Expanding the original commutator on the LHS

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Homework Statement

Using [x,e^iap]=-ħae^iap show that xⁿe^iap = e^iap(x-ħa)ⁿ

Homework Equations

[x,e^iap]=-ħae^iap
From which it follows that,
xe^iap = e^iap(x-ħa)

The Attempt at a Solution

[xⁿ,e^iap] = [xx^n-1,e^iap]
= [x,e^iap]x^n-1 + x[x^n-1,e^iap]
= -ħae^iapx^n-1 + x(x^n-1e^iap-e^iapx^n-1)
= -ħae^iapx^n-1 + xⁿe^iap - e^iap(x-ħa)x^n-1

Expanding the original commutator on the LHS and moving the second term to the RHS gives,

xⁿe^iap = -ħae^iapx^n-1 + xⁿe^iap - e^iap(x-ħa)x^n-1 + e^iapxⁿ
= -ħae^iapx^n-1 + xⁿe^iap + ħae^iapx^n-1
xⁿe^iap = xⁿe^iap Grrrrrrrrrrrrrrrr

 

[George Jones]

From which it follows that,
xe^iap = e^iap(x-ħa)

So,
$$\left( X - \hbar a \right) = e^{-iaP} X e^{iaP}.$$

Now, raise both sides to the power of ##n##.

 

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So,
$$\left( X - \hbar a \right) = e^{-iaP} X e^{iaP}.$$

Now, raise both sides to the power of ##n##.

$$\left(X - \hbar a \right)^n = e^{-niaP} X^n e^{niap}.$$

 

[DrClaude]

$$\left(X - \hbar a \right)^n = e^{-niaP} X^n e^{niap}.$$

Really? Try a simple case first: ##(e^{-iaP} X e^{iaP})^2##.

 

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Does this constitute a legitimate proof by induction?:

$$Xe^{iaP} = e^{iaP}(X-\hbar a)$$
$$X^ne^{iaP} = X^{n-1}Xe^{iap} = X^{n-1}e^{iap}(X-\hbar a) = X^{n-2}Xe^{iap}(X-\hbar a) = X^{n-2}e^{iap}(X-\hbar a)^2$$
$$= \dots$$
$$=X^{0}e^{iap}(X-\hbar a)^{n} = e^{iap}(X-\hbar a)^n$$

 

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Really? Try a simple case first: ##(e^{-iaP} X e^{iaP})^2##.

Thanks for making me feel stupid by asking if I was serious.

 

[DrClaude]

Thanks for making me feel stupid by asking if I was serious.

Saying "Really?" was a way to point out politely that you had made a mistake. It surely is better than "Wrong!", no?

 

[George Jones]

$$\left(X - \hbar a \right)^n = e^{-niaP} X^n e^{niap}.$$

Because ##X## and ##P## don't commute, this isn't correct. For example, consider ##\left(AB\right)^2 = ABAB##. If ##A## and ##B## commute, then the ##BA## in the middle can be written as ##AB##, and, consequently, ##\left(AB\right)^2 = A^2 B^2##. If ##A## and ##B## do not commute, then this step is not justified.

@DrClaude made the good suggestion of trying ##n = 2##, but let's try ##n = 3##.
$$\left( X - \hbar a \right)^3 = \left( e^{-iaP} X e^{iaP} \right)^3 = \left( e^{-iaP} X e^{iaP} \right)\left( e^{-iaP} X e^{iaP} \right)\left( e^{-iaP} X e^{iaP} \right).$$
Now, remove the backets.

Does this constitute a legitimate proof by induction?:

$$Xe^{iaP} = e^{iaP}(X-\hbar a)$$
$$X^ne^{iaP} = X^{n-1}Xe^{iap} = X^{n-1}e^{iap}(X-\hbar a) = X^{n-2}Xe^{iap}(X-\hbar a) = X^{n-2}e^{iap}(X-\hbar a)^2$$
$$= \dots$$
$$=X^{0}e^{iap}(X-\hbar a)^{n} = e^{iap}(X-\hbar a)^n$$

A more standard form of an inductive argument would be:

Let ##P\left(n\right)## be the statement ##X^n e^{iaP} = e^{iaP} \left( X - \hbar a \right)^n##. Show that ##P\left(1\right)## is true. Show that if ##P\left(k\right)## is true, then ##P\left(k+1\right)## is true.

Usually, an ellipsis does not appear in an argument that uses mathematical induction.

 

[cdot]

Because ##X## and ##P## don't commute, this isn't correct. For example, consider ##\left(AB\right)^2 = ABAB##. If ##A## and ##B## commute, then the ##BA## in the middle can be written as ##AB##, and, consequently, ##\left(AB\right)^2 = A^2 B^2##. If ##A## and ##B## do not commute, then this step is not justified.

@DrClaude made the good suggestion of trying ##n = 2##, but let's try ##n = 3##.
$$\left( X - \hbar a \right)^3 = \left( e^{-iaP} X e^{iaP} \right)^3 = \left( e^{-iaP} X e^{iaP} \right)\left( e^{-iaP} X e^{iaP} \right)\left( e^{-iaP} X e^{iaP} \right).$$
Now, remove the backets.
A more standard form of an inductive argument would be:

Let ##P\left(n\right)## be the statement ##X^n e^{iaP} = e^{iaP} \left( X - \hbar a \right)^n##. Show that ##P\left(1\right)## is true. Show that if ##P\left(k\right)## is true, then ##P\left(k+1\right)## is true.

Usually, an ellipsis does not appear in an argument that uses mathematical induction.

Ahhh that makes sense now. For practice with proof by induction as you explained it, does this work?

If ##[X,P] = i\hbar## show that ##[X,P^n] = nP^{n-1}i\hbar##

Let ##C(n) = [X,P^n]##
##~~~~~~C(1) = [X,P] = i\hbar##
## \text{If } C(k) = kP^{k-1}i\hbar \text{ is true,} ##
## C(k+1) = [X,P^{k+1}] = [X,P^k]P + P^k[X,P]##
## ~~~~~~~~~~~~~~~= (kP^{k-1}[X,P])P + P^k[X,P]##
## ~~~~~~~~~~~~~~~= kP^ki\hbar + P^ki\hbar##
## C(k+1) = (k+1)P^ki\hbar##