Proof of identity using polynomial operator and commutators
Suppose x is a real number, and $hat{A}$ and $hat{B}$ are two non-commuting operators.
A polynomial operator is defined by
$$g(hat{B}) = sum_{n=0}^infty a_n hat{B}^n,$$
where $a_n$ are real numbers.
If [$hat{A}$ , $hat{B}$] = c, where c is a complex number, prove that
$$exp(xhat{A})g(hat{B})exp(-xhat{A}) = g(hat{B} + cx).$$
A previous proof yielded the taylor expansion of a similar expression, which led me to believe it could be the key, thus i started at
$$exp(xhat{A})g(hat{B})exp(-xhat{A}) = sum_{n=0}^infty [hat{A},g(hat{B})]_n frac{x^n}{n!}.$$
Here, $[hat{A} , hat{B}]_n$ := $[hat{A},[hat{A},hat{B}]]_{n-1}$, describes the nested commutator, with $[hat{A},hat{B}]_0 := hat{B}.$
Sadly, I could not make it work as of now. Am I missing an obvious or simple solution? Any help is greatly appreciated.
polynomials operator-theory quantum-mechanics
asked Mar 17 '18 at 0:14
Fred Frog
111
add a comment |
Suppose x is a real number, and $hat{A}$ and $hat{B}$ are two non-commuting operators.
A polynomial operator is defined by
$$g(hat{B}) = sum_{n=0}^infty a_n hat{B}^n,$$
where $a_n$ are real numbers.
If [$hat{A}$ , $hat{B}$] = c, where c is a complex number, prove that
$$exp(xhat{A})g(hat{B})exp(-xhat{A}) = g(hat{B} + cx).$$
A previous proof yielded the taylor expansion of a similar expression, which led me to believe it could be the key, thus i started at
$$exp(xhat{A})g(hat{B})exp(-xhat{A}) = sum_{n=0}^infty [hat{A},g(hat{B})]_n frac{x^n}{n!}.$$
Here, $[hat{A} , hat{B}]_n$ := $[hat{A},[hat{A},hat{B}]]_{n-1}$, describes the nested commutator, with $[hat{A},hat{B}]_0 := hat{B}.$
Sadly, I could not make it work as of now. Am I missing an obvious or simple solution? Any help is greatly appreciated.
polynomials operator-theory quantum-mechanics
asked Mar 17 '18 at 0:14
Fred Frog
111
add a comment |
Suppose x is a real number, and $hat{A}$ and $hat{B}$ are two non-commuting operators.
A polynomial operator is defined by
$$g(hat{B}) = sum_{n=0}^infty a_n hat{B}^n,$$
where $a_n$ are real numbers.
If [$hat{A}$ , $hat{B}$] = c, where c is a complex number, prove that
$$exp(xhat{A})g(hat{B})exp(-xhat{A}) = g(hat{B} + cx).$$
A previous proof yielded the taylor expansion of a similar expression, which led me to believe it could be the key, thus i started at
$$exp(xhat{A})g(hat{B})exp(-xhat{A}) = sum_{n=0}^infty [hat{A},g(hat{B})]_n frac{x^n}{n!}.$$
Here, $[hat{A} , hat{B}]_n$ := $[hat{A},[hat{A},hat{B}]]_{n-1}$, describes the nested commutator, with $[hat{A},hat{B}]_0 := hat{B}.$
Sadly, I could not make it work as of now. Am I missing an obvious or simple solution? Any help is greatly appreciated.
polynomials operator-theory quantum-mechanics
asked Mar 17 '18 at 0:14
Fred Frog
111
Suppose x is a real number, and $hat{A}$ and $hat{B}$ are two non-commuting operators.
A polynomial operator is defined by
$$g(hat{B}) = sum_{n=0}^infty a_n hat{B}^n,$$
where $a_n$ are real numbers.
If [$hat{A}$ , $hat{B}$] = c, where c is a complex number, prove that
$$exp(xhat{A})g(hat{B})exp(-xhat{A}) = g(hat{B} + cx).$$
A previous proof yielded the taylor expansion of a similar expression, which led me to believe it could be the key, thus i started at
$$exp(xhat{A})g(hat{B})exp(-xhat{A}) = sum_{n=0}^infty [hat{A},g(hat{B})]_n frac{x^n}{n!}.$$
Here, $[hat{A} , hat{B}]_n$ := $[hat{A},[hat{A},hat{B}]]_{n-1}$, describes the nested commutator, with $[hat{A},hat{B}]_0 := hat{B}.$
Sadly, I could not make it work as of now. Am I missing an obvious or simple solution? Any help is greatly appreciated.
polynomials operator-theory quantum-mechanics
polynomials operator-theory quantum-mechanics
asked Mar 17 '18 at 0:14
Fred Frog
111
asked Mar 17 '18 at 0:14
Fred Frog
111
asked Mar 17 '18 at 0:14
Fred Frog
111
asked Mar 17 '18 at 0:14
Fred Frog
111
asked Mar 17 '18 at 0:14
Fred Frog
111
111
add a comment |
add a comment |
1 Answer
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You are essentially there!
Observe $hat A/c$ acts like a derivative operator on powers of $hat B$, and so $g(hat B)$ as well,
$$
[hat A, g(hat B)]= c g^{(1)}(hat B), qquad Longrightarrow \
[hat A, g(hat B)]_n= c^n g^{(n)} (hat B),
$$
where the superscript in parenthesis denotes number of derivatives;
so that, substituting in your final (Duhamel) expression readily yields the Taylor expansion at $hat B$,
$$exp(xhat{A})g(hat{B})exp(-xhat{A}) = sum_{n=0}^infty [hat{A},g(hat{B})]_n ~ frac{x^n}{n!}\
= sum_{n=0}^infty g^{(n)} (hat B) ~frac{c^n x^n}{n!} = g(hat B + cx)
~ .$$
(Historically, this was Baker's minority contribution to the CBH expansion algorithm.)
answered Dec 26 '18 at 17:08
Cosmas Zachos
1,543520
add a comment |
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1 Answer
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1 Answer
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active
oldest
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You are essentially there!
Observe $hat A/c$ acts like a derivative operator on powers of $hat B$, and so $g(hat B)$ as well,
$$
[hat A, g(hat B)]= c g^{(1)}(hat B), qquad Longrightarrow \
[hat A, g(hat B)]_n= c^n g^{(n)} (hat B),
$$
where the superscript in parenthesis denotes number of derivatives;
so that, substituting in your final (Duhamel) expression readily yields the Taylor expansion at $hat B$,
$$exp(xhat{A})g(hat{B})exp(-xhat{A}) = sum_{n=0}^infty [hat{A},g(hat{B})]_n ~ frac{x^n}{n!}\
= sum_{n=0}^infty g^{(n)} (hat B) ~frac{c^n x^n}{n!} = g(hat B + cx)
~ .$$
(Historically, this was Baker's minority contribution to the CBH expansion algorithm.)
answered Dec 26 '18 at 17:08
Cosmas Zachos
1,543520
add a comment |
You are essentially there!
Observe $hat A/c$ acts like a derivative operator on powers of $hat B$, and so $g(hat B)$ as well,
$$
[hat A, g(hat B)]= c g^{(1)}(hat B), qquad Longrightarrow \
[hat A, g(hat B)]_n= c^n g^{(n)} (hat B),
$$
where the superscript in parenthesis denotes number of derivatives;
so that, substituting in your final (Duhamel) expression readily yields the Taylor expansion at $hat B$,
$$exp(xhat{A})g(hat{B})exp(-xhat{A}) = sum_{n=0}^infty [hat{A},g(hat{B})]_n ~ frac{x^n}{n!}\
= sum_{n=0}^infty g^{(n)} (hat B) ~frac{c^n x^n}{n!} = g(hat B + cx)
~ .$$
(Historically, this was Baker's minority contribution to the CBH expansion algorithm.)
answered Dec 26 '18 at 17:08
Cosmas Zachos
1,543520
add a comment |
You are essentially there!
Observe $hat A/c$ acts like a derivative operator on powers of $hat B$, and so $g(hat B)$ as well,
$$
[hat A, g(hat B)]= c g^{(1)}(hat B), qquad Longrightarrow \
[hat A, g(hat B)]_n= c^n g^{(n)} (hat B),
$$
where the superscript in parenthesis denotes number of derivatives;
so that, substituting in your final (Duhamel) expression readily yields the Taylor expansion at $hat B$,
$$exp(xhat{A})g(hat{B})exp(-xhat{A}) = sum_{n=0}^infty [hat{A},g(hat{B})]_n ~ frac{x^n}{n!}\
= sum_{n=0}^infty g^{(n)} (hat B) ~frac{c^n x^n}{n!} = g(hat B + cx)
~ .$$
(Historically, this was Baker's minority contribution to the CBH expansion algorithm.)
answered Dec 26 '18 at 17:08
Cosmas Zachos
1,543520
You are essentially there!
Observe $hat A/c$ acts like a derivative operator on powers of $hat B$, and so $g(hat B)$ as well,
$$
[hat A, g(hat B)]= c g^{(1)}(hat B), qquad Longrightarrow \
[hat A, g(hat B)]_n= c^n g^{(n)} (hat B),
$$
where the superscript in parenthesis denotes number of derivatives;
so that, substituting in your final (Duhamel) expression readily yields the Taylor expansion at $hat B$,
$$exp(xhat{A})g(hat{B})exp(-xhat{A}) = sum_{n=0}^infty [hat{A},g(hat{B})]_n ~ frac{x^n}{n!}\
= sum_{n=0}^infty g^{(n)} (hat B) ~frac{c^n x^n}{n!} = g(hat B + cx)
~ .$$
(Historically, this was Baker's minority contribution to the CBH expansion algorithm.)
answered Dec 26 '18 at 17:08
Cosmas Zachos
1,543520
edited Dec 26 '18 at 20:49
answered Dec 26 '18 at 17:08
Cosmas Zachos
1,543520
answered Dec 26 '18 at 17:08
Cosmas Zachos
1,543520
answered Dec 26 '18 at 17:08
Cosmas Zachos
1,543520
1,543520
add a comment |
add a comment |
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