3
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ Rather than using images to express your question could you please type it up using Mathjax? The reason is that if the image link ever dies then the question becomes unreadable. That being said, $|j\rangle$ usually refers to the $j$-th element of some orthonormal basis. from the expression is appears you are working with a bipartite system with local dimensions of $4$. $\endgroup$ – Rammus Dec 6 '20 at 14:02
4
$\begingroup$

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:

$$ \frac{1}{\sqrt{4}} \left( |00\rangle|00\rangle + |01\rangle|01\rangle + |10\rangle|10\rangle + |11\rangle|11\rangle \right) $$ $$ \frac{1}{\sqrt{4}} \left( |0000\rangle + |0101\rangle + |1010\rangle + |1111\rangle \right) $$

$\endgroup$
2
  • 1
    $\begingroup$ So for a system of dimension $3$, you always think of it as a subspace of a two-qubit system? $\endgroup$ – Rammus Dec 6 '20 at 14:33
  • 1
    $\begingroup$ As @Rammus pointed out, each |j> may not be a multiqubit state. It could be a 'qudit'. i.e. a single system with 'd' dimensions. Then you can generalize. $\endgroup$ – QuestionEverything Dec 6 '20 at 18:13
3
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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:

  • qutrit has a three-dimensional Hilbert space and its computational basis states are often labeled $|0\rangle, |1\rangle, |2\rangle$,
  • a system of two qubits has a Hilbert space with four dimensions and even though the basis states are often labeled as $|00\rangle, |01\rangle, |10\rangle, |11\rangle$, one could also named them $|0\rangle, |1\rangle, |2\rangle, |3\rangle$,
  • a system of $n$ qubits has a Hilbert space with dimension $2^n$,
  • quantum harmonic oscillator has countably infinite dimensional Hilbert space.

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.

$\endgroup$
0

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.