Quantum Toffoli gate equation Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Does quantum control allow to implement any gate?Obtaining gate $e^-iDelta t Z$ from elementary gatesExplicit Conversion Between Universal Gate SetsUnderstanding the Group Leaders Optimization AlgorithmMatrix representation and CX gateComposing the CNOT gate as a tensor product of two level matricesRewrite circuit with measurements with unitariesHow to understand the operators for watermarking schemes?Implementing these $N×N$ matrices on $log N$ qubitsCalculating entries of unitary transformation
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Quantum Toffoli gate equation
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Does quantum control allow to implement any gate?Obtaining gate $e^-iDelta t Z$ from elementary gatesExplicit Conversion Between Universal Gate SetsUnderstanding the Group Leaders Optimization AlgorithmMatrix representation and CX gateComposing the CNOT gate as a tensor product of two level matricesRewrite circuit with measurements with unitariesHow to understand the operators for watermarking schemes?Implementing these $N×N$ matrices on $log N$ qubitsCalculating entries of unitary transformation
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
I was reading a research article on quantum computing and didn't understand the tensor notations for the unitary operations. The article defined two controlled gates.
Let $U_2^m$ be a $2^m times 2^m$ unitary matrix, $I_2^m$ be a $2^m times 2^m$ identity matrix. Then, controlled gates $C_n^j(U_2^m)$ and $V_n^j(U_2^m)$ with $n$ control qubits and $m$ target qubits are defined by $$ C_n^j(U_2^m)=(|jrangle langle j|) otimes U_2^m+ sum_i=0,i neq j^2^n-1((|irangle langle i| otimes I_2^m$$
$$ V_n^j(U_2^m) = U_2^m otimes (|jrangle langle j|) + sum_i=0,i neq j^2^n-1( I_2^m otimes (|irangle langle i| ))$$
Then they say that $C_2^j(X)$ and $V_2^j(X) $are toffoli gates.
Can someone explain the equations that are given
and how does this special case be a Toffoli?
quantum-gate tensor-product
$endgroup$
add a comment |
$begingroup$
I was reading a research article on quantum computing and didn't understand the tensor notations for the unitary operations. The article defined two controlled gates.
Let $U_2^m$ be a $2^m times 2^m$ unitary matrix, $I_2^m$ be a $2^m times 2^m$ identity matrix. Then, controlled gates $C_n^j(U_2^m)$ and $V_n^j(U_2^m)$ with $n$ control qubits and $m$ target qubits are defined by $$ C_n^j(U_2^m)=(|jrangle langle j|) otimes U_2^m+ sum_i=0,i neq j^2^n-1((|irangle langle i| otimes I_2^m$$
$$ V_n^j(U_2^m) = U_2^m otimes (|jrangle langle j|) + sum_i=0,i neq j^2^n-1( I_2^m otimes (|irangle langle i| ))$$
Then they say that $C_2^j(X)$ and $V_2^j(X) $are toffoli gates.
Can someone explain the equations that are given
and how does this special case be a Toffoli?
quantum-gate tensor-product
$endgroup$
add a comment |
$begingroup$
I was reading a research article on quantum computing and didn't understand the tensor notations for the unitary operations. The article defined two controlled gates.
Let $U_2^m$ be a $2^m times 2^m$ unitary matrix, $I_2^m$ be a $2^m times 2^m$ identity matrix. Then, controlled gates $C_n^j(U_2^m)$ and $V_n^j(U_2^m)$ with $n$ control qubits and $m$ target qubits are defined by $$ C_n^j(U_2^m)=(|jrangle langle j|) otimes U_2^m+ sum_i=0,i neq j^2^n-1((|irangle langle i| otimes I_2^m$$
$$ V_n^j(U_2^m) = U_2^m otimes (|jrangle langle j|) + sum_i=0,i neq j^2^n-1( I_2^m otimes (|irangle langle i| ))$$
Then they say that $C_2^j(X)$ and $V_2^j(X) $are toffoli gates.
Can someone explain the equations that are given
and how does this special case be a Toffoli?
quantum-gate tensor-product
$endgroup$
I was reading a research article on quantum computing and didn't understand the tensor notations for the unitary operations. The article defined two controlled gates.
Let $U_2^m$ be a $2^m times 2^m$ unitary matrix, $I_2^m$ be a $2^m times 2^m$ identity matrix. Then, controlled gates $C_n^j(U_2^m)$ and $V_n^j(U_2^m)$ with $n$ control qubits and $m$ target qubits are defined by $$ C_n^j(U_2^m)=(|jrangle langle j|) otimes U_2^m+ sum_i=0,i neq j^2^n-1((|irangle langle i| otimes I_2^m$$
$$ V_n^j(U_2^m) = U_2^m otimes (|jrangle langle j|) + sum_i=0,i neq j^2^n-1( I_2^m otimes (|irangle langle i| ))$$
Then they say that $C_2^j(X)$ and $V_2^j(X) $are toffoli gates.
Can someone explain the equations that are given
and how does this special case be a Toffoli?
quantum-gate tensor-product
quantum-gate tensor-product
edited Apr 12 at 18:30
Sanchayan Dutta♦
6,68641556
6,68641556
asked Apr 12 at 13:16
UpstartUpstart
1657
1657
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here $i$ and $j$ are bit strings of size $n$. Correspondingly, $|irangle$, $|jrangle$ are some basis vectors in $2^n$-dimensional space, that corresponds to $n$-qubit register.
Those controlled operations $C$ and $V$ act on $(n+m)$-qubit space. You can consider first $n$ qubits as control register and last $m$ qubits as target register. Now, $C_n^j(U_2^m)$ applies unitary operation $U_2^m$ on the target register if control register is in the state $|jrangle$ and applies $I_2^m$ (i.e. do nothing) otherwise. You can see this by applying $C_n^j(U_2^m)$ on some vector $|xrangle|yrangle$ from the $(n+m)$-qubit space, where $x$ is some $n$-bit string:
$$
C_n^j(U_2^m) |xrangle|yrangle = (|jrangle langle j|xrangle) otimes U_2^m |yrangle+ sum_i=0,i neq j^2^n-1((|irangle langle i| x rangle) otimes |yrangle)
$$
Here $|irangle langle i|xrangle = 0$ if $xneq i$ and it equals $|irangle$ if $x=i$.
Hence
$$C_n^j(U_2^m) |xrangle|yrangle = |jrangle otimes U_2^m |yrangle + 0 = |xrangle otimes U_2^m |yrangle ~~textif~~ x=j$$
and
$$C_n^j(U_2^m) |xrangle|yrangle = 0 + |xrangle|yrangle = |xrangle|yrangle ~~textif~~ xneq j.$$
Gate $V_n^j(U_2^m)$ is basically the same as $C_n^j(U_2^m)$, though we consider first $m$ qubits as target and last $n$ qubits as control register in this case.
Now, if $j=11$ then $C_2^j(X)$ is exactly CCNOT gate on 3 qubits. Because we apply $X$ (i.e. negating the value) on the last qubit only if two first qubits are in $|11rangle$ state.
$endgroup$
$begingroup$
$y$ is an m bit string ? hence $|y rangle$ lies in a$2^m$ dimensional hilbert space?
$endgroup$
– Upstart
Apr 12 at 16:32
$begingroup$
yes, that is it.
$endgroup$
– Danylo Y
Apr 12 at 16:46
$begingroup$
why is $langle i|xrangle=0$ if $xneq i$ i see that it is an inner product between them but how is it zero because two binary strings dot product can still be non zero if they are not equal
$endgroup$
– Upstart
Apr 12 at 16:51
$begingroup$
$langle a | b rangle = langle a_1 | b_1 rangle langle a_2 | b_2 rangle ... langle a_n | b_n rangle$. This is zero if $a_i neq b_i$ at least for some $i$.
$endgroup$
– Danylo Y
Apr 12 at 16:56
$begingroup$
i read that is $= a_1b_1+ a_2b_2+....+a_nb_n$
$endgroup$
– Upstart
Apr 12 at 17:02
|
show 5 more comments
Your Answer
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1 Answer
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1 Answer
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votes
$begingroup$
Here $i$ and $j$ are bit strings of size $n$. Correspondingly, $|irangle$, $|jrangle$ are some basis vectors in $2^n$-dimensional space, that corresponds to $n$-qubit register.
Those controlled operations $C$ and $V$ act on $(n+m)$-qubit space. You can consider first $n$ qubits as control register and last $m$ qubits as target register. Now, $C_n^j(U_2^m)$ applies unitary operation $U_2^m$ on the target register if control register is in the state $|jrangle$ and applies $I_2^m$ (i.e. do nothing) otherwise. You can see this by applying $C_n^j(U_2^m)$ on some vector $|xrangle|yrangle$ from the $(n+m)$-qubit space, where $x$ is some $n$-bit string:
$$
C_n^j(U_2^m) |xrangle|yrangle = (|jrangle langle j|xrangle) otimes U_2^m |yrangle+ sum_i=0,i neq j^2^n-1((|irangle langle i| x rangle) otimes |yrangle)
$$
Here $|irangle langle i|xrangle = 0$ if $xneq i$ and it equals $|irangle$ if $x=i$.
Hence
$$C_n^j(U_2^m) |xrangle|yrangle = |jrangle otimes U_2^m |yrangle + 0 = |xrangle otimes U_2^m |yrangle ~~textif~~ x=j$$
and
$$C_n^j(U_2^m) |xrangle|yrangle = 0 + |xrangle|yrangle = |xrangle|yrangle ~~textif~~ xneq j.$$
Gate $V_n^j(U_2^m)$ is basically the same as $C_n^j(U_2^m)$, though we consider first $m$ qubits as target and last $n$ qubits as control register in this case.
Now, if $j=11$ then $C_2^j(X)$ is exactly CCNOT gate on 3 qubits. Because we apply $X$ (i.e. negating the value) on the last qubit only if two first qubits are in $|11rangle$ state.
$endgroup$
$begingroup$
$y$ is an m bit string ? hence $|y rangle$ lies in a$2^m$ dimensional hilbert space?
$endgroup$
– Upstart
Apr 12 at 16:32
$begingroup$
yes, that is it.
$endgroup$
– Danylo Y
Apr 12 at 16:46
$begingroup$
why is $langle i|xrangle=0$ if $xneq i$ i see that it is an inner product between them but how is it zero because two binary strings dot product can still be non zero if they are not equal
$endgroup$
– Upstart
Apr 12 at 16:51
$begingroup$
$langle a | b rangle = langle a_1 | b_1 rangle langle a_2 | b_2 rangle ... langle a_n | b_n rangle$. This is zero if $a_i neq b_i$ at least for some $i$.
$endgroup$
– Danylo Y
Apr 12 at 16:56
$begingroup$
i read that is $= a_1b_1+ a_2b_2+....+a_nb_n$
$endgroup$
– Upstart
Apr 12 at 17:02
|
show 5 more comments
$begingroup$
Here $i$ and $j$ are bit strings of size $n$. Correspondingly, $|irangle$, $|jrangle$ are some basis vectors in $2^n$-dimensional space, that corresponds to $n$-qubit register.
Those controlled operations $C$ and $V$ act on $(n+m)$-qubit space. You can consider first $n$ qubits as control register and last $m$ qubits as target register. Now, $C_n^j(U_2^m)$ applies unitary operation $U_2^m$ on the target register if control register is in the state $|jrangle$ and applies $I_2^m$ (i.e. do nothing) otherwise. You can see this by applying $C_n^j(U_2^m)$ on some vector $|xrangle|yrangle$ from the $(n+m)$-qubit space, where $x$ is some $n$-bit string:
$$
C_n^j(U_2^m) |xrangle|yrangle = (|jrangle langle j|xrangle) otimes U_2^m |yrangle+ sum_i=0,i neq j^2^n-1((|irangle langle i| x rangle) otimes |yrangle)
$$
Here $|irangle langle i|xrangle = 0$ if $xneq i$ and it equals $|irangle$ if $x=i$.
Hence
$$C_n^j(U_2^m) |xrangle|yrangle = |jrangle otimes U_2^m |yrangle + 0 = |xrangle otimes U_2^m |yrangle ~~textif~~ x=j$$
and
$$C_n^j(U_2^m) |xrangle|yrangle = 0 + |xrangle|yrangle = |xrangle|yrangle ~~textif~~ xneq j.$$
Gate $V_n^j(U_2^m)$ is basically the same as $C_n^j(U_2^m)$, though we consider first $m$ qubits as target and last $n$ qubits as control register in this case.
Now, if $j=11$ then $C_2^j(X)$ is exactly CCNOT gate on 3 qubits. Because we apply $X$ (i.e. negating the value) on the last qubit only if two first qubits are in $|11rangle$ state.
$endgroup$
$begingroup$
$y$ is an m bit string ? hence $|y rangle$ lies in a$2^m$ dimensional hilbert space?
$endgroup$
– Upstart
Apr 12 at 16:32
$begingroup$
yes, that is it.
$endgroup$
– Danylo Y
Apr 12 at 16:46
$begingroup$
why is $langle i|xrangle=0$ if $xneq i$ i see that it is an inner product between them but how is it zero because two binary strings dot product can still be non zero if they are not equal
$endgroup$
– Upstart
Apr 12 at 16:51
$begingroup$
$langle a | b rangle = langle a_1 | b_1 rangle langle a_2 | b_2 rangle ... langle a_n | b_n rangle$. This is zero if $a_i neq b_i$ at least for some $i$.
$endgroup$
– Danylo Y
Apr 12 at 16:56
$begingroup$
i read that is $= a_1b_1+ a_2b_2+....+a_nb_n$
$endgroup$
– Upstart
Apr 12 at 17:02
|
show 5 more comments
$begingroup$
Here $i$ and $j$ are bit strings of size $n$. Correspondingly, $|irangle$, $|jrangle$ are some basis vectors in $2^n$-dimensional space, that corresponds to $n$-qubit register.
Those controlled operations $C$ and $V$ act on $(n+m)$-qubit space. You can consider first $n$ qubits as control register and last $m$ qubits as target register. Now, $C_n^j(U_2^m)$ applies unitary operation $U_2^m$ on the target register if control register is in the state $|jrangle$ and applies $I_2^m$ (i.e. do nothing) otherwise. You can see this by applying $C_n^j(U_2^m)$ on some vector $|xrangle|yrangle$ from the $(n+m)$-qubit space, where $x$ is some $n$-bit string:
$$
C_n^j(U_2^m) |xrangle|yrangle = (|jrangle langle j|xrangle) otimes U_2^m |yrangle+ sum_i=0,i neq j^2^n-1((|irangle langle i| x rangle) otimes |yrangle)
$$
Here $|irangle langle i|xrangle = 0$ if $xneq i$ and it equals $|irangle$ if $x=i$.
Hence
$$C_n^j(U_2^m) |xrangle|yrangle = |jrangle otimes U_2^m |yrangle + 0 = |xrangle otimes U_2^m |yrangle ~~textif~~ x=j$$
and
$$C_n^j(U_2^m) |xrangle|yrangle = 0 + |xrangle|yrangle = |xrangle|yrangle ~~textif~~ xneq j.$$
Gate $V_n^j(U_2^m)$ is basically the same as $C_n^j(U_2^m)$, though we consider first $m$ qubits as target and last $n$ qubits as control register in this case.
Now, if $j=11$ then $C_2^j(X)$ is exactly CCNOT gate on 3 qubits. Because we apply $X$ (i.e. negating the value) on the last qubit only if two first qubits are in $|11rangle$ state.
$endgroup$
Here $i$ and $j$ are bit strings of size $n$. Correspondingly, $|irangle$, $|jrangle$ are some basis vectors in $2^n$-dimensional space, that corresponds to $n$-qubit register.
Those controlled operations $C$ and $V$ act on $(n+m)$-qubit space. You can consider first $n$ qubits as control register and last $m$ qubits as target register. Now, $C_n^j(U_2^m)$ applies unitary operation $U_2^m$ on the target register if control register is in the state $|jrangle$ and applies $I_2^m$ (i.e. do nothing) otherwise. You can see this by applying $C_n^j(U_2^m)$ on some vector $|xrangle|yrangle$ from the $(n+m)$-qubit space, where $x$ is some $n$-bit string:
$$
C_n^j(U_2^m) |xrangle|yrangle = (|jrangle langle j|xrangle) otimes U_2^m |yrangle+ sum_i=0,i neq j^2^n-1((|irangle langle i| x rangle) otimes |yrangle)
$$
Here $|irangle langle i|xrangle = 0$ if $xneq i$ and it equals $|irangle$ if $x=i$.
Hence
$$C_n^j(U_2^m) |xrangle|yrangle = |jrangle otimes U_2^m |yrangle + 0 = |xrangle otimes U_2^m |yrangle ~~textif~~ x=j$$
and
$$C_n^j(U_2^m) |xrangle|yrangle = 0 + |xrangle|yrangle = |xrangle|yrangle ~~textif~~ xneq j.$$
Gate $V_n^j(U_2^m)$ is basically the same as $C_n^j(U_2^m)$, though we consider first $m$ qubits as target and last $n$ qubits as control register in this case.
Now, if $j=11$ then $C_2^j(X)$ is exactly CCNOT gate on 3 qubits. Because we apply $X$ (i.e. negating the value) on the last qubit only if two first qubits are in $|11rangle$ state.
edited Apr 12 at 15:08
answered Apr 12 at 14:53
Danylo YDanylo Y
55116
55116
$begingroup$
$y$ is an m bit string ? hence $|y rangle$ lies in a$2^m$ dimensional hilbert space?
$endgroup$
– Upstart
Apr 12 at 16:32
$begingroup$
yes, that is it.
$endgroup$
– Danylo Y
Apr 12 at 16:46
$begingroup$
why is $langle i|xrangle=0$ if $xneq i$ i see that it is an inner product between them but how is it zero because two binary strings dot product can still be non zero if they are not equal
$endgroup$
– Upstart
Apr 12 at 16:51
$begingroup$
$langle a | b rangle = langle a_1 | b_1 rangle langle a_2 | b_2 rangle ... langle a_n | b_n rangle$. This is zero if $a_i neq b_i$ at least for some $i$.
$endgroup$
– Danylo Y
Apr 12 at 16:56
$begingroup$
i read that is $= a_1b_1+ a_2b_2+....+a_nb_n$
$endgroup$
– Upstart
Apr 12 at 17:02
|
show 5 more comments
$begingroup$
$y$ is an m bit string ? hence $|y rangle$ lies in a$2^m$ dimensional hilbert space?
$endgroup$
– Upstart
Apr 12 at 16:32
$begingroup$
yes, that is it.
$endgroup$
– Danylo Y
Apr 12 at 16:46
$begingroup$
why is $langle i|xrangle=0$ if $xneq i$ i see that it is an inner product between them but how is it zero because two binary strings dot product can still be non zero if they are not equal
$endgroup$
– Upstart
Apr 12 at 16:51
$begingroup$
$langle a | b rangle = langle a_1 | b_1 rangle langle a_2 | b_2 rangle ... langle a_n | b_n rangle$. This is zero if $a_i neq b_i$ at least for some $i$.
$endgroup$
– Danylo Y
Apr 12 at 16:56
$begingroup$
i read that is $= a_1b_1+ a_2b_2+....+a_nb_n$
$endgroup$
– Upstart
Apr 12 at 17:02
$begingroup$
$y$ is an m bit string ? hence $|y rangle$ lies in a$2^m$ dimensional hilbert space?
$endgroup$
– Upstart
Apr 12 at 16:32
$begingroup$
$y$ is an m bit string ? hence $|y rangle$ lies in a$2^m$ dimensional hilbert space?
$endgroup$
– Upstart
Apr 12 at 16:32
$begingroup$
yes, that is it.
$endgroup$
– Danylo Y
Apr 12 at 16:46
$begingroup$
yes, that is it.
$endgroup$
– Danylo Y
Apr 12 at 16:46
$begingroup$
why is $langle i|xrangle=0$ if $xneq i$ i see that it is an inner product between them but how is it zero because two binary strings dot product can still be non zero if they are not equal
$endgroup$
– Upstart
Apr 12 at 16:51
$begingroup$
why is $langle i|xrangle=0$ if $xneq i$ i see that it is an inner product between them but how is it zero because two binary strings dot product can still be non zero if they are not equal
$endgroup$
– Upstart
Apr 12 at 16:51
$begingroup$
$langle a | b rangle = langle a_1 | b_1 rangle langle a_2 | b_2 rangle ... langle a_n | b_n rangle$. This is zero if $a_i neq b_i$ at least for some $i$.
$endgroup$
– Danylo Y
Apr 12 at 16:56
$begingroup$
$langle a | b rangle = langle a_1 | b_1 rangle langle a_2 | b_2 rangle ... langle a_n | b_n rangle$. This is zero if $a_i neq b_i$ at least for some $i$.
$endgroup$
– Danylo Y
Apr 12 at 16:56
$begingroup$
i read that is $= a_1b_1+ a_2b_2+....+a_nb_n$
$endgroup$
– Upstart
Apr 12 at 17:02
$begingroup$
i read that is $= a_1b_1+ a_2b_2+....+a_nb_n$
$endgroup$
– Upstart
Apr 12 at 17:02
|
show 5 more comments
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StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
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Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown