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Suppose I had 3 entangled qubits and I wanted to apply a SWAP gate on the first and third qubit.

Because it's entangled I can't decompose it into individual states, and because the qubits are not adjacent I can't simply take a Kronecker product with an identity matrix.

How would I go about creating the matrix for applying this transformation? Also is there a generalization for swapping 2 arbitrary qubits in an n qubit system?

glS
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Bob Person
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2 Answers2

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Gates don't depend on states they act on. So it doesn't matter if the state is entangled or not. The gate you've described (let's call it $U$) acts on a standard basis by this rules $$ U |0x0\rangle = |0x0\rangle $$ $$ U |0x1\rangle = |1x0\rangle $$ $$ U |1x0\rangle = |0x1\rangle $$ $$ U |1x1\rangle = |1x1\rangle $$ for each $x=0$ and $x=1$.

You can write this matrix as $$ U=\sum_{x=0}^1 \big(|0x0\rangle\langle 0x0 |+|1x0\rangle\langle 0x1 |+|0x1\rangle\langle 1x0 |+|1x1\rangle\langle 1x1 | \big) = $$ $$ = |0\rangle\langle 0| \otimes I \otimes |0\rangle\langle 0| + |1\rangle\langle 0| \otimes I \otimes |0\rangle\langle 1| + |0\rangle\langle 1| \otimes I \otimes |1\rangle\langle 0| + |1\rangle\langle 1| \otimes I \otimes |1\rangle\langle 1| $$

Danylo Y
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  • Can you explain why $|0x0 \rangle \langle 0x0 | = | 0 \rangle \langle 0 | \otimes I \otimes | 0 \rangle \langle 0 |$?

    (And why this is true for the other outer products in the summation.

     – wavosa Apr 02 '23 at 15:58
  • $|0x0 \rangle \langle 0x0 | = | 0 \rangle \langle 0 | \otimes | x \rangle \langle x | \otimes | 0 \rangle \langle 0 |$ by the mixed-product property of tensor products. It's the sum over $x=0$ and $x=1$ gives $| 0 \rangle \langle 0 | \otimes I \otimes | 0 \rangle \langle 0 |$. Similarly for others. – Danylo Y Apr 02 '23 at 16:35
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If I may, I wanna share an algorithmic generalization for the inspired @Danylo Y answer. This considers also The Algorithmic Method in this answer here related to the CNOT gate.

In order to algorithmically build a 2-qubit SWAP gate to operate swap within n-qubits, one could define an array containing n ID gates, i.e. ID=((1,0),(0,1)) matrices, named here as M, and overwrite both elements, the one subscripted as qba_idx with |i><j|, and the one as qbb_idx with |j><i|, considering $i,j \in {0,1}$.

As a reference, |0><0| = ((1,0),(0,0)), |0><1|=((0,1),(0,0)), |1><0|=((0,0),(1,0)), |1><1|=((0,0),(0,1)).

So, to build the n-qubit SWAP gate, one can perform operations as:

// 'state' is the current (2^n)×1 column statevector for n-qubits
M = [ID] * n // list with n matrices of ((1,0),(0,1))
final_state = zeros((2^n, 1)) // column vector with all zeros, (2^n)×1
for i in [0,1]:
    for j in [0,1]:
        M[qba_idx] = ket_bra(i,j) //|i><j|
        M[qbb_idx] = ket_bra(j,i) //|j><i|
        swap_gate = M[0]
        for m in M[1:end]:
            swap_gate = kron_product(swap_gate, m)
        final_state = final_state + matrix_multiply(swap_gate, state)
// final_state now has the resulting statevector with 
// both qubits (qba_idx and qbb_idx) swapped

Don't forget that n-qubits will require a swap_gate matrix with size $2^n\times2^n$. For example, 10 qubits will require a $1024\times1024$ = 1,048,576 elements. If they are of double precision, this means 67108864 bits, or 8 Mb of memory. In the other hand 20 qubits will require something around 8 Tb of memory for swap_gate variable only. If we consider each element as a complex number, these memory requirement just doubles. Mind that when coding your gates.

iperetta
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