- #1
NoPhysicsGenius
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The following is problem number 7 of section 2 in the book Introductory Quantum Mechanics by Vladimir Rojansky:
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.
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].
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.
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