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

• RedMech
Thank you for your patience and help. I appreciate it.In summary, the problem statement asks to show that the operator relation e^(ipa/ħ)xe^(-ipa/ħ) = x+a holds. By substituting x=iħd/dp and using the representation x=iħd/dp, it can be shown that [e^(ipa/ħ)xe^(-ipa/ħ)]f(p) = af(p)+cf'(p) for any function f(p) and a numerical factor c. By rewriting the last term in terms of the x operator, it can be shown that e^(ipa/ħ)xe^(-ipa/ħ) is equivalent to x+a, thus
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.

RedMech said:
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?

Last edited:
RedMech said:
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.

RedMech said:
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.

TSny said:
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?

TSny said:
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.

RedMech said:
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.

TSny said:
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.

C=iħ. Giving us af(p) + iħd/dpf(p).

[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?

Yes, I think that's right.

TSny said:
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.

## 1. What is the significance of the power series representation in showing the relation of e^(ipa/ħ)xe^(-ipa/ħ)=x+a?

The power series representation is a useful tool in mathematics for understanding the behavior of functions. In this case, it allows us to break down the complex exponential function into a more manageable form, making it easier to see the relation between the two sides of the equation.

## 2. How does the use of power series help us understand the mathematical properties of e^(ipa/ħ)xe^(-ipa/ħ)=x+a?

By expanding the exponential functions into their respective power series, we can see that the coefficients of each term on both sides of the equation are equal. This demonstrates the equality of the two expressions and helps us understand the relationship between them.

## 3. Can the power series representation be used for other mathematical equations or just for e^(ipa/ħ)xe^(-ipa/ħ)=x+a?

The power series representation can be used for a wide range of mathematical equations, not just for e^(ipa/ħ)xe^(-ipa/ħ)=x+a. It is a common tool in calculus and is often used to approximate functions that cannot be easily integrated or differentiated.

## 4. What is the process for deriving the power series representation of e^(ipa/ħ)xe^(-ipa/ħ)=x+a?

The power series representation of e^(ipa/ħ)xe^(-ipa/ħ)=x+a can be derived using the Maclaurin series, which is a special case of the Taylor series. This involves finding the derivatives of the function and evaluating them at 0, then using the general formula for the Maclaurin series to express the function as a sum of infinitely many terms.

## 5. How is the power series representation of e^(ipa/ħ)xe^(-ipa/ħ)=x+a useful in physics and other scientific fields?

The power series representation of this equation is commonly used in quantum mechanics and other areas of physics to understand the behavior of particles and systems. It allows for more accurate calculations and predictions, and also provides a deeper understanding of the underlying mathematical principles involved.

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