# Controlling high-dimensional Hilbert spaces with a single qubit

In superdense coding, you can use one qubit to control the Hilbert space of two qubits and steer it into 4 mutually orthogonal states, so that measurement of both qubits together will not have a probabilistic outcome.

I think the idea is cool. I want to understand a more general question that arises from this result.

What if I have 100 qubits (or N qubits, if you like), entangled by a neutral party, and 99 are sent to my friend and 1 is sent to me. How much control do I have over the composite quantum state? How many mutually orthogonal states can one steer the global state into?

Thanks.

EDIT:

Thank you for your interest in this question. I can't seem to figure out who is right, so let's consider a slightly different question.

There are a total of 5 qubits. I have 2, and my friend has 3. Clearly the upper bound of classical bits I can send to my friend is 5. But is it possible to find 5 unitary operations I can perform on my two qubits to steer the state into 5 orthogonal states.

If you can answer that, then how does this generalize to N and M qubits?

Hopefully this comes closer to the real question I want to ask.

• Do you know about the Schmidt decomposition? – Norbert Schuch Nov 10 '18 at 12:10
• @NorbertSchuch, but that can explain well only the bipartite case. – Siddhant Singh Nov 10 '18 at 12:42
• @SiddhāntSingh The OP and their friend - this is the bipartite case. – Norbert Schuch Nov 10 '18 at 13:02
• @psitae It tells you that all what matters is your reduced state. Any pure state with the same (1-qubit) reduced state on your side can be transformed into each other by an operation of the purifying party (i.e., your friend). So the space you have access to is independent of the system your friend holds. Differently speaking, you can still only prepare 4 orthogonal states. – Norbert Schuch Nov 10 '18 at 13:03
• @psitae No matter how many qubits you use, you are still working with a bipartite system with your friend. So it totally depends on the dimension of your subsystem. If you only keep 1 qubit at your side, then still at most you can send 2 bits to your friend, – XXDD Nov 11 '18 at 3:26

There are a total of 5 qubits. I have 2, and my friend has 3. Clearly the upper bound of classical bits I can send to my friend is 5. But is it possible to find 5 unitary operations I can perform on my two qubits to steer the state into 5 orthogonal states.

You can simply set up two standard superdense coding runs. So, this starts with each qubit of yours being maximally entangled with a qubit of your friend's. You do the typical things with each qubit, send them over, and your friend can extract 2 bits from each, giving a total of 4 bits.

You know that you cannot do better than this because of the idea of Schmidt coefficients. Whatever (pure) quantum state you and your friend hold, there is a unitary that your friend can apply that changes it into a pure state on just two of his/her qubits.

If you can answer that, then how does this generalize to N and M qubits?

In general, if you have $$n$$ qubits, and your friend has $$m>n$$ qubits, you can communicate $$2n$$ classical bits by sending the $$n$$ qubits.

If you want to go up to different dimensions of spin, and on your side, you have a spin of dimension $$d$$, and your friend has a spin of dimension $$d'>d$$, I believe you can always send $$2\log_2(d)$$ classical bits. Again, you set up a $$d$$-dimensional maximally entangled state between the two spins, and you perform the correct unitary options. If memory serves, these are comprised of the power 0 to $$d-1$$ of the two unitaries $$\tilde X$$ and $$\tilde Z$$. These are both $$d^{th}$$ roots of the identity matrix, but in different bases. We have $$\tilde Z=\sum_{i=0}^{d-1}\omega^i|i\rangle\langle i|\qquad \tilde X=\sum_{i=0}^{d-2}|i\rangle\langle i+1|+|d-1\rangle\langle 0|.$$

Let me call you A and your friend B.

You initially share a state $$|\psi\rangle$$. W.l.o.g., we can write this in its Schmidt decomposition: $$|\psi\rangle = \sum_{i=1}^d \lambda_i |i\rangle_A|i\rangle_B.$$ You are now asking how many (orthogonal) states A can implement by acting with a unitary on their side of the state.

This number is upper bounded by the total number of basis states involved on $$B$$'s side, which is $$d$$, times the dimension of the space of Alice, $$d_A$$. So one can prepare at most $$d_Ad$$ states.

In your case, $$d_A=d=2$$, so you can prepare at most $$4$$ orthogonal states. This is exactly achieved if your state is maximally entangled ($$\lambda_1=\lambda_2=1/\sqrt2$$); this is dense coding.

(You can easily see that the same idea as in dense coding also applies if $$d_A$$ is larger. Then, you can accordingly encode more states, but A also has to send a larger $$d_A$$-dimensional system -- the amount of saving in the information you send is always given by $$d$$.)

So the bottomline is that the number of states you can steer the system to is always given by (i) the dimension of your system and (ii) the amount of entanglement, but entirely independent of the system size of B.

• This is true only for 2 qudits. Not even for 3 qubits. His friend has a lot more degrees of freedom as other N-2 qubits, which you are not taking into account. – Siddhant Singh Nov 10 '18 at 14:26
• @SiddhāntSingh Do you mind giving an example where this fails? N qubits are just a huge 2^N-level system, unless you distribute them between several friends. --- If it is about the disbalance in dimension, you should be able to give a counterexample for a qubit-qutrit system. – Norbert Schuch Nov 10 '18 at 14:30
• The fact that his friend has other N-1 qubits implies that they are not just some ancillary qubits, they can be manipulated over by his friend differently for each about he possess. It does not matter whether they are distributed over a single person or multiple. This case of N qubits is always a multipartite case. This is reason why Schmidt decomposition cannot give you the complete picture because there will be correlations among the qubits his friend possess too, which are not addressed by your case because you assumed all his qubits are a single system. – Siddhant Singh Nov 10 '18 at 14:41
• @SiddhāntSingh But the OP wants to steer the joint state by acting on their part (the qubit), not on their friend's share. Their friends part is left untouched. (And if it would be other way round, the result would be the same, with the dimensions exchanged.) --- In either case, there is nothing more complicated here than in the bipartite case, because it is the bipartite case. – Norbert Schuch Nov 10 '18 at 14:43
• @SiddhāntSingh To quote the OP (highlighting mine): "99 are sent to my friend and 1 is sent to me. How much control do I have over the quantum state? " – Norbert Schuch Nov 10 '18 at 14:46

The question, as to me, it seems that the person has tried to make the analogy of a bipartite entanglement distribution with a $$N$$-partite distribution, which are entirely different cases. When explaining the effect of one entity in a system, entangled with more than one other constituent, various factors have to be taken into account, such as witnesses of entanglement which might (in their own sense), allow a measure of entanglememt to be distributed over multiparty. And mainly the Monogamous nature of quantum entanglement (see CKW Inequality).

The answer the question precisely, it has first to be defined to which class the composite entangled state belongs to. The multipartite entanglement can be addressed to this date, only for few specific states such as the GHZ states or W-states most broadly. A general framework is not yet known to answer this question. These two class of states saturate the CKW inequality in two extremes and hence can be formulated in a quantifiable manner in protocols.

Especially for GHZ states, this question can be answered and it is definitely possible as is executed in various protocols of Multiparty Secret Sharing where GHZ states are distributed and later on the members have to distinguish the resulting orthogonal states by appropriate projectors.

No mechanism, in general, is known for your query, but a particular class of states can make this possible.