Kdjykuj

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?

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
$$C_n^j(U_{2^m}) |xrangle|yrangle = |jrangle otimes U_{2^m} |yrangle + 0 = |xrangle otimes U_{2^m} |yrangle ~~text{if}~~ x=j$$
and
$$C_n^j(U_{2^m}) |xrangle|yrangle = 0 + |xrangle|yrangle = |xrangle|yrangle ~~text{if}~~ xneq j.$$

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.