Showing Relation of e^(ipa/ħ)xe^(-ipa/ħ)=x+a Using Power Series

[RedMech]

1. The problem statement:

Show that if the operator relation

e^(ipa/ħ)xe^(-ipa/ħ) = x+a

holds. The operator e^A is defined to the
∞
e^A= Ʃ(A^n)/n!
n=0

[Hint: Calculate e^(ipa/ħ)xe^(-ipa/ħ)f(p) where f(p)is any function of p, and use the representation x=iħd/dp]


2. Homework Equations :

I am not entirely sure but I think those presented in the problem statement are sufficient.


3. The Attempt at a Solution :

I calculated e^(ipa/ħ)xe^(-ipa/ħ)f(p) by replacing x=iħd/dp to get;

e^(ipa/ħ)iħd/dp[e^(-ipa/ħ)f(p)]= e^(ipa/ħ){iħ[(-ia/ħ)e^(-ipa/ħ)f(p)+e^(-ipa/ħ)f'(p)]}
= e^(ipa/ħ){ae^(-ipa/ħ)f(p)+e^(-ipa/ħ)f'(p)]}
= af(p)+f'(p).
∴ af(p)+f'(p) = x + a.

From this point I become entirely confused, I don't know how everything is suppose to tie into the power series.

 

[TSny]

1. The problem statement:

Show that if the operator relation

e^(ipa/ħ)xe^(-ipa/ħ) = x+a

holds. The operator e^A is defined to the
∞
e^A= Ʃ(A^n)/n!
n=0

Hello RedMech. From the wording of the problem, I'm not sure what the question is. Could you restate what it is you are suppose to show?

 

[RedMech]

Hello RedMech. From the wording of the problem, I'm not sure what the question is. Could you restate what it is you are suppose to show?

I copied it word for word.
It's question 8 on this page in this book,
http://books.google.co.zm/books?id=...ulate the N needed to normalize ψ (x)&f=false

 

[TSny]

I copied it word for word.

Well, not quite. You added the word "if" that threw me a bit. But, no problem. Your link cleared it up.

I calculated e^(ipa/ħ)xe^(-ipa/ħ)f(p) by replacing x=iħd/dp to get;

e^(ipa/ħ)iħd/dp[e^(-ipa/ħ)f(p)]= e^(ipa/ħ){iħ[(-ia/ħ)e^(-ipa/ħ)f(p)+e^(-ipa/ħ)f'(p)]}
= e^(ipa/ħ){ae^(-ipa/ħ)f(p)+e^(-ipa/ħ)f'(p)]}
= af(p)+f'(p).

This looks good except for a missing numerical factor for the part shown in red above.
Your result will then have the form af(p)+cf'(p) where c is the missing factor. See what happens if you then rewrite f'(p) in terms of the x operator acting on f(p).

From this point I become entirely confused, I don't know how everything is suppose to tie into the power series.

You don't need to use the power series. I think the problem statement threw that in just to state the meaning of e^A.

 

[RedMech]

Well, not quite. You added the word "if" that threw me a bit. But, no problem. Your link cleared it up.



This looks good except for a missing numerical factor for the part shown in red above.
Your result will then have the form af(p)+cf'(p) where c is the missing factor. See what happens if you then rewrite f'(p) in terms of the x operator acting on f(p).



You don't need to use the power series. I think the problem statement threw that in just to state the meaning of e^A.

So af(p)+cf'(p), would be my final solution. I pay no attention to the power series?

 

[RedMech]

See what happens if you then rewrite f'(p) in terms of the x operator acting on f(p).

Not sure I know what you mean by this. The phrasing of the question has thrown me off so much that I don't know what I am looking for.

 

[TSny]

So af(p)+cf'(p), would be my final solution. I pay no attention to the power series?

No, that's not the final solution. You are asked to show that e^(ipa/ħ)xe^(-ipa/ħ) = x+a. So far you have shown that [e^(ipa/ħ)xe^(-ipa/ħ)]f(p) = af(p)+cf'(p) for any function f(p) and where c is the factor you dropped (you'll need to go back and find the value of c). You still need to show that e^(ipa/ħ)xe^(-ipa/ħ) is equivalent to x+a. But, you are almost there. You just need to rewrite the last term of af(p)+cf'(p) in terms of the operator x.

You will not need the power series.

 

[RedMech]

You still need to show that e^(ipa/ħ)xe^(-ipa/ħ) is equivalent to x+a. But, you are almost there. You just need to rewrite the last term of af(p)+cf'(p) in terms of the operator x.

Okay, I'm with you now. Let me see what I can do and I'll come back with my result.

 

[RedMech]

C=iħ. Giving us af(p) + iħd/dpf(p).

Leading to
[e^(ipa/ħ)xe^(-ipa/ħ)]f(p)= [a + iħd/dp]f(p).

∴e^(ipa/ħ)xe^(-ipa/ħ)=a + iħd/dp as required. Does this seem right?

 

[TSny]

Yes, I think that's right.

 

[RedMech]

Yes, I think that's right.

Thank you so much. This makes sense. I see why I needed to bring in the f(p) function. My careless differentiation mistake and that power series confused me.