What is the meaning of the state $|1\rangle-|1\rangle$?

Question

I don't understand it. But suppose you have a state of two qubits in superposition and it looks like $|10\rangle-|11\rangle$. Now you have in your circuits two gates one is a controlled-NOT from the first qubit to the second and the other gate following is the Hadamard gate on the first qubit.

If I applied first the CNOT I got, $|11\rangle-|10\rangle$, right? (I am not entirely sure why it's allowed to work like this but ok.)

Now I was trying two approaches yielding different results but I am actually not sure about what is the meaning of the first qubit.

1. So one approach was to take the original state $|10\rangle-|11\rangle$ and apply to it $\operatorname{H}$ gate multiplied by CNOT gate.

2. The second approach is to take the state after the CNOT $|11\rangle-|10\rangle$ and just as I did with the CNOT I would apply $\operatorname{H}$ gate only on the first qubit which is $|1\rangle-|1\rangle$, right? (Could be there is some coefficient there like $\frac{1}{\sqrt{2}}$ but I don't care about it now). But the thing I don't understand is how do I apply $\operatorname{H}$ gate on $|1\rangle-|1\rangle$? It isn't even clear to me what it means. Doesn't $|1\rangle-|1\rangle$ equal nothing? I mean, it isn't even a state, no?

Anyways it is strange, the first approach would yead a vector like $(-1,1,1,-1)$ I am not sure what it means. But it looks like I can never get such a state in the second approach, simply because I can't get more than two states combined because the Hadamard is acting only on the first qubit while the second qubit should remain unchanged. But mainly I am not sure what's $|1\rangle-|1\rangle$?

Answer 1

The second approach is to take the state after the CNOT $|11\rangle-|10\rangle$ and just as I did with the CNOT I would apply $\operatorname{H}$ gate only on the first qubit which is $|1\rangle-|1\rangle$, right?

The state of the first qubit is not $|1\rangle-|1\rangle$. It is simply $|1\rangle$.

The tensor product follows a distributive law. See this for a proof.

In your case, the composite state of two qubits (let's call them $A$ and $B$) is $|10\rangle-|11\rangle$. It may be more explicitly written as $$|1\rangle_A\otimes |0\rangle_B - |1\rangle_A \otimes|1\rangle_B$$ $$=|1\rangle_A\otimes (|0\rangle_B - |1\rangle_B).$$ This follows from the distributive law I linked. From here it should be clear that the state of the first qubit A is $|1\rangle_A$ and not $|1\rangle_A - |1\rangle_A$. And the state of the second qubit B is $|0\rangle_B-|1\rangle_B$. Can you take it from here?

Answer 2

Sanchayan Dutta's answer correctly points out that $|1\rangle-|1\rangle$ doesn't actually arise in your example problem, but it doesn't answer the question in the title: what is $|1\rangle-|1\rangle$?

$|1\rangle-|1\rangle = 0$, the zero vector in the Hilbert space. It has no interpretation as a quantum state because it isn't normalizable. It isn't even handwavily quasi-normalizable like the position and momentum eigenstates of a free particle in continuous QM.

However, it's a perfectly good vector, and can appear in calculations, as long as it isn't the entire state (or a factor of the entire state, as you thought it was). For example, this is a correct calculation:

$$\frac{|{+}\rangle + |{-}\rangle}{\sqrt2} = \frac{|0\rangle+|1\rangle+|0\rangle-|1\rangle}{2} = \frac{|0\rangle+|0\rangle}{2} + \frac{|1\rangle-|1\rangle}{2} = |0\rangle + 0 = |0\rangle$$