# Maximally entangled state definition, and orthonormal basis of maximally entangled state

My questions are probably more about details but the answer will help me to precisely understand the mathematical structure.

My questions are related to page 14 of this pdf.

Mathematical context:

We have $$H=H_A \otimes H_B$$ with $$d = \mathrm{min}(\dim(H_A), \dim(H_B))$$

We call $$|\Omega\rangle = \frac{1}{\sqrt{d}} \sum_i |ii \rangle$$ the maximally entangled state.

First question:

For me this state is ill-defined in general. If $$H_A$$ and $$H_B$$ represent the same physical system (a qubit for example), then it is well defined because if I choose $$|i\rangle = | 0 \rangle$$ for the first system, then as the second one is the same, I know that $$|ii\rangle = |00\rangle$$.

However, if the two systems are different, for example, a qubit and a harmonic oscillator how do I know how to choose the kets?

I could have (first space = qubit, second one = harmonic oscillator):

$$|\Omega\rangle = \frac{1}{\sqrt{2}} \left(|+0\rangle + |-1\rangle \right)$$

Or:

$$|\Omega\rangle = \frac{1}{\sqrt{2}} \left(|00\rangle + |11\rangle \right)$$

Would you agree with me that $$|\Omega\rangle$$ is an ill-defined state in general? But maybe that it doesn't matter for practical purposes? One of the confusion it gives me is the property that any maximal entangled state is of the form $$\mathbb{I} \otimes U | \Omega \rangle$$, $$U$$ being unitary. How to prove it properly if $$|\Omega \rangle$$ is ill-defined ?

Second question:

We can take a basis $$\{ U_i \}$$ of $$\mathcal{M}_d(\mathbb{C})$$ composed of $$d^2$$ unitary matrices orthogonal with respect to the Hilbert-Schmidt scalar product on matrices.

Then the family $$\left( U_i \otimes \mathbb{I} \right)| \Omega \rangle$$ is composed of $$d^2$$ orthonormal vectors.

In the document, he says that it is a basis of entangled states of $$\mathbb{C}^d \otimes \mathbb{C}^d$$. I agree that it matches in terms of dimensions, but why is it a basis of this space precisely? For example, could we say as well it is a basis of $$\mathbb{C}^{d^2}$$? In short: is there something to understand in his statement or not really?

• Every maximally entangled state is of the form $\phi = (\Bbb I \times U)|\Omega\rangle$ and not just $|\Omega\rangle$. Anyway, it's simple to construct it using Schmidt decomposition (cf. this answer). However, note that the Hilbert space for a quantum harmonic oscillator is infinite-dimensional i.e. $L^2(\Bbb R)$ and things get complicated there. Werner et al. apparently have a paper on this topic. – Sanchayan Dutta Dec 18 '19 at 10:58
• [cont.] The author writes $\Bbb C^d \otimes C^d$ and not $\Bbb C^{d^2}$ in order to emphasize that a basis is being determined for a composite system. Of course, there exists an isomorphism between the two. The idea is to choose a subspace of a common dimension for both $H_A$ and $H_B$. – Sanchayan Dutta Dec 18 '19 at 10:59
• @SanchayanDutta thank you for your answer. Forget about the infinite dimension problem, restrict the harmonic oscillator to a finite dimension $N$ for example. My point was to say that the state $|ii\rangle$ is ill defined, and thus $|\Omega \rangle$ as well. If I choose $|0\rangle$ for the first system, what should I take for the second one : $|0>$, $|1\rangle$, $|2\rangle$ ? (assuming I have $N>3$ for the second system). – StarBucK Dec 18 '19 at 11:04
• Then should I understand the property more in the sense: I call $|\Omega\rangle$ any fixed maximally entangled state, and using local unitaries I can construct any maximally entangled state from $|\Omega \rangle$ ? – StarBucK Dec 18 '19 at 11:05
• Glad to help! I'll try to write this as a proper answer tomorrow and delete the comments, as comments tend to be transient (I ended up writing way more than what I initially expected.) :) – Sanchayan Dutta Dec 18 '19 at 12:19

Let's say that you're working in the energy eigenbasis of a quantum harmonic oscillator $$\dagger$$ and label the basis states as $$\{|0⟩_H,|1⟩_H,|2⟩_H,…\}$$, and you have the basis $$\{|0⟩_Q,|1⟩_Q\}$$ for your qubit. $$d$$ would be $$\text{min}(𝑁,2)=2$$ in this case. Then say you choose the basis $$\{|0_Q0_H⟩,|0_Q2_H⟩,|1_Q0_H⟩,|1_Q2_H⟩\}$$ for $$\Bbb C^2\otimes \Bbb C^2$$. Or you could choose the $$\{|0_Q+_H⟩,|0_Q−_H⟩,|1_H+_Q⟩,|1_H−_H⟩\}$$ basis, where $$|+⟩=\frac{1}{\sqrt 2}(|0⟩+|1⟩)$$ and $$|−⟩=\frac{1}{\sqrt 2}(|0⟩+|1⟩)$$. Both are fine.

A valid $$|\Omega⟩$$ would be $$\frac{1}{\sqrt 2}(|0_Q+_H⟩+|1_Q−_H⟩)$$. Another valid $$|\Omega⟩$$ would be $$\frac{1}{\sqrt 2}(|0_Q0_H⟩+|1_Q2_H⟩)$$. Both the $$|\Omega⟩$$'s are perfectly good examples of maximally mixed states! I'm not sure "how do I know how to choose the kets?" is the right question.

You're right that any maximally entangled state can be constructed by applied local unitaries on some (non-unique) $$|\Omega⟩$$. "Ill-defined" is not the right term here; it's just "non-uniquely defined". The point is that any maximally mixed state is of the form $$(I\otimes U)|\Omega⟩$$ and can be expressed as some (constrained) combination of (non-unique) basis states of some (non-unique) subspace $$\Bbb C^d\otimes \Bbb C^d$$ of $$H_A\otimes H_B$$.

Perhaps you were confused because you were thinking $$|\Omega\rangle$$ is some unique state. That's certainly not the case! The description of $$|\Omega⟩$$ simply provides the mathematical structure that we expect of a legitimate maximally entangled state.

The author writes $$\Bbb C^d \otimes \Bbb C^d$$ and not $$\Bbb C^{d^2}$$ in order to emphasize that a basis is being determined for a composite system. Of course, there exists an isomorphism between the two. The idea is to choose a subspace of a common dimension for both $$H_A$$ and $$H_B$$. I do not think there's any other deeper reason for why the author chooses to write $$\Bbb{C}^{d}\otimes \Bbb{C}^d$$ rather than $$\Bbb C^{d^2}$$.

$$\dagger$$: In general, the Hilbert space for a quantum harmonic oscillator is infinite-dimensional i.e. $$L^2(\Bbb R)$$ and things get complicated. Keyl et al. apparently have a paper (arXiv:quant-ph/0212014) on this topic.

For me this state is ill-defined in general

It's not "ill-defined", just not uniquely defined. If the dimensions are different you can indeed define different inequivalent states that all have the same right to be called "maximally entangled".

Note that the "maximally entangled state" is not fully characterised even in the case in which the dimensions are the same: you can define it as $$\sum_i |u_i,v_i\rangle$$ for every pair of orthonormal bases $$(u_i)_i, (v_i)_i$$ for the two spaces. Usually when these states are used this "undefinedness" does not matter (think e.g. of the definition of the Choi matrix using maximally entangled states). Although it is true that when the dimensions are the same these states can all be mapped one into the other via local operations, which is not the case for "maximally entangled states" in spaces with different dimensions.