Mixing
Exponential function of a Hermitian matrix
Clash Royale CLAN TAG#URR8PPP
up vote
2
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Given
$$H = beginpmatrixsin theta & 0 & cos theta \ 0 & 1 & 0 \ cos theta & 0 & -sin theta endpmatrix$$
where $theta=pi/6$, then what is $exp left( i fracpi2 H right)$?
I tried to calculate in the following way
$e^(ipi H)/2=[e^(ipi/2)]^H=i^H$. I do not know how to proceed.
linear-algebra matrices matrix-exponential
edited Jul 25 at 15:55
Rodrigo de Azevedo
12.6k41751
asked Jul 25 at 15:49
Purushothaman
1866
add a comment |Â
up vote
2
down vote
favorite
Given
$$H = beginpmatrixsin theta & 0 & cos theta \ 0 & 1 & 0 \ cos theta & 0 & -sin theta endpmatrix$$
where $theta=pi/6$, then what is $exp left( i fracpi2 H right)$?
I tried to calculate in the following way
$e^(ipi H)/2=[e^(ipi/2)]^H=i^H$. I do not know how to proceed.
linear-algebra matrices matrix-exponential
edited Jul 25 at 15:55
Rodrigo de Azevedo
12.6k41751
asked Jul 25 at 15:49
Purushothaman
1866
1
You need to diagonalize $H,$ if possible. Then you can use the Taylor series expansion of the exponential to finish.
– Adrian Keister
Jul 25 at 15:51
@AdrianKeister As it's a reflection matrix, it's quite easy to diagonalise, even over $Bbb R$.
– Lord Shark the Unknown
Jul 25 at 15:57
@LordSharktheUnknown: Right most indubitably ho.
– Adrian Keister
Jul 25 at 16:04
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Given
$$H = beginpmatrixsin theta & 0 & cos theta \ 0 & 1 & 0 \ cos theta & 0 & -sin theta endpmatrix$$
where $theta=pi/6$, then what is $exp left( i fracpi2 H right)$?
I tried to calculate in the following way
$e^(ipi H)/2=[e^(ipi/2)]^H=i^H$. I do not know how to proceed.
linear-algebra matrices matrix-exponential
edited Jul 25 at 15:55
Rodrigo de Azevedo
12.6k41751
asked Jul 25 at 15:49
Purushothaman
1866
Given
$$H = beginpmatrixsin theta & 0 & cos theta \ 0 & 1 & 0 \ cos theta & 0 & -sin theta endpmatrix$$
where $theta=pi/6$, then what is $exp left( i fracpi2 H right)$?
I tried to calculate in the following way
$e^(ipi H)/2=[e^(ipi/2)]^H=i^H$. I do not know how to proceed.
linear-algebra matrices matrix-exponential
edited Jul 25 at 15:55
Rodrigo de Azevedo
12.6k41751
asked Jul 25 at 15:49
Purushothaman
1866
edited Jul 25 at 15:55
Rodrigo de Azevedo
12.6k41751
edited Jul 25 at 15:55
Rodrigo de Azevedo
12.6k41751
edited Jul 25 at 15:55
Rodrigo de Azevedo
12.6k41751
12.6k41751
asked Jul 25 at 15:49
Purushothaman
1866
asked Jul 25 at 15:49
Purushothaman
1866
asked Jul 25 at 15:49
Purushothaman
1866
1866
1
You need to diagonalize $H,$ if possible. Then you can use the Taylor series expansion of the exponential to finish.
– Adrian Keister
Jul 25 at 15:51
@AdrianKeister As it's a reflection matrix, it's quite easy to diagonalise, even over $Bbb R$.
– Lord Shark the Unknown
Jul 25 at 15:57
@LordSharktheUnknown: Right most indubitably ho.
– Adrian Keister
Jul 25 at 16:04
add a comment |Â
1
You need to diagonalize $H,$ if possible. Then you can use the Taylor series expansion of the exponential to finish.
– Adrian Keister
Jul 25 at 15:51
@AdrianKeister As it's a reflection matrix, it's quite easy to diagonalise, even over $Bbb R$.
– Lord Shark the Unknown
Jul 25 at 15:57
@LordSharktheUnknown: Right most indubitably ho.
– Adrian Keister
Jul 25 at 16:04
1
1
You need to diagonalize $H,$ if possible. Then you can use the Taylor series expansion of the exponential to finish.
– Adrian Keister
Jul 25 at 15:51
You need to diagonalize $H,$ if possible. Then you can use the Taylor series expansion of the exponential to finish.
– Adrian Keister
Jul 25 at 15:51
@AdrianKeister As it's a reflection matrix, it's quite easy to diagonalise, even over $Bbb R$.
– Lord Shark the Unknown
Jul 25 at 15:57
@AdrianKeister As it's a reflection matrix, it's quite easy to diagonalise, even over $Bbb R$.
– Lord Shark the Unknown
Jul 25 at 15:57
@LordSharktheUnknown: Right most indubitably ho.
– Adrian Keister
Jul 25 at 16:04
@LordSharktheUnknown: Right most indubitably ho.
– Adrian Keister
Jul 25 at 16:04
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
3
down vote
accepted
The eigenvalues of this matrix are $1$ (double) and $-1$ (simple).
As it is real symmetric, it is diagonalisable. If $v$ is an eigenvector
then $Hv=pm v$. Therefore
$$exp(pi i H/2)v=exp(pmpi i/2)v
=pm iv=iHv.$$
Hence $exp(pi iH/2)=iH$.
edited Jul 25 at 16:07
Rodrigo de Azevedo
12.6k41751
answered Jul 25 at 16:02
Lord Shark the Unknown
84.9k950111
add a comment |Â
up vote
2
down vote
Note that $H^2=I$, thus $$ e^tH = I + tH +t^2/2!I +t^3/3!H +... $$
$$= (1+ t^2/2! +t^4/4!+...)I + (t +t^3/3! + t^5/5!+...)H $$
$$= (sinh t)I + (cosh t)H$$
For $t=(pi/2) i$ we get $$ sinh t=0 $$ and $$ cosh t=i$$ thus $$e^(pi/2) iH = iH$$
answered Jul 25 at 17:37
Mohammad Riazi-Kermani
27.4k41852
add a comment |Â
up vote
1
down vote
The eigenvalues of $H$ are $lambda =1,1,-1$
According to Cayley-Hamilton Theorem $$ e^tH=alpha I +beta H + gamma H^2$$ where $ alpha,beta,gamma$ are functions of t to be found by the equation $$ e^tlambda=alpha +beta lambda + gamma lambda^2 $$ and its derivative with repect to $lambda.$
We find $$ beta = (e^t+e^-t)/2, gamma = 1/2te^t - beta /2, alpha = e^t-beta - gamma $$
For $t=ipi /2$ we have $$ beta = i, gamma = -pi /4, alpha = pi/4 $$
Thus we get $$e^(ipi /2) H=iH$$
answered Jul 25 at 17:01
Mohammad Riazi-Kermani
27.4k41852
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The eigenvalues of this matrix are $1$ (double) and $-1$ (simple).
As it is real symmetric, it is diagonalisable. If $v$ is an eigenvector
then $Hv=pm v$. Therefore
$$exp(pi i H/2)v=exp(pmpi i/2)v
=pm iv=iHv.$$
Hence $exp(pi iH/2)=iH$.
edited Jul 25 at 16:07
Rodrigo de Azevedo
12.6k41751
answered Jul 25 at 16:02
Lord Shark the Unknown
84.9k950111
add a comment |Â
up vote
3
down vote
accepted
The eigenvalues of this matrix are $1$ (double) and $-1$ (simple).
As it is real symmetric, it is diagonalisable. If $v$ is an eigenvector
then $Hv=pm v$. Therefore
$$exp(pi i H/2)v=exp(pmpi i/2)v
=pm iv=iHv.$$
Hence $exp(pi iH/2)=iH$.
edited Jul 25 at 16:07
Rodrigo de Azevedo
12.6k41751
answered Jul 25 at 16:02
Lord Shark the Unknown
84.9k950111
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The eigenvalues of this matrix are $1$ (double) and $-1$ (simple).
As it is real symmetric, it is diagonalisable. If $v$ is an eigenvector
then $Hv=pm v$. Therefore
$$exp(pi i H/2)v=exp(pmpi i/2)v
=pm iv=iHv.$$
Hence $exp(pi iH/2)=iH$.
edited Jul 25 at 16:07
Rodrigo de Azevedo
12.6k41751
answered Jul 25 at 16:02
Lord Shark the Unknown
84.9k950111
The eigenvalues of this matrix are $1$ (double) and $-1$ (simple).
As it is real symmetric, it is diagonalisable. If $v$ is an eigenvector
then $Hv=pm v$. Therefore
$$exp(pi i H/2)v=exp(pmpi i/2)v
=pm iv=iHv.$$
Hence $exp(pi iH/2)=iH$.
edited Jul 25 at 16:07
Rodrigo de Azevedo
12.6k41751
answered Jul 25 at 16:02
Lord Shark the Unknown
84.9k950111
edited Jul 25 at 16:07
Rodrigo de Azevedo
12.6k41751
edited Jul 25 at 16:07
Rodrigo de Azevedo
12.6k41751
edited Jul 25 at 16:07
Rodrigo de Azevedo
12.6k41751
12.6k41751
answered Jul 25 at 16:02
Lord Shark the Unknown
84.9k950111
answered Jul 25 at 16:02
Lord Shark the Unknown
84.9k950111
answered Jul 25 at 16:02
Lord Shark the Unknown
84.9k950111
84.9k950111
add a comment |Â
add a comment |Â
up vote
2
down vote
Note that $H^2=I$, thus $$ e^tH = I + tH +t^2/2!I +t^3/3!H +... $$
$$= (1+ t^2/2! +t^4/4!+...)I + (t +t^3/3! + t^5/5!+...)H $$
$$= (sinh t)I + (cosh t)H$$
For $t=(pi/2) i$ we get $$ sinh t=0 $$ and $$ cosh t=i$$ thus $$e^(pi/2) iH = iH$$
answered Jul 25 at 17:37
Mohammad Riazi-Kermani
27.4k41852
add a comment |Â
up vote
2
down vote
Note that $H^2=I$, thus $$ e^tH = I + tH +t^2/2!I +t^3/3!H +... $$
$$= (1+ t^2/2! +t^4/4!+...)I + (t +t^3/3! + t^5/5!+...)H $$
$$= (sinh t)I + (cosh t)H$$
For $t=(pi/2) i$ we get $$ sinh t=0 $$ and $$ cosh t=i$$ thus $$e^(pi/2) iH = iH$$
answered Jul 25 at 17:37
Mohammad Riazi-Kermani
27.4k41852
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Note that $H^2=I$, thus $$ e^tH = I + tH +t^2/2!I +t^3/3!H +... $$
$$= (1+ t^2/2! +t^4/4!+...)I + (t +t^3/3! + t^5/5!+...)H $$
$$= (sinh t)I + (cosh t)H$$
For $t=(pi/2) i$ we get $$ sinh t=0 $$ and $$ cosh t=i$$ thus $$e^(pi/2) iH = iH$$
answered Jul 25 at 17:37
Mohammad Riazi-Kermani
27.4k41852
Note that $H^2=I$, thus $$ e^tH = I + tH +t^2/2!I +t^3/3!H +... $$
$$= (1+ t^2/2! +t^4/4!+...)I + (t +t^3/3! + t^5/5!+...)H $$
$$= (sinh t)I + (cosh t)H$$
For $t=(pi/2) i$ we get $$ sinh t=0 $$ and $$ cosh t=i$$ thus $$e^(pi/2) iH = iH$$
answered Jul 25 at 17:37
Mohammad Riazi-Kermani
27.4k41852
answered Jul 25 at 17:37
Mohammad Riazi-Kermani
27.4k41852
answered Jul 25 at 17:37
Mohammad Riazi-Kermani
27.4k41852
answered Jul 25 at 17:37
Mohammad Riazi-Kermani
27.4k41852
27.4k41852
add a comment |Â
add a comment |Â
up vote
1
down vote
The eigenvalues of $H$ are $lambda =1,1,-1$
According to Cayley-Hamilton Theorem $$ e^tH=alpha I +beta H + gamma H^2$$ where $ alpha,beta,gamma$ are functions of t to be found by the equation $$ e^tlambda=alpha +beta lambda + gamma lambda^2 $$ and its derivative with repect to $lambda.$
We find $$ beta = (e^t+e^-t)/2, gamma = 1/2te^t - beta /2, alpha = e^t-beta - gamma $$
For $t=ipi /2$ we have $$ beta = i, gamma = -pi /4, alpha = pi/4 $$
Thus we get $$e^(ipi /2) H=iH$$
answered Jul 25 at 17:01
Mohammad Riazi-Kermani
27.4k41852
add a comment |Â
up vote
1
down vote
The eigenvalues of $H$ are $lambda =1,1,-1$
According to Cayley-Hamilton Theorem $$ e^tH=alpha I +beta H + gamma H^2$$ where $ alpha,beta,gamma$ are functions of t to be found by the equation $$ e^tlambda=alpha +beta lambda + gamma lambda^2 $$ and its derivative with repect to $lambda.$
We find $$ beta = (e^t+e^-t)/2, gamma = 1/2te^t - beta /2, alpha = e^t-beta - gamma $$
For $t=ipi /2$ we have $$ beta = i, gamma = -pi /4, alpha = pi/4 $$
Thus we get $$e^(ipi /2) H=iH$$
answered Jul 25 at 17:01
Mohammad Riazi-Kermani
27.4k41852
add a comment |Â
up vote
1
down vote
up vote
1
down vote
The eigenvalues of $H$ are $lambda =1,1,-1$
According to Cayley-Hamilton Theorem $$ e^tH=alpha I +beta H + gamma H^2$$ where $ alpha,beta,gamma$ are functions of t to be found by the equation $$ e^tlambda=alpha +beta lambda + gamma lambda^2 $$ and its derivative with repect to $lambda.$
We find $$ beta = (e^t+e^-t)/2, gamma = 1/2te^t - beta /2, alpha = e^t-beta - gamma $$
For $t=ipi /2$ we have $$ beta = i, gamma = -pi /4, alpha = pi/4 $$
Thus we get $$e^(ipi /2) H=iH$$
answered Jul 25 at 17:01
Mohammad Riazi-Kermani
27.4k41852
The eigenvalues of $H$ are $lambda =1,1,-1$
According to Cayley-Hamilton Theorem $$ e^tH=alpha I +beta H + gamma H^2$$ where $ alpha,beta,gamma$ are functions of t to be found by the equation $$ e^tlambda=alpha +beta lambda + gamma lambda^2 $$ and its derivative with repect to $lambda.$
We find $$ beta = (e^t+e^-t)/2, gamma = 1/2te^t - beta /2, alpha = e^t-beta - gamma $$
For $t=ipi /2$ we have $$ beta = i, gamma = -pi /4, alpha = pi/4 $$
Thus we get $$e^(ipi /2) H=iH$$
answered Jul 25 at 17:01
Mohammad Riazi-Kermani
27.4k41852
answered Jul 25 at 17:01
Mohammad Riazi-Kermani
27.4k41852
answered Jul 25 at 17:01
Mohammad Riazi-Kermani
27.4k41852
answered Jul 25 at 17:01
Mohammad Riazi-Kermani
27.4k41852
27.4k41852
add a comment |Â
add a comment |Â
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1
You need to diagonalize $H,$ if possible. Then you can use the Taylor series expansion of the exponential to finish.
– Adrian Keister
Jul 25 at 15:51
@AdrianKeister As it's a reflection matrix, it's quite easy to diagonalise, even over $Bbb R$.
– Lord Shark the Unknown
Jul 25 at 15:57
@LordSharktheUnknown: Right most indubitably ho.
– Adrian Keister
Jul 25 at 16:04