# QM - Operator Algebra Problem 2.7 Rojansky

1. Nov 24, 2005

### NoPhysicsGenius

The following is problem number 7 of section 2 in the book Introductory Quantum Mechanics by Vladimir Rojansky:
Show by operator algebra that, if $$\alpha\beta - \beta\alpha = 1$$, then $$\alpha \beta^2 - \beta^2 \alpha = 2 \beta$$, $$\alpha \beta^3 - \beta^3 \alpha = 3 \beta^2$$, and, in general, $$\alpha \beta^n - \beta^n \alpha = n \beta^{n-1}$$.
I have no problems proving the first two statements.
To prove the first statement, we do the following:
$$\alpha \beta - \beta \alpha = 1$$
$$\Rightarrow ( \alpha \beta - \beta \alpha ) \beta = 1 (\beta)$$
$$\Rightarrow \alpha \beta^2 - \beta \alpha \beta = \beta$$
$$\Rightarrow \alpha \beta^2 - \beta ( 1 + \beta \alpha ) = \beta$$
$$\Rightarrow \alpha \beta^2 - \beta - \beta^2 \alpha = \beta$$
$$\Rightarrow \alpha \beta^2 - \beta^2 \alpha = 2 \beta$$
$$\square$$
To prove the second statement, we start with the conclusion of the first statement:
$$\alpha \beta^2 - \beta^2 \alpha = 2 \beta$$
$$\Rightarrow ( \alpha \beta^2 - \beta^2 \alpha ) \beta = 2 \beta ( \beta )$$
$$\Rightarrow \alpha \beta^3 - \beta^2 \alpha \beta = 2 \beta^2$$
$$\Rightarrow \alpha \beta^3 - \beta^2 ( 1 + \beta \alpha ) = 2 \beta^2$$
$$\Rightarrow \alpha \beta^3 - \beta^2 - \beta^3 \alpha = 2 \beta^2$$
$$\Rightarrow \alpha \beta^3 - \beta^3 \alpha = 3 \beta^2$$
$$\square$$
However, I am having problems proving the final statement. Here's what I've got:
$$\alpha \beta - \beta \alpha = 1$$
$$\Rightarrow ( \alpha \beta - \beta \alpha ) \beta^{n-1} = \beta^{n-1}$$
$$\Rightarrow \alpha \beta^n - \beta \alpha \beta^{n-1} = \beta^{n-1}$$
$$\Rightarrow \alpha \beta^n - \beta \alpha \beta \beta^{n-2} = \beta^{n-1}$$
$$\Rightarrow \alpha \beta^n - \beta ( 1 + \beta \alpha ) \beta^{n-2} = \beta^{n-1}$$
$$\Rightarrow \alpha \beta^n - \beta \beta^{n-2} - \beta^2 \alpha \beta^{n-2} = \beta^{n-1}$$
$$\Rightarrow \alpha \beta^n - \beta^2 \alpha \beta^{n-2} = 2 \beta^{n-1}$$
I feel like I'm going in circles! I'm clearly missing some sort of mathematical technique here ... (Is it sort of like induction?)

Last edited: Nov 24, 2005
2. Nov 24, 2005

### Physics Monkey

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 $$\alpha \beta^n - \beta^n \alpha = n \beta^{n-1}$$. 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. Nov 24, 2005

### NoPhysicsGenius

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 $$P(n)$$ be a statement about the positive integer $$n$$. Suppose that
1. $$P(1)$$ is true.
2. Whenever $$k$$ is a positive integer for which $$P(k)$$ is true, then $$P(k + 1)$$ is also true.​
Then $$P(n)$$ is true for every positive integer $$n$$.
Therefore, let $$P(n)$$ be the statement $$\alpha \beta^n - \beta^n \alpha = n \beta^{n-1}$$.
We note that the statement $$P(1)$$, which is $$\alpha \beta - \beta \alpha = 1$$, is true, since it was given.
Suppose now that $$k$$ is a positive integer for which $$P(k)$$ is true. Then $$\alpha \beta^k - \beta^k \alpha = k \beta^{k-1}$$. (This is the induction hypothesis.)
We now want to establish that $$P(k + 1)$$ is true, which is equivalent to showing that:
$$\alpha \beta^{k+1} - \beta^{k+1} \alpha = (k+1) \beta^k$$.
Multiplying both sides of the induction hypothesis $$P(k)$$ by $$\beta$$, we get:
$$( \alpha \beta^k - \beta^k \alpha ) \beta = ( k \beta^{k-1} ) \beta$$
$$\Rightarrow \alpha \beta^{k+1} - \beta^k \alpha \beta = k \beta^k$$
$$\Rightarrow \alpha \beta^{k+1} - \beta^k ( 1 + \beta \alpha ) = k \beta^k$$
$$\Rightarrow \alpha \beta^{k+1} - \beta^k - \beta^k \beta \alpha = k \beta^k$$
$$\Rightarrow \alpha \beta^{k+1} - \beta^{k+1} \alpha = k \beta^k + \beta^k$$
$$\Rightarrow \alpha \beta^{k+1} - \beta^{k+1} \alpha = (k+1) \beta^k$$
Hence $$P(k+1)$$ is true. It now follows by induction that $$P(n)$$ is true for every positive integer $$n$$.
There ... That wasn't so bad!