Taylor series expansion of an exponential generates Hermite

[castrodisastro]

Homework Statement

"Show that the Hermite polynomials generated in the Taylor series expansion
e^{(2ξt - t2)} = ∑(H_n(ξ)/n!)tⁿ (starting from n=0 to ∞)
are the same as generated in 7.58*."

2. Homework Equations
*7.58 is an equation in the book "Introductory Quantum Mechanics" by Richard Liboff.
(ξ-(∂/∂ξ))ⁿ × e^-(ξ2/2) = H_n(ξ) × e^-(ξ2/2)

∑ƒⁿ(a)/n! × (x-a)ⁿ

The Attempt at a Solution

To check if I am doing things correctly, I chose n=2 and according to the book I should get

A₂(4ξ² – 2)e^-(ξ2)/2

where A₂ is a normalization constant.

I am told to Taylor Expand e^{(2ξt - t2)}
Now the Right Hand Side tells me that H_n is a function of ξ so I believe I am supposed to to apply H_n to e^(2ξt-t2) with respect to ξ.

So from the equation 7.58 from the book, if I choose n=2 then I get H₂
(ξ-(∂/∂ξ))² which equals (ξ² - ξ(∂/∂ξ) - (∂/∂ξ)(ξ) - (∂²/∂ξ²)) so if I now perform the operation
[(ξ² - ξ(∂/∂ξ) - (∂/∂ξ)(ξ) - (∂²/∂ξ²)) × e^{(2ξt - t2)}](t²/2!)

I should get A₂(4ξ² – 2) × e^-(ξ2)/2 but with e^{(2ξt - t2)} in place of e^-(ξ2)/2 correct? Well I do not.

Instead I get

(ξ²e^{(2ξt - t2)} - ξ(2t)e^{(2ξt - t2)} - e^{(2ξt - t2)} - ξ(2t)e^{(2ξt - t2)} + (2t)²e^{(2ξt - t2)}) × t²/2

I can factor out e^{(2ξt - t2)} but it doesn't do anything that would lead me to an answer. I mean, it's obvious this is incorrect since I have the variable t but it doesn't show up anywhere in the table that the Taylor expansion is supposed to correspond to.

I have looked up videos and checked textbooks for performing a Taylor expansion but they just show me how to evaluate a polynomial at a point a on the function ƒ(x) but I am explicitly given the Taylor expansion to have tⁿ instead of (t-a)ⁿ so I don't think I should just pick a random point.

I also tried using (t-(∂/∂t))ⁿ to see if maybe I was supposed to infer a change of variable to t instead of ξ but that just ended in a huge long equation that did not seem to simplify. Is my approach wrong? Please let me know.

Also, please, please, please, please...do not be rude. To say I love physics is a complete understatement. I have been and will continue to put in the work to learn as much as possible. I do not like to take shortcuts, but my calculus knowledge is lacking because I didn't realize that physics was what I wanted to do until a little bit later in life. So I ask anyone who may help me to not treat me like someone that doesn't value the process of learning by saying something along the lines of "Just look at the definition! Did you even read the book??" or one that I have seen here many times "why would you even do that?!" I would not like to get berated by those whom I automatically respect because of their knowledge of physics.

Thank you in advance.

 

[Ray Vickson]

Homework Statement

"Show that the Hermite polynomials generated in the Taylor series expansion
e^{(2ξt - t2)} = ∑(H_n(ξ)/n!)tⁿ (starting from n=0 to ∞)
are the same as generated in 7.58*."

2. Homework Equations
*7.58 is an equation in the book "Introductory Quantum Mechanics" by Richard Liboff.
(ξ-(∂/∂ξ))ⁿ × e^-(ξ2/2) = H_n(ξ) × e^-(ξ2/2)

∑ƒⁿ(a)/n! × (x-a)ⁿ

The Attempt at a Solution

To check if I am doing things correctly, I chose n=2 and according to the book I should get

A₂(4ξ² – 2)e^-(ξ2)/2

where A₂ is a normalization constant.

I am told to Taylor Expand e^{(2ξt - t2)}
Now the Right Hand Side tells me that H_n is a function of ξ so I believe I am supposed to to apply H_n to e^(2ξt-t2) with respect to ξ.

So from the equation 7.58 from the book, if I choose n=2 then I get H₂
(ξ-(∂/∂ξ))² which equals (ξ² - ξ(∂/∂ξ) - (∂/∂ξ)(ξ) - (∂²/∂ξ²)) so if I now perform the operation
[(ξ² - ξ(∂/∂ξ) - (∂/∂ξ)(ξ) - (∂²/∂ξ²)) × e^{(2ξt - t2)}](t²/2!)

******************
You are confusing the two types of calculation.

(x - \partial / \partial x)^2 f(x) = (x - \partial / \partial x) ( x f(x) - \partial f(x)/\partial x)\\<br /> = x ( x f(x) - \partial f(x)/\partial x) - \partial /\partial x \: (x f(x) - \partial f(x)/\partial x)) = \cdots
gives the $n=2$ form of 7.58 when you apply it to $f(x) = e^{-x^2/2}$.

Apply
\left(\frac{\partial}{\partial t} \right)^2 \left. e^{2 x t - t^2} \right|_{t=0}
to get the $n = 2$ term in the Taylor expansion of $e^{2 x t - t^2}$ around $t = 0$. (Evaluate the derivative first, then set $t = 0$.)

********************************I should get A₂(4ξ² – 2) × e^-(ξ2)/2 but with e^{(2ξt - t2)} in place of e^-(ξ2)/2 correct? Well I do not.

Instead I get

(ξ²e^{(2ξt - t2)} - ξ(2t)e^{(2ξt - t2)} - e^{(2ξt - t2)} - ξ(2t)e^{(2ξt - t2)} + (2t)²e^{(2ξt - t2)}) × t²/2

I can factor out e^{(2ξt - t2)} but it doesn't do anything that would lead me to an answer. I mean, it's obvious this is incorrect since I have the variable t but it doesn't show up anywhere in the table that the Taylor expansion is supposed to correspond to.

I have looked up videos and checked textbooks for performing a Taylor expansion but they just show me how to evaluate a polynomial at a point a on the function ƒ(x) but I am explicitly given the Taylor expansion to have tⁿ instead of (t-a)ⁿ so I don't think I should just pick a random point.

I also tried using (t-(∂/∂t))ⁿ to see if maybe I was supposed to infer a change of variable to t instead of ξ but that just ended in a huge long equation that did not seem to simplify. Is my approach wrong? Please let me know.

Also, please, please, please, please...do not be rude. To say I love physics is a complete understatement. I have been and will continue to put in the work to learn as much as possible. I do not like to take shortcuts, but my calculus knowledge is lacking because I didn't realize that physics was what I wanted to do until a little bit later in life. So I ask anyone who may help me to not treat me like someone that doesn't value the process of learning by saying something along the lines of "Just look at the definition! Did you even read the book??" or one that I have seen here many times "why would you even do that?!" I would not like to get berated by those whom I automatically respect because of their knowledge of physics.

******************************
You have a very odd concept of what constitutes "rudeness". To tell someone to look at the definition is not rude; you would be surprised at how many times students have trouble just because they do NOT look at, or pay attention to, definitions. Sometimes such a reminder is all it takes for them to be able to finish the problem themselves. The question about whether the student has read the book is NOT misplaced; many students do not read the book, and getting them to actually read it is often the very best learning advice that can be offered. Sometimes students come back and say it is an "internet" course, or something similar, and that there is no book, but we are not to know that unless they tell us. I have never, ever seen anyone say "why would you even do that?"

*****************************Thank you in advance.

 

[vela]

To check if I am doing things correctly, I chose n=2 and according to the book I should get

A₂(4ξ² – 2)e^-(ξ2)/2

where A₂ is a normalization constant.

If you set n=2 in 7.58, you get
$$\left(\xi - \frac{\partial}{\partial \xi}\right)^2 e^{-\xi^2/2} = H_2(\zeta)e^{-\xi^2/2}.$$ Calculate out the left hand side and compare it to the righthand size to identify what $H_2(\xi)$ is equal to.

I am told to Taylor Expand e^{(2ξt - t2)}
Now the Right Hand Side tells me that H_n is a function of ξ so I believe I am supposed to apply H_n to e^(2ξt-t2) with respect to ξ.

I'm not sure what you mean when you say "apply $H_n$ to … with respect to $\xi$ , but I suspect it's the reason for your confusion.

You're being asked to expand $f(t) = e^{2\xi t-t^2}$ in a Taylor series about t=0. Don't let the presence of $\xi$ derail you. Just treat it like a constant while finding the expansion.

Also, please, please, please, please...do not be rude. To say I love physics is a complete understatement. I have been and will continue to put in the work to learn as much as possible. I do not like to take shortcuts, but my calculus knowledge is lacking because I didn't realize that physics was what I wanted to do until a little bit later in life. So I ask anyone who may help me to not treat me like someone that doesn't value the process of learning by saying something along the lines of "Just look at the definition! Did you even read the book??" or one that I have seen here many times "why would you even do that?!" I would not like to get berated by those whom I automatically respect because of their knowledge of physics.

I suspect your lack of confidence in your math skills is causing you to focus on and read more into some comments than is really there. Students are urged to look up definitions and to consult their textbooks, not to berate them, but because it's what they should be doing at a minimum. As hard as it may be to believe, many students don't realize that these are things they can and should do on their own.

 

[castrodisastro]

You're being asked to expand $f(t) = e^{2\xi t-t^2}$ in a Taylor series about t=0. Don't let the presence of $\xi$ derail you. Just treat it like a constant while finding the expansion.

Ah ok I got it!

I must have been making mistake when taking numerous derivatives.

Thank both of you.As for the comment I had made about being berated, I think that because I am forced to type this out, as opposed to speaking, it is difficult to imitate the rude way that some tutors speak to other students. Sure there are ways to misinterpret what people say but some words used by tutors leave no other interpretation.

A tutor could have said "did you even read the book?" but instead wrote a paragraph about how students want to take shortcuts and use wikipedia and/or not put in effort. The tutor even expressed frustration at "this generation that doesn't appreciate the learning process". Now if I were a student that was putting in some effort, and I hit a wall and turn to a community for help, it doesn't help (it almost chips away at the student's confidence) for someone to assume that the effort you put in is so pitiful that the only conclusion a person can make is that you must have just glanced at wikipedia.