What is the difference between 3 qubits, 2 qutrits and a 6th level qunit? Are they equivalent? Why / why not?

Can 6 classical bits be super-densely coded into each?


The Hilbert space dimension of $n$ qudits is $d^n$, where $d$ is the dimension of the qudit ($d=2$ for qubit, $d=3$ for qutrit, etc). So three qubits have an $8$ dimensional space, two qutrits have a $9$ dimensional space, and one $d=6$ qudit has a six dimensional space. As such, we cannot regard them as equivalent.

I guess you meant to compare situations with equal total Hilbert space dimension. Such as a comparing a pair of qubits with a $d=4$ system. In this case, there is mathematically no distinction. You could choose to relabel the basis states $|00\rangle$, $|01\rangle$, $|10\rangle$ and $|11\rangle$ as the qudit basis states $|0\rangle$, $|1\rangle$, $|2\rangle$ and $|3\rangle$. Then any qudit operations defined with the qudit basis could be equivalently defined with the qubits, and vice-versa. You could also use other mappings between basis states, this was just an example.

We could also use a subspace of a larger space to simulate a smaller one. For example, suppose you want to simulate a spin-$1$ particle, which is a 3 level system. You could do this using a pair of qubits (a four level system) and identifying three basis states of the former with three of the latter (such as $|-1\rangle$, $|0\rangle$ and $|1\rangle$ with $|00\rangle$, $|01\rangle$ and $|10\rangle$, for example). As long as you implement your spin-$1$ operations correctly, you'll always avoid the $|11\rangle$ state, and your two qubits effectively become a qutrit.

You might also be interested in my answer to the question What is the most economical and preferred basis for the qudit?

  • $\begingroup$ As per usual, excellent answer! $\endgroup$ – user820789 Jun 21 '18 at 9:44

They are not equivalent. It can be seen by the fact that the system of $3$ qubits acts on a $8$ dimensional Hilbert space, the 2 qutrit system acts on a $9$ dimensional Hilbert space, and the 6 level qunit acts on a $6$ dimensional Hilbert space. Consequently, the nature of the states defined by each of the quantum systems is different.

This dimension argument comes from the fact that a k-level n-qunit system acts on a state space of dimension $k^n$.

For the superdense-coding, I am aware that Bell pairs are used in order to obtain the desired coding, and as you do not consider such entangled qubits, I am not sure to answer such question.

  • $\begingroup$ Could you elaborate on "the nature of the states defined by each of the quantum systems is different."? $\endgroup$ – user820789 Jun 21 '18 at 8:55
  • $\begingroup$ With that I mean that indeed the quantum states that you present they are not equivalent. $\endgroup$ – Josu Etxezarreta Martinez Jun 21 '18 at 10:00
  • $\begingroup$ I understood that! Just looking for more explanation into how they differ (: $\endgroup$ – user820789 Jun 21 '18 at 10:09
  • $\begingroup$ I was stating a similar reasoning as @JamesWootton, implying that the states are not equivalent because the state spaces of the quantum systems differ in dimension. $\endgroup$ – Josu Etxezarreta Martinez Jun 21 '18 at 10:32

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