tex]

[tex]\Rightarrow \alpha \beta^3 - \beta^3 \alpha = 3 \beta^2[/tex]

[tex]\square[/tex]

However, I am having problems proving the final statement. Here's what I've got:

[tex]\alpha \beta - \beta \alpha = 1[/tex]

[tex]\Rightarrow ( \alpha \beta - \beta \alpha ) \beta^{n-1} = \beta^{n-1}[/tex]

[tex]\Rightarrow \alpha \beta^n - \beta \alpha \beta^{n-1} = \beta^{n-1}[/tex]

[tex]\Rightarrow \alpha \beta^n - \beta \alpha \beta \beta^{n-2} = \beta^{n-1}[/tex]

[tex]\Rightarrow \alpha \beta^n - \beta ( 1 + \beta \alpha ) \beta^{n-2} = \beta^{n-1}[/tex]

[tex]\Rightarrow \alpha \beta^n - \beta \beta^{n-2} - \beta^2 \alpha \beta^{n-2} = \beta^{n-1}[/tex]

[tex]\Rightarrow \alpha \beta^n - \beta^2 \alpha \beta^{n-2} = 2 \beta^{n-1}[/tex]

I feel like I'm going in circles! I'm clearly missing some sort of mathematical technique here ... (Is it sort of like induction?)

Please help. Thank you.

__Introductory Quantum Mechanics__by Vladimir Rojansky:**Show by operator algebra that, if [tex]\alpha\beta - \beta\alpha = 1[/tex], then [tex]\alpha \beta^2 - \beta^2 \alpha = 2 \beta[/tex], [tex]\alpha \beta^3 - \beta^3 \alpha = 3 \beta^2[/tex], and, in general, [tex]\alpha \beta^n - \beta^n \alpha = n \beta^{n-1}[/tex].**

To prove the first statement, we do the following:

[tex]\alpha \beta - \beta \alpha = 1[/tex]

[tex]\Rightarrow ( \alpha \beta - \beta \alpha ) \beta = 1 (\beta)[/tex]

[tex]\Rightarrow \alpha \beta^2 - \beta \alpha \beta = \beta[/tex]

[tex]\Rightarrow \alpha \beta^2 - \beta ( 1 + \beta \alpha ) = \beta[/tex]

[tex]\Rightarrow \alpha \beta^2 - \beta - \beta^2 \alpha = \beta[/tex]

[tex]\Rightarrow \alpha \beta^2 - \beta^2 \alpha = 2 \beta [/tex]

[tex]\square[/tex]

To prove the second statement, we start with the conclusion of the first statement:

[tex] \alpha \beta^2 - \beta^2 \alpha = 2 \beta [/tex]

[tex]\Rightarrow ( \alpha \beta^2 - \beta^2 \alpha ) \beta = 2 \beta ( \beta )[/tex]

[tex]\Rightarrow \alpha \beta^3 - \beta^2 \alpha \beta = 2 \beta^2[/tex]

[tex]\Rightarrow \alpha \beta^3 - \beta^2 ( 1 + \beta \alpha ) = 2 \beta^2[/tex]

[tex]\Rightarrow \alpha \beta^3 - \beta^2 - \beta^3 \alpha = 2 \beta^2[/tex]

[tex]\Rightarrow \alpha \beta^3 - \beta^3 \alpha = 3 \beta^2[/tex]

[tex]\square[/tex]

However, I am having problems proving the final statement. Here's what I've got:

[tex]\alpha \beta - \beta \alpha = 1[/tex]

[tex]\Rightarrow ( \alpha \beta - \beta \alpha ) \beta^{n-1} = \beta^{n-1}[/tex]

[tex]\Rightarrow \alpha \beta^n - \beta \alpha \beta^{n-1} = \beta^{n-1}[/tex]

[tex]\Rightarrow \alpha \beta^n - \beta \alpha \beta \beta^{n-2} = \beta^{n-1}[/tex]

[tex]\Rightarrow \alpha \beta^n - \beta ( 1 + \beta \alpha ) \beta^{n-2} = \beta^{n-1}[/tex]

[tex]\Rightarrow \alpha \beta^n - \beta \beta^{n-2} - \beta^2 \alpha \beta^{n-2} = \beta^{n-1}[/tex]

[tex]\Rightarrow \alpha \beta^n - \beta^2 \alpha \beta^{n-2} = 2 \beta^{n-1}[/tex]

I feel like I'm going in circles! I'm clearly missing some sort of mathematical technique here ... (Is it sort of like induction?)

Please help. Thank you.

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