Quantum Computing Stack Exchange is a question and answer site for engineers, scientists, programmers, and computing professionals interested in quantum computing. It only takes a minute to sign up.
Sign up to join this community
I am trying to understand the state given here:
$$ \frac{1}{\sqrt{n}} \sum_{j=0}^{n-1} |j\rangle |j\rangle $$
Suppose $n = 4$, I would interpret this state as being $1/2(|00\rangle + |11\rangle + |22\rangle + |33\rangle)$. Is this correct? And if so, what does $|2\rangle$ or $|3\rangle$ mean?
Each $|j\rangle$ is a multi-qbit state. So if $n = 4$, $|2\rangle$ and $|3\rangle$ are $|10\rangle$ and $|11\rangle$ respectively. For $n = 4$ your final formula would be:
$$ \frac{1}{\sqrt{4}} \left( |00\rangle|00\rangle + |01\rangle|01\rangle + |10\rangle|10\rangle + |11\rangle|11\rangle \right) $$ $$ \frac{1}{\sqrt{4}} \left( |0000\rangle + |0101\rangle + |1010\rangle + |1111\rangle \right) $$
The dimension of the Hilbert space associated to a quantum system does not have to be two. Quantum systems whose Hilbert space is two-dimensional - known as qubits - are the most popular, but they do not exhaust all possibilities. For example:
Answering your question more directly: Your expansion of the sum in the question is correct and $|2\rangle, |3\rangle$ simply refer to the third and fourth states in the computational basis of the four-dimensional Hilbert space.
The thing that you write inside a ket is just a label. It can actually correspond to anything that you want it to. By convention, labels 0, 1, 2, 3... correspond to orthogonal states. So, you should interpret this as whatever system you're using has enough capacity to have a set of 4 orthogonal states, i.e. the Hilbert space dimension is at least 4. You're probably used to just seeing $|0\rangle$ and $|1\rangle$ because most of the time in quantum information, we deal with a qubit, which has Hilbert space dimension 2. But that's an assumption of convenience. Physically, it's far more common to have multiple levels.
