Mixing
Time evolution of a finite dim. quantum system
$begingroup$
We will study the time-evolution of a finite dimensional quantum system. To this end, let us consider a quantum mechanical system with the Hilbert space $mathbb{C}^2$. We denote by $left . left | 0 right rangleright .$ and $left . left | 1 right rangleright .$ the standard basis elements $(1,0)^T$ and $(0,1)^T$. Let the Hamiltonian of the system in this basis be given by
$$
H=begin{pmatrix}
0 &-i \
-i &0
end{pmatrix}
$$
and assume that for $t=0$ the state of the system is just given by $psi(t=0)=left . left | 0 right rangleright .$. In the following, we also assume natural units in which $hbar=1$.
Problems:
a) Determine the eigenvalues $lambda_i$ and the normalized eigenvectors $f_i$ of $H$.
b) Compute the time evolution operator $U(t)=e^{-iHt}$ of the system, according to $f(H)=sum_{i=1}^{2}f(lambda_i)left . left | e_i right rangleright . left langle left . e_i right | right .$
for $f$ and analytic function for all $lambda_i$. Compute the time evolved state $psi(t)=U(t)psi(t=0)$.
I do not understand what to do for problem b). I need help for this one. For a), I found out that the eigenvalues $lambda_i$ are $pm 1$, and the normalized eigenvectors $f_i$ are $frac{1}{sqrt 2}begin{pmatrix}
i\
1
end{pmatrix}$ and $frac{1}{sqrt 2}begin{pmatrix}
-i\
1
end{pmatrix}$
hilbert-spaces mathematical-physics quantum-mechanics
asked Jan 31 at 16:25
UnknownWUnknownW
1,045922
$endgroup$
|
show 6 more comments
$begingroup$
We will study the time-evolution of a finite dimensional quantum system. To this end, let us consider a quantum mechanical system with the Hilbert space $mathbb{C}^2$. We denote by $left . left | 0 right rangleright .$ and $left . left | 1 right rangleright .$ the standard basis elements $(1,0)^T$ and $(0,1)^T$. Let the Hamiltonian of the system in this basis be given by
$$
H=begin{pmatrix}
0 &-i \
-i &0
end{pmatrix}
$$
and assume that for $t=0$ the state of the system is just given by $psi(t=0)=left . left | 0 right rangleright .$. In the following, we also assume natural units in which $hbar=1$.
Problems:
a) Determine the eigenvalues $lambda_i$ and the normalized eigenvectors $f_i$ of $H$.
b) Compute the time evolution operator $U(t)=e^{-iHt}$ of the system, according to $f(H)=sum_{i=1}^{2}f(lambda_i)left . left | e_i right rangleright . left langle left . e_i right | right .$
for $f$ and analytic function for all $lambda_i$. Compute the time evolved state $psi(t)=U(t)psi(t=0)$.
I do not understand what to do for problem b). I need help for this one. For a), I found out that the eigenvalues $lambda_i$ are $pm 1$, and the normalized eigenvectors $f_i$ are $frac{1}{sqrt 2}begin{pmatrix}
i\
1
end{pmatrix}$ and $frac{1}{sqrt 2}begin{pmatrix}
-i\
1
end{pmatrix}$
hilbert-spaces mathematical-physics quantum-mechanics
asked Jan 31 at 16:25
UnknownWUnknownW
1,045922
$endgroup$
$begingroup$
You need to diagonalize $H$. If you can write $H=PDP^{-1},$ with $D$ diagonal and $P$ orthogonal, then it's much more straight-forward to take the matrix exponential $e^{-iHt}:;e^{-iHt}=e^{-iPDP^{-1}t}=Pe^{-iDt}P^{-1}.$
$endgroup$
– Adrian Keister
Jan 31 at 16:37
$begingroup$
@AdrianKeister Thank you for your comment. I figured out that $$ H=begin{pmatrix}-1&1\ 1&1end{pmatrix}begin{pmatrix}i&0\ 0&-iend{pmatrix}begin{pmatrix}-frac{1}{2}&frac{1}{2}\ frac{1}{2}&frac{1}{2}end{pmatrix} $$
$endgroup$
– UnknownW
Jan 31 at 17:20
$begingroup$
Looks like you're on the right track!
$endgroup$
– Adrian Keister
Jan 31 at 17:26
$begingroup$
@AdrianKeister That's good to know. I'm still confused to what exactly I am doing. What connection does it have with the $f(H)$? I apologize if the questions seem easy as I am kind of new to this subject.
$endgroup$
– UnknownW
Jan 31 at 18:15
$begingroup$
The formula given for $f(H)$ corresponds to my $e^{-iHt}=Pe^{-iDt}P^{-1},$ because you can essentially act on the eigenvalues of the matrix instead of on the matrix itself, once you've diagonalized it. The entries on the diagonal matrix $D$ are the eigenvalues.
$endgroup$
– Adrian Keister
Jan 31 at 18:20
|
show 6 more comments
$begingroup$
We will study the time-evolution of a finite dimensional quantum system. To this end, let us consider a quantum mechanical system with the Hilbert space $mathbb{C}^2$. We denote by $left . left | 0 right rangleright .$ and $left . left | 1 right rangleright .$ the standard basis elements $(1,0)^T$ and $(0,1)^T$. Let the Hamiltonian of the system in this basis be given by
$$
H=begin{pmatrix}
0 &-i \
-i &0
end{pmatrix}
$$
and assume that for $t=0$ the state of the system is just given by $psi(t=0)=left . left | 0 right rangleright .$. In the following, we also assume natural units in which $hbar=1$.
Problems:
a) Determine the eigenvalues $lambda_i$ and the normalized eigenvectors $f_i$ of $H$.
b) Compute the time evolution operator $U(t)=e^{-iHt}$ of the system, according to $f(H)=sum_{i=1}^{2}f(lambda_i)left . left | e_i right rangleright . left langle left . e_i right | right .$
for $f$ and analytic function for all $lambda_i$. Compute the time evolved state $psi(t)=U(t)psi(t=0)$.
I do not understand what to do for problem b). I need help for this one. For a), I found out that the eigenvalues $lambda_i$ are $pm 1$, and the normalized eigenvectors $f_i$ are $frac{1}{sqrt 2}begin{pmatrix}
i\
1
end{pmatrix}$ and $frac{1}{sqrt 2}begin{pmatrix}
-i\
1
end{pmatrix}$
hilbert-spaces mathematical-physics quantum-mechanics
asked Jan 31 at 16:25
UnknownWUnknownW
1,045922
$endgroup$
We will study the time-evolution of a finite dimensional quantum system. To this end, let us consider a quantum mechanical system with the Hilbert space $mathbb{C}^2$. We denote by $left . left | 0 right rangleright .$ and $left . left | 1 right rangleright .$ the standard basis elements $(1,0)^T$ and $(0,1)^T$. Let the Hamiltonian of the system in this basis be given by
$$
H=begin{pmatrix}
0 &-i \
-i &0
end{pmatrix}
$$
and assume that for $t=0$ the state of the system is just given by $psi(t=0)=left . left | 0 right rangleright .$. In the following, we also assume natural units in which $hbar=1$.
Problems:
a) Determine the eigenvalues $lambda_i$ and the normalized eigenvectors $f_i$ of $H$.
b) Compute the time evolution operator $U(t)=e^{-iHt}$ of the system, according to $f(H)=sum_{i=1}^{2}f(lambda_i)left . left | e_i right rangleright . left langle left . e_i right | right .$
for $f$ and analytic function for all $lambda_i$. Compute the time evolved state $psi(t)=U(t)psi(t=0)$.
I do not understand what to do for problem b). I need help for this one. For a), I found out that the eigenvalues $lambda_i$ are $pm 1$, and the normalized eigenvectors $f_i$ are $frac{1}{sqrt 2}begin{pmatrix}
i\
1
end{pmatrix}$ and $frac{1}{sqrt 2}begin{pmatrix}
-i\
1
end{pmatrix}$
hilbert-spaces mathematical-physics quantum-mechanics
hilbert-spaces mathematical-physics quantum-mechanics
asked Jan 31 at 16:25
UnknownWUnknownW
1,045922
asked Jan 31 at 16:25
UnknownWUnknownW
1,045922
asked Jan 31 at 16:25
UnknownWUnknownW
1,045922
asked Jan 31 at 16:25
UnknownWUnknownW
1,045922
asked Jan 31 at 16:25
UnknownWUnknownW
1,045922
1,045922
$begingroup$
You need to diagonalize $H$. If you can write $H=PDP^{-1},$ with $D$ diagonal and $P$ orthogonal, then it's much more straight-forward to take the matrix exponential $e^{-iHt}:;e^{-iHt}=e^{-iPDP^{-1}t}=Pe^{-iDt}P^{-1}.$
$endgroup$
– Adrian Keister
Jan 31 at 16:37
$begingroup$
@AdrianKeister Thank you for your comment. I figured out that $$ H=begin{pmatrix}-1&1\ 1&1end{pmatrix}begin{pmatrix}i&0\ 0&-iend{pmatrix}begin{pmatrix}-frac{1}{2}&frac{1}{2}\ frac{1}{2}&frac{1}{2}end{pmatrix} $$
$endgroup$
– UnknownW
Jan 31 at 17:20
$begingroup$
Looks like you're on the right track!
$endgroup$
– Adrian Keister
Jan 31 at 17:26
$begingroup$
@AdrianKeister That's good to know. I'm still confused to what exactly I am doing. What connection does it have with the $f(H)$? I apologize if the questions seem easy as I am kind of new to this subject.
$endgroup$
– UnknownW
Jan 31 at 18:15
$begingroup$
The formula given for $f(H)$ corresponds to my $e^{-iHt}=Pe^{-iDt}P^{-1},$ because you can essentially act on the eigenvalues of the matrix instead of on the matrix itself, once you've diagonalized it. The entries on the diagonal matrix $D$ are the eigenvalues.
$endgroup$
– Adrian Keister
Jan 31 at 18:20
|
show 6 more comments
$begingroup$
You need to diagonalize $H$. If you can write $H=PDP^{-1},$ with $D$ diagonal and $P$ orthogonal, then it's much more straight-forward to take the matrix exponential $e^{-iHt}:;e^{-iHt}=e^{-iPDP^{-1}t}=Pe^{-iDt}P^{-1}.$
$endgroup$
– Adrian Keister
Jan 31 at 16:37
$begingroup$
@AdrianKeister Thank you for your comment. I figured out that $$ H=begin{pmatrix}-1&1\ 1&1end{pmatrix}begin{pmatrix}i&0\ 0&-iend{pmatrix}begin{pmatrix}-frac{1}{2}&frac{1}{2}\ frac{1}{2}&frac{1}{2}end{pmatrix} $$
$endgroup$
– UnknownW
Jan 31 at 17:20
$begingroup$
Looks like you're on the right track!
$endgroup$
– Adrian Keister
Jan 31 at 17:26
$begingroup$
@AdrianKeister That's good to know. I'm still confused to what exactly I am doing. What connection does it have with the $f(H)$? I apologize if the questions seem easy as I am kind of new to this subject.
$endgroup$
– UnknownW
Jan 31 at 18:15
$begingroup$
The formula given for $f(H)$ corresponds to my $e^{-iHt}=Pe^{-iDt}P^{-1},$ because you can essentially act on the eigenvalues of the matrix instead of on the matrix itself, once you've diagonalized it. The entries on the diagonal matrix $D$ are the eigenvalues.
$endgroup$
– Adrian Keister
Jan 31 at 18:20
$begingroup$
You need to diagonalize $H$. If you can write $H=PDP^{-1},$ with $D$ diagonal and $P$ orthogonal, then it's much more straight-forward to take the matrix exponential $e^{-iHt}:;e^{-iHt}=e^{-iPDP^{-1}t}=Pe^{-iDt}P^{-1}.$
$endgroup$
– Adrian Keister
Jan 31 at 16:37
$begingroup$
You need to diagonalize $H$. If you can write $H=PDP^{-1},$ with $D$ diagonal and $P$ orthogonal, then it's much more straight-forward to take the matrix exponential $e^{-iHt}:;e^{-iHt}=e^{-iPDP^{-1}t}=Pe^{-iDt}P^{-1}.$
$endgroup$
– Adrian Keister
Jan 31 at 16:37
$begingroup$
@AdrianKeister Thank you for your comment. I figured out that $$ H=begin{pmatrix}-1&1\ 1&1end{pmatrix}begin{pmatrix}i&0\ 0&-iend{pmatrix}begin{pmatrix}-frac{1}{2}&frac{1}{2}\ frac{1}{2}&frac{1}{2}end{pmatrix} $$
$endgroup$
– UnknownW
Jan 31 at 17:20
$begingroup$
@AdrianKeister Thank you for your comment. I figured out that $$ H=begin{pmatrix}-1&1\ 1&1end{pmatrix}begin{pmatrix}i&0\ 0&-iend{pmatrix}begin{pmatrix}-frac{1}{2}&frac{1}{2}\ frac{1}{2}&frac{1}{2}end{pmatrix} $$
$endgroup$
– UnknownW
Jan 31 at 17:20
$begingroup$
Looks like you're on the right track!
$endgroup$
– Adrian Keister
Jan 31 at 17:26
$begingroup$
Looks like you're on the right track!
$endgroup$
– Adrian Keister
Jan 31 at 17:26
$begingroup$
@AdrianKeister That's good to know. I'm still confused to what exactly I am doing. What connection does it have with the $f(H)$? I apologize if the questions seem easy as I am kind of new to this subject.
$endgroup$
– UnknownW
Jan 31 at 18:15
$begingroup$
@AdrianKeister That's good to know. I'm still confused to what exactly I am doing. What connection does it have with the $f(H)$? I apologize if the questions seem easy as I am kind of new to this subject.
$endgroup$
– UnknownW
Jan 31 at 18:15
$begingroup$
The formula given for $f(H)$ corresponds to my $e^{-iHt}=Pe^{-iDt}P^{-1},$ because you can essentially act on the eigenvalues of the matrix instead of on the matrix itself, once you've diagonalized it. The entries on the diagonal matrix $D$ are the eigenvalues.
$endgroup$
– Adrian Keister
Jan 31 at 18:20
$begingroup$
The formula given for $f(H)$ corresponds to my $e^{-iHt}=Pe^{-iDt}P^{-1},$ because you can essentially act on the eigenvalues of the matrix instead of on the matrix itself, once you've diagonalized it. The entries on the diagonal matrix $D$ are the eigenvalues.
$endgroup$
– Adrian Keister
Jan 31 at 18:20
|
show 6 more comments
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$begingroup$
You need to diagonalize $H$. If you can write $H=PDP^{-1},$ with $D$ diagonal and $P$ orthogonal, then it's much more straight-forward to take the matrix exponential $e^{-iHt}:;e^{-iHt}=e^{-iPDP^{-1}t}=Pe^{-iDt}P^{-1}.$
$endgroup$
– Adrian Keister
Jan 31 at 16:37
$begingroup$
@AdrianKeister Thank you for your comment. I figured out that $$ H=begin{pmatrix}-1&1\ 1&1end{pmatrix}begin{pmatrix}i&0\ 0&-iend{pmatrix}begin{pmatrix}-frac{1}{2}&frac{1}{2}\ frac{1}{2}&frac{1}{2}end{pmatrix} $$
$endgroup$
– UnknownW
Jan 31 at 17:20
$begingroup$
Looks like you're on the right track!
$endgroup$
– Adrian Keister
Jan 31 at 17:26
$begingroup$
@AdrianKeister That's good to know. I'm still confused to what exactly I am doing. What connection does it have with the $f(H)$? I apologize if the questions seem easy as I am kind of new to this subject.
$endgroup$
– UnknownW
Jan 31 at 18:15
$begingroup$
The formula given for $f(H)$ corresponds to my $e^{-iHt}=Pe^{-iDt}P^{-1},$ because you can essentially act on the eigenvalues of the matrix instead of on the matrix itself, once you've diagonalized it. The entries on the diagonal matrix $D$ are the eigenvalues.
$endgroup$
– Adrian Keister
Jan 31 at 18:20