# What does a zero state combined state vector for 2 qubits mean?

I was fiddling with the quantum circuit used for Deutsch's algorithm and I was led to a zero state as a result which is bizarre. I don't know how to explain this result.

***I'm using Nielsen and Chuang to study QC

• You don't explain why you think the terms cancel out. Without that, it's hard for us to help... – DaftWullie Jul 2 '19 at 15:25

$$\frac{1}{2} (|0\rangle|0 \oplus f(0)\rangle - |0\rangle|1 \oplus f(0)\rangle + |1\rangle|0 \oplus f(1)\rangle - |1\rangle|1 \oplus f(1)\rangle)$$
If $$f(0) \neq f(1)$$, consider the first two terms (the only ones which can cancel with each other, since the state of the first qubit is $$|0\rangle$$): the state of the second qubit in them is $$|0 \oplus f(0)\rangle$$ and $$|1 \oplus f(0)\rangle$$, which are different states no matter what the value of $$f(0)$$ is.
It will be helpful for you to consider the cases for $$f(x) = x$$ and $$f(x) = 1-x$$ separately to convince yourself that the cancellation doesn't happen.