Show that $e^{frac{-hbar}{2}partial_{a}partial_{b}}e^{frac{-ab}{hbar}}=2e^{frac{-2ab}{hbar}}$












1












$begingroup$


I think this question could be answered by only using mathematics (it relates to physics). Where, $partial_{x}f$ is denoted as partial derivative of $f$ w.r.t $x$, and first exponential term behaves like operator at some sense.



My idea is expanding both sides into power series form (Taylor series).

LHS is
$$sum_{n=0}^{infty}left(frac{-hbar}{2}right)^{n}frac{1}{n!}partial_{a}^{n}partial_{b}^{n}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}$$
Expand the first sum to calculate derivative, it will be:
$$sum_{k=0}^{infty}frac{1}{2^{k}k!}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{(l+k)!}{(l!)^{2}}$$
While, RHS is
$$2sum_{n=0}^{infty}frac{1}{n!}left(frac{-2ab}{hbar}right)^n$$
And I don't see if it is right. Or, the considered equation is not even right. If the equation is correct (by mean "equation"), is there any way to prove it?



Originally, this equation is from a physics paper by A. C. Hirshfeld



It is equation 5.20. In the paper, $a=a$ and $bar{a}=b$.



EDIT
Below is what I have done:
$$sum_{n}^{infty}left(frac{-hbar}{2}right)^{n}frac{1}{n!}partial_{a}^{n}partial_{b}^{n}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}=sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+frac{-h}{2}partial_{a}partial_{b}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+\
+left(frac{-hbar}{2}right)^{2}partial_{a}^{2}partial_{b}^{2}frac{1}{2!}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+...$$
$$=sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+frac{-hbar}{2}sum_{l=1}^{infty}left(frac{-1}{hbar}right)^{l}a^{l-1}b^{l-1}frac{l^{2}}{l!}+\
+left(frac{-hbar}{2}right)^{2}frac{1}{2!}sum_{l=2}^{infty}left(frac{-1}{hbar}right)^{l}a^{l-2}b^{l-2}frac{l^{2}(l-1)^2}{l!}+...$$
$$=sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+frac{-hbar}{2}sum_{l=1}^{infty}left(frac{-1}{hbar}right)left(frac{-1}{hbar}right)^{l-1}a^{l-1}b^{l-1}frac{l}{(l-1)!}+\
+left(frac{-hbar}{2}right)^{2}frac{1}{2!}sum_{l=2}^{infty}left(frac{-1}{hbar}right)^{2}left(frac{-1}{hbar}right)^{l-2}a^{l-2}b^{l-2}frac{l(l-1)}{(l-2)!}+...$$
$$=sum_{k=0}^{infty}frac{1}{2^{k}}frac{1}{k!}sum_{l=0}^{infty}left(frac{-1}{hbar}right)^{l}a^{l}b^{l}frac{(l+k)!}{l!^{2}}$$






power-series formal-power-series deformation-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It should be sufficient to combinatorially prove the $(-ab/hbar)^n/n!$ coefficient is the same in both expressions. I think you've made a mistake somewhere, because if you haven't we'd need $2^{n+1}=sum_{kge 0}frac{binom{n}{k}}{2^k}$. But by the binomial theorem, the right-hand side is $1.5^n$.
    $endgroup$
    – J.G.
    Jan 12 at 13:07










  • $begingroup$
    I don't know, but I checked, there is no mistake. I added what I have done.
    $endgroup$
    – Duong H.D Hoang
    Jan 13 at 6:31
















1












$begingroup$


I think this question could be answered by only using mathematics (it relates to physics). Where, $partial_{x}f$ is denoted as partial derivative of $f$ w.r.t $x$, and first exponential term behaves like operator at some sense.



My idea is expanding both sides into power series form (Taylor series).

LHS is
$$sum_{n=0}^{infty}left(frac{-hbar}{2}right)^{n}frac{1}{n!}partial_{a}^{n}partial_{b}^{n}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}$$
Expand the first sum to calculate derivative, it will be:
$$sum_{k=0}^{infty}frac{1}{2^{k}k!}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{(l+k)!}{(l!)^{2}}$$
While, RHS is
$$2sum_{n=0}^{infty}frac{1}{n!}left(frac{-2ab}{hbar}right)^n$$
And I don't see if it is right. Or, the considered equation is not even right. If the equation is correct (by mean "equation"), is there any way to prove it?



Originally, this equation is from a physics paper by A. C. Hirshfeld



It is equation 5.20. In the paper, $a=a$ and $bar{a}=b$.



EDIT
Below is what I have done:
$$sum_{n}^{infty}left(frac{-hbar}{2}right)^{n}frac{1}{n!}partial_{a}^{n}partial_{b}^{n}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}=sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+frac{-h}{2}partial_{a}partial_{b}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+\
+left(frac{-hbar}{2}right)^{2}partial_{a}^{2}partial_{b}^{2}frac{1}{2!}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+...$$
$$=sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+frac{-hbar}{2}sum_{l=1}^{infty}left(frac{-1}{hbar}right)^{l}a^{l-1}b^{l-1}frac{l^{2}}{l!}+\
+left(frac{-hbar}{2}right)^{2}frac{1}{2!}sum_{l=2}^{infty}left(frac{-1}{hbar}right)^{l}a^{l-2}b^{l-2}frac{l^{2}(l-1)^2}{l!}+...$$
$$=sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+frac{-hbar}{2}sum_{l=1}^{infty}left(frac{-1}{hbar}right)left(frac{-1}{hbar}right)^{l-1}a^{l-1}b^{l-1}frac{l}{(l-1)!}+\
+left(frac{-hbar}{2}right)^{2}frac{1}{2!}sum_{l=2}^{infty}left(frac{-1}{hbar}right)^{2}left(frac{-1}{hbar}right)^{l-2}a^{l-2}b^{l-2}frac{l(l-1)}{(l-2)!}+...$$
$$=sum_{k=0}^{infty}frac{1}{2^{k}}frac{1}{k!}sum_{l=0}^{infty}left(frac{-1}{hbar}right)^{l}a^{l}b^{l}frac{(l+k)!}{l!^{2}}$$






power-series formal-power-series deformation-theory







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It should be sufficient to combinatorially prove the $(-ab/hbar)^n/n!$ coefficient is the same in both expressions. I think you've made a mistake somewhere, because if you haven't we'd need $2^{n+1}=sum_{kge 0}frac{binom{n}{k}}{2^k}$. But by the binomial theorem, the right-hand side is $1.5^n$.
    $endgroup$
    – J.G.
    Jan 12 at 13:07










  • $begingroup$
    I don't know, but I checked, there is no mistake. I added what I have done.
    $endgroup$
    – Duong H.D Hoang
    Jan 13 at 6:31














1












1








1


0



$begingroup$


I think this question could be answered by only using mathematics (it relates to physics). Where, $partial_{x}f$ is denoted as partial derivative of $f$ w.r.t $x$, and first exponential term behaves like operator at some sense.



My idea is expanding both sides into power series form (Taylor series).

LHS is
$$sum_{n=0}^{infty}left(frac{-hbar}{2}right)^{n}frac{1}{n!}partial_{a}^{n}partial_{b}^{n}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}$$
Expand the first sum to calculate derivative, it will be:
$$sum_{k=0}^{infty}frac{1}{2^{k}k!}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{(l+k)!}{(l!)^{2}}$$
While, RHS is
$$2sum_{n=0}^{infty}frac{1}{n!}left(frac{-2ab}{hbar}right)^n$$
And I don't see if it is right. Or, the considered equation is not even right. If the equation is correct (by mean "equation"), is there any way to prove it?



Originally, this equation is from a physics paper by A. C. Hirshfeld



It is equation 5.20. In the paper, $a=a$ and $bar{a}=b$.



EDIT
Below is what I have done:
$$sum_{n}^{infty}left(frac{-hbar}{2}right)^{n}frac{1}{n!}partial_{a}^{n}partial_{b}^{n}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}=sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+frac{-h}{2}partial_{a}partial_{b}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+\
+left(frac{-hbar}{2}right)^{2}partial_{a}^{2}partial_{b}^{2}frac{1}{2!}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+...$$
$$=sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+frac{-hbar}{2}sum_{l=1}^{infty}left(frac{-1}{hbar}right)^{l}a^{l-1}b^{l-1}frac{l^{2}}{l!}+\
+left(frac{-hbar}{2}right)^{2}frac{1}{2!}sum_{l=2}^{infty}left(frac{-1}{hbar}right)^{l}a^{l-2}b^{l-2}frac{l^{2}(l-1)^2}{l!}+...$$
$$=sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+frac{-hbar}{2}sum_{l=1}^{infty}left(frac{-1}{hbar}right)left(frac{-1}{hbar}right)^{l-1}a^{l-1}b^{l-1}frac{l}{(l-1)!}+\
+left(frac{-hbar}{2}right)^{2}frac{1}{2!}sum_{l=2}^{infty}left(frac{-1}{hbar}right)^{2}left(frac{-1}{hbar}right)^{l-2}a^{l-2}b^{l-2}frac{l(l-1)}{(l-2)!}+...$$
$$=sum_{k=0}^{infty}frac{1}{2^{k}}frac{1}{k!}sum_{l=0}^{infty}left(frac{-1}{hbar}right)^{l}a^{l}b^{l}frac{(l+k)!}{l!^{2}}$$






power-series formal-power-series deformation-theory







share|cite|improve this question











$endgroup$




I think this question could be answered by only using mathematics (it relates to physics). Where, $partial_{x}f$ is denoted as partial derivative of $f$ w.r.t $x$, and first exponential term behaves like operator at some sense.



My idea is expanding both sides into power series form (Taylor series).

LHS is
$$sum_{n=0}^{infty}left(frac{-hbar}{2}right)^{n}frac{1}{n!}partial_{a}^{n}partial_{b}^{n}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}$$
Expand the first sum to calculate derivative, it will be:
$$sum_{k=0}^{infty}frac{1}{2^{k}k!}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{(l+k)!}{(l!)^{2}}$$
While, RHS is
$$2sum_{n=0}^{infty}frac{1}{n!}left(frac{-2ab}{hbar}right)^n$$
And I don't see if it is right. Or, the considered equation is not even right. If the equation is correct (by mean "equation"), is there any way to prove it?



Originally, this equation is from a physics paper by A. C. Hirshfeld



It is equation 5.20. In the paper, $a=a$ and $bar{a}=b$.



EDIT
Below is what I have done:
$$sum_{n}^{infty}left(frac{-hbar}{2}right)^{n}frac{1}{n!}partial_{a}^{n}partial_{b}^{n}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}=sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+frac{-h}{2}partial_{a}partial_{b}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+\
+left(frac{-hbar}{2}right)^{2}partial_{a}^{2}partial_{b}^{2}frac{1}{2!}sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+...$$
$$=sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+frac{-hbar}{2}sum_{l=1}^{infty}left(frac{-1}{hbar}right)^{l}a^{l-1}b^{l-1}frac{l^{2}}{l!}+\
+left(frac{-hbar}{2}right)^{2}frac{1}{2!}sum_{l=2}^{infty}left(frac{-1}{hbar}right)^{l}a^{l-2}b^{l-2}frac{l^{2}(l-1)^2}{l!}+...$$
$$=sum_{l=0}^{infty}left(frac{-ab}{hbar}right)^{l}frac{1}{l!}+frac{-hbar}{2}sum_{l=1}^{infty}left(frac{-1}{hbar}right)left(frac{-1}{hbar}right)^{l-1}a^{l-1}b^{l-1}frac{l}{(l-1)!}+\
+left(frac{-hbar}{2}right)^{2}frac{1}{2!}sum_{l=2}^{infty}left(frac{-1}{hbar}right)^{2}left(frac{-1}{hbar}right)^{l-2}a^{l-2}b^{l-2}frac{l(l-1)}{(l-2)!}+...$$
$$=sum_{k=0}^{infty}frac{1}{2^{k}}frac{1}{k!}sum_{l=0}^{infty}left(frac{-1}{hbar}right)^{l}a^{l}b^{l}frac{(l+k)!}{l!^{2}}$$






power-series formal-power-series deformation-theory




power-series formal-power-series deformation-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 13 at 6:28







Duong H.D Hoang

















asked Jan 12 at 11:47









Duong H.D HoangDuong H.D Hoang

556




556








  • 1




    $begingroup$
    It should be sufficient to combinatorially prove the $(-ab/hbar)^n/n!$ coefficient is the same in both expressions. I think you've made a mistake somewhere, because if you haven't we'd need $2^{n+1}=sum_{kge 0}frac{binom{n}{k}}{2^k}$. But by the binomial theorem, the right-hand side is $1.5^n$.
    $endgroup$
    – J.G.
    Jan 12 at 13:07










  • $begingroup$
    I don't know, but I checked, there is no mistake. I added what I have done.
    $endgroup$
    – Duong H.D Hoang
    Jan 13 at 6:31














  • 1




    $begingroup$
    It should be sufficient to combinatorially prove the $(-ab/hbar)^n/n!$ coefficient is the same in both expressions. I think you've made a mistake somewhere, because if you haven't we'd need $2^{n+1}=sum_{kge 0}frac{binom{n}{k}}{2^k}$. But by the binomial theorem, the right-hand side is $1.5^n$.
    $endgroup$
    – J.G.
    Jan 12 at 13:07










  • $begingroup$
    I don't know, but I checked, there is no mistake. I added what I have done.
    $endgroup$
    – Duong H.D Hoang
    Jan 13 at 6:31








1




1




$begingroup$
It should be sufficient to combinatorially prove the $(-ab/hbar)^n/n!$ coefficient is the same in both expressions. I think you've made a mistake somewhere, because if you haven't we'd need $2^{n+1}=sum_{kge 0}frac{binom{n}{k}}{2^k}$. But by the binomial theorem, the right-hand side is $1.5^n$.
$endgroup$
– J.G.
Jan 12 at 13:07




$begingroup$
It should be sufficient to combinatorially prove the $(-ab/hbar)^n/n!$ coefficient is the same in both expressions. I think you've made a mistake somewhere, because if you haven't we'd need $2^{n+1}=sum_{kge 0}frac{binom{n}{k}}{2^k}$. But by the binomial theorem, the right-hand side is $1.5^n$.
$endgroup$
– J.G.
Jan 12 at 13:07












$begingroup$
I don't know, but I checked, there is no mistake. I added what I have done.
$endgroup$
– Duong H.D Hoang
Jan 13 at 6:31




$begingroup$
I don't know, but I checked, there is no mistake. I added what I have done.
$endgroup$
– Duong H.D Hoang
Jan 13 at 6:31










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