3.2.3 Parallel operations

If we apply one operation on the first qubit and another on the second qubit then the order of these two operations does not matter. That is, if $U$ and $V$ are arbitrary single-qubit operations then

We can verify this intuitive fact by using Eqs. 3.56 and 3.57. For every basis state $\left|a,b\right\rangle$ where $a,b\in\{0,1\}$,

so Eq. 3.58 follows by linearity.

In fact, since the two operations act on different qubits, we could even do them in parallel! This suggests introducing a new notation for the operation in Eq. 3.58:

We are re-using the tensor product symbol that we originally introduced for combining two independent single-qubit states into a two-qubit state. The same meaning extends also to quantum operations: $U\otimes V$ denotes the combined operation that consists of applying $U$ and $V$ to two different subsystems of a larger system. This notation is particularly convenient since it interplays nicely with the original tensor product for states:

This equation simply says that if you have two independent states and you apply an operation that acts independently on each of the two states, then you just end up applying each of the two operations on the corresponding state. We will call $U\otimes V$ a tensor product of two quantum operations or a parallel operation.

Quirky allows us to apply local quantum operations in parallel. For example, we could have written the sequence of operations in Eq. 3.54 as

where the $\mathrm{NOT}$ operation and the Hadamard operation are now applied in parallel.

Let us discuss another example. What happens if we apply $H$ to both qubits? This operation is denoted by $H\otimes H$ and according to Eqs. 2.34 and 3.59 acts as follows:

The resulting state is called the uniform superposition on two qubits since it contains each of the two-qubit basis vectors with an equal amplitude. If we perform a measurement on this state we obtain all four outcomes with equal probability. We can readily verify this using Quirky:

Eq. 3.59 shows that if we apply a parallel operation to a product state then we get another product state. In fact, starting from $\left|00\right\rangle$ we can get any product state in this way. This gives an intuitive interpretation to the product states: they are exactly those states that we can get by applying an arbitrary parallel operation to two qubits initialized in state $\left|00\right\rangle$.

We close our discussion of local operations with some general remarks. First, it is not hard to check that

for any four single-qubit operations $U$, $U^{\prime}$, $V$, $V^{\prime}$. Can you draw a picture to visualize what is going on here?

We can now make use of the identity operation $I$ from Eq. 2.32 which acts as $I\left|\psi\right\rangle=\left|\psi\right\rangle$ (this trivially extends to $I_{1}$ and $I_{2}$ on two qubits). It is rather useless on its own, but can be conveniently used in tensor product notation. For example, it allows us to write

which makes it very clear that, e.g., $U_{1}$ applies the $U$-operation on the first qubit and nothing on the second qubit. For example, the identity $U_{1}V_{2}=V_{2}U_{1}$ from Eq. 3.58 now becomes

which is rather intuitive.

You might wonder if the order of operations actually matters if we apply them to the same qubit? If we consider two rotations then it follows from Eq. 2.29 that their order is not important since

However, if $U$ and $V$ are two arbitrary single-qubit operations (in particular, one of them is a reflection) then their composition generally depends on the order (see 3.11 below). That is,

This issue of course persists also when you have another qubit around, but are still acting with both operations on the same qubit. For example,

We can also write this as $U_{1}V_{1}\neq V_{1}U_{1}$ (and similarly when we apply both operations to the second qubit).

To understand intuitively the difference between Eqs. 3.61 and 3.62, imagine that $U$ means “putting on a sock” and $V$ means “putting on a shoe”. Clearly, when $U$ and $V$ are applied to the same foot, you will get different results depending on the order! However, if you apply $U$ and $V$ to different feet (e.g., you use $U\otimes I$ and $I\otimes V$) then you will get the same result irrespectively of the order of the two operations. Either way, if you want to be properly dressed, you should apply $(U\otimes U)$ and then $(V\otimes V)$.