QM - Operator Algebra Problem 2.7 Rojansky

In summary, the first two statements of the problem are true, but the third statement is not. The proof of the third statement is missing.
  • #1
NoPhysicsGenius
58
0
The following is problem number 7 of section 2 in the book 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].
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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  • #2
It's exactly like induction. For some reason you tried to prove the general statement by going all the way back to the first statement, so in some sense you are going in circles! Do you like you did for the others and start with [tex] \alpha \beta^n - \beta^n \alpha = n \beta^{n-1} [/tex]. Follow the standard induction proof and try to show that the n+1 statement is true. You should use the same couple of steps that you used in your previous two calculations.
 
  • #3
Physics Monkey said:
It's exactly like induction. For some reason you tried to prove the general statement by going all the way back to the first statement, so in some sense you are going in circles! Do you like you did for the others and start with [tex] \alpha \beta^n - \beta^n \alpha = n \beta^{n-1} [/tex]. Follow the standard induction proof and try to show that the n+1 statement is true. You should use the same couple of steps that you used in your previous two calculations.
According to An Introduction To Abstract Mathematics by Robert J. Bond and William J. Keane, the First Principle of Mathematical Induction is as follows:
Let [tex]P(n)[/tex] be a statement about the positive integer [tex]n[/tex]. Suppose that
1. [tex]P(1)[/tex] is true.
2. Whenever [tex]k[/tex] is a positive integer for which [tex]P(k)[/tex] is true, then [tex]P(k + 1)[/tex] is also true.​
Then [tex]P(n)[/tex] is true for every positive integer [tex]n[/tex].
Therefore, let [tex]P(n)[/tex] be the statement [tex]\alpha \beta^n - \beta^n \alpha = n \beta^{n-1}[/tex].
We note that the statement [tex]P(1)[/tex], which is [tex]\alpha \beta - \beta \alpha = 1[/tex], is true, since it was given.
Suppose now that [tex]k[/tex] is a positive integer for which [tex]P(k)[/tex] is true. Then [tex]\alpha \beta^k - \beta^k \alpha = k \beta^{k-1}[/tex]. (This is the induction hypothesis.)
We now want to establish that [tex]P(k + 1)[/tex] is true, which is equivalent to showing that:
[tex]\alpha \beta^{k+1} - \beta^{k+1} \alpha = (k+1) \beta^k[/tex].
Multiplying both sides of the induction hypothesis [tex]P(k)[/tex] by [tex]\beta[/tex], we get:
[tex]( \alpha \beta^k - \beta^k \alpha ) \beta = ( k \beta^{k-1} ) \beta[/tex]
[tex]\Rightarrow \alpha \beta^{k+1} - \beta^k \alpha \beta = k \beta^k[/tex]
[tex]\Rightarrow \alpha \beta^{k+1} - \beta^k ( 1 + \beta \alpha ) = k \beta^k[/tex]
[tex]\Rightarrow \alpha \beta^{k+1} - \beta^k - \beta^k \beta \alpha = k \beta^k[/tex]
[tex]\Rightarrow \alpha \beta^{k+1} - \beta^{k+1} \alpha = k \beta^k + \beta^k[/tex]
[tex]\Rightarrow \alpha \beta^{k+1} - \beta^{k+1} \alpha = (k+1) \beta^k[/tex]
Hence [tex]P(k+1)[/tex] is true. It now follows by induction that [tex]P(n)[/tex] is true for every positive integer [tex]n[/tex].
There ... That wasn't so bad!
Thanks for your help.
 
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1. What is QM - Operator Algebra Problem 2.7 Rojansky?

QM - Operator Algebra Problem 2.7 Rojansky is a specific problem in quantum mechanics that involves using operator algebra to solve a mathematical equation. It is commonly used as a practice problem in physics and applied mathematics courses.

2. What is operator algebra?

Operator algebra is a branch of mathematics that deals with the manipulation and properties of operators, which are mathematical functions that act on a mathematical space. In quantum mechanics, operators represent physical observables, such as position, momentum, and energy.

3. What is the goal of QM - Operator Algebra Problem 2.7 Rojansky?

The goal of QM - Operator Algebra Problem 2.7 Rojansky is to use operator algebra techniques to find the eigenstates and eigenvalues of a given operator. This allows us to solve for the physical observables of a quantum system, giving us a better understanding of its behavior.

4. What are some common applications of operator algebra?

Operator algebra is used extensively in quantum mechanics, as well as in other fields such as functional analysis, differential equations, and control theory. It is also applied in engineering, physics, and computer science to solve problems related to linear and nonlinear systems.

5. Are there any resources available to help with QM - Operator Algebra Problem 2.7 Rojansky?

Yes, there are many online resources, textbooks, and practice problems available to help with QM - Operator Algebra Problem 2.7 Rojansky. Additionally, seeking help from a professor or tutor can also be beneficial in understanding the concepts and techniques involved in solving this type of problem.

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