# Prove that there exist tripartite $|\psi\rangle$ which cannot be written as $|\psi\rangle=\sum_i\lambda_i|i_A\rangle|i_B\rangle|i_C\rangle$

Exercise 2.77 in Nielsen and Chuang asks to show by example that there exist tripartite states $$| \psi \rangle_{ABC}$$ which cannot be written as

$$| \psi \rangle = \sum_i \lambda_i | i_A \rangle | i_B \rangle | i_C \rangle .$$

In this unofficial solution set, they consider the state

$$| \psi \rangle = | 0 \rangle_A \otimes | \Phi_+ \rangle_{BC} = |0 \rangle_A \otimes \bigg[ \frac{1}{\sqrt{2}} \Big( |00 \rangle_{BC} + | 11 \rangle_{BC} \Big) \bigg].$$

Then they perform the arbitrary (unitary) change of basis

$$|0\rangle = \alpha | \phi_0 \rangle + \beta | \phi_1 \rangle , \\ |1\rangle = \gamma | \phi_0 \rangle + \delta | \phi_1 \rangle$$

obtaining

$$| \psi \rangle = \Big( \alpha | \phi_0 \rangle_A + \beta | \phi_1 \rangle_B \Big) \otimes \bigg[ \frac{1}{\sqrt{2}} \Big( |\phi_0 \phi_0 \rangle_{BC} + | \phi_1 \phi_1 \rangle_{BC} \Big) \bigg].$$

Since the cross terms cannot vanish, the proof is finished. However, I don't see the last step. Why do you have $$|00 \rangle_{BC} + | 11 \rangle_{BC} = |\phi_0 \phi_0 \rangle_{BC} + | \phi_1 \phi_1 \rangle_{BC}$$ ? I mean, for instance, the unitarity of the change of basis doesn't buy us the cancellation of the terms $$| \phi_0 \phi_1 \rangle$$ and $$| \phi_1 \phi_0 \rangle$$ (I think).

EDIT: Note that:

1) The coefficients $$\alpha$$, $$\beta$$, $$\gamma$$ and $$\delta$$ are not, in general, real. Since we want to show that there is no way of rewritting $$| \psi \rangle$$ in Schmidt form, we have to consider the most general case where these coefficients are complex.

2) The change of basis is, in the most general case, different for each party. I mean, we should rather write

$$|0 \rangle_X = \alpha_X \, | \phi_0 \rangle_X + \beta_X \, | \phi_1 \rangle_X \, , \\ | 1 \rangle_X = \gamma_X \, | \phi_0 \rangle_X + \delta_X \, | \phi_1 \rangle_X \, ,$$

where $$X= A, B, C$$. Unitarity conditions are

$$| \alpha_X |^2 + |\beta_X |^2 = | \gamma_X |^2 + |\delta_X |^2 = 1 \, ,\\ \alpha_X^* \gamma_X + \beta_X^* \delta_X = 0$$

for each $$X$$. I don't see how you can cancel the cross terms using this.

In case you are interested in an alternative proof...

For a state $$|\psi\rangle$$ as given, the reduced density matrices of each of the three qubits must all be of the form $$\sum_i\lambda_i|i_X\rangle\langle i_X|$$, where the basis $$|i_X\rangle$$ is orthonormal. That means that all three density matrices have the same spectrum (eigenvalues $$\lambda_i$$).

Now, consider a state of the form $$|0\rangle(|00\rangle+|11\rangle)/\sqrt{2}$$ The first qubit is separable, so the eigenvalues of the reduced density matrix are $$\lbrace 1,0\rbrace$$. The other two qubits are maximally mixed due to the entanglement, with eigenvalues $$\lbrace 1/2,1/2\rbrace$$.

• I don't see it. For a bipartite state you can use the Schmidt decomposition $|\psi \rangle= \sum_i \lambda_i |i_A \rangle |i_B \rangle$ to see that both density matrices $\rho_A = \sum_i \lambda_i^2 | i_A \rangle \langle i_A |$ and $\rho_B = \sum_i \lambda_i^2 | i_B \rangle \langle i_B |$ have the same spectrum (same eigenvalues $\lambda_i^2$). However, for a tripartite state, since apparently you don't always have Schmidt decomposition, how do you know all density matrices have the same spectrum? Feb 27 '20 at 11:07
• We're trying to make the argument oppositely. If all states could be written in the form $|\psi\rangle$, the reduced density matrices would have the same spectra. However, since we can show that three are 3-qubit states where the reduced density matrices do not have the same spectra, we conclude that not all states can be written in the form $|\psi\rangle$. Feb 27 '20 at 12:02
• How do you ensure that partial traces do not change under unitary transformations? May 7 at 9:46
• @user2820579 Local unitary transformations only change the individual reduced density matrices and do not change their spectra. This is a property of the partial trace (effectively, trace is basis independent). May 7 at 12:11
• I was not thinking in local unitary transformations, but in entangling ones. What you say is that if $|\Psi\rangle$ is not separable, then it cannot be separable under any unitary right? Is there a theorem that proves this? May 7 at 12:39

It is not true that $$|00 \rangle + | 11 \rangle = |\phi_0 \phi_0 \rangle + | \phi_1 \phi_1 \rangle$$ for an arbitrary change of basis. For example, let $$|\phi_0 \rangle = |0\rangle$$ and $$|\phi_1 \rangle = i|1 \rangle$$.

You can, however, pick $$\phi_1$$ so that this equality is true, given $$\phi_0$$. If $$|\phi_0 \rangle = \kappa | 0 \rangle + \mu | 1 \rangle$$, pick $$| \phi_1 \rangle = \mu | 0 \rangle - \kappa | 1 \rangle$$ .

• But you want to consider the most general case, since you want to show that $| \psi \rangle$ has no Schmidt form. It's not enough to show that it has no Schmidt form when you pick an appropriate $\phi_1$. Feb 25 '20 at 20:19

the trick is to show the relation $$\alpha \beta$$ = - $$\gamma \delta$$ so the cross terms cancels out. It's not so straightforward in my opinion.

Let's develop the term (dropping the subscript here) $$|00⟩+|11⟩$$ in term of the new basis :

$$|00⟩+|11⟩ = (\alpha^2+\gamma^2) |\phi_0 \phi_0⟩ + (\beta^2+\delta^2) |\phi_1 \phi_1⟩ + \alpha \beta |\phi_0 \phi_1⟩ + \alpha \beta |\phi_1 \phi_0⟩ + \gamma \delta |\phi_0 \phi_1⟩ + \gamma \delta |\phi_1 \phi_0⟩$$

If we can show that $$\alpha \beta$$ = - $$\gamma \delta$$ then the above expression reduces to

$$|00⟩+|11⟩ =(\alpha^2+\gamma^2) |\phi_0 \phi_0⟩ + (\beta^2+\delta^2) |\phi_1 \phi_1⟩$$

and we would only need to prove $$(\alpha^2+\gamma^2) = (\beta^2+\delta^2) = 1$$.

Let's show that indeed $$\alpha \beta$$ = - $$\gamma \delta$$. First remind that a unitary transform preserves orthogonality and since $$|0⟩$$ and $$|1⟩$$ are orthogonal, so are $$\alpha |\phi_0⟩ + \beta |\phi_1⟩$$ and $$\gamma |\phi_0⟩ + \delta |\phi_1⟩$$

writing the scalar product between the two and using orthogonality we find the relation :

$$\alpha \gamma + \beta \delta = 0$$

Not yet what we want. We also know from definition of a qubit that

$$\alpha^2 +\beta^2 =1$$

$$\gamma^2 + \delta^2 = 1$$

combining those three equalities and with some help from wolphram I found that this system had solutions all implying the equality $$\alpha \beta = -\gamma \delta$$. I'm not writing the solutions here but you can use this link to wolphram to check it yourself. Hence the cross term do cancel.

To show that $$(\alpha^2+\gamma^2) = (\beta^2+\delta^2) = 1$$, we can use the result we just showed and the equality using orthogonality :

$$\alpha \gamma = - \beta \delta$$

$$\alpha \beta = -\gamma \delta$$

so $$\gamma = -\beta$$ and $$\alpha^2 + \beta^2 = \alpha^2 + \gamma^2 = 1$$. proceed the same way to show $$\beta^2 + \delta^2 = 1$$.

• Two objections: 1) I think you are assuming the coefficients $\alpha$, $\beta$, $\gamma$ and $\delta$ are real. 2) I think you are assuming that the change of basis is the same for all parties, and I think this need not be so. I mean that the orthonormal bases in the Schmidt decomposition $|i_A \rangle$, $| i_B \rangle$ and $| i_C \rangle$ need not equal (they need not be the same linear combination of $| \phi_0 \rangle$ and $| \phi_1 \rangle$ ). Feb 25 '20 at 17:26
• it's valid objections, the proof would need to be reworked to includes them Feb 25 '20 at 17:50