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

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


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

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.


3 Answers 3


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 $\lbrace 1/2,1/2\rbrace$.

  • 1
    $\begingroup$ 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? $\endgroup$
    – MBolin
    Commented Feb 27, 2020 at 11:07
  • 3
    $\begingroup$ 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 $|\psi\rangle$. $\endgroup$
    – DaftWullie
    Commented Feb 27, 2020 at 12:02
  • $\begingroup$ How do you ensure that partial traces do not change under unitary transformations? $\endgroup$ Commented May 7, 2021 at 9:46
  • $\begingroup$ @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). $\endgroup$
    – DaftWullie
    Commented May 7, 2021 at 12:11
  • $\begingroup$ 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? $\endgroup$ Commented May 7, 2021 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 $ .

  • $\begingroup$ 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 $\phi_1$. $\endgroup$
    – MBolin
    Commented Feb 25, 2020 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 $\beta^2 + \delta^2 = 1$.

  • 1
    $\begingroup$ 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 $| \phi_1 \rangle$ ). $\endgroup$
    – MBolin
    Commented Feb 25, 2020 at 17:26
  • $\begingroup$ it's valid objections, the proof would need to be reworked to includes them $\endgroup$ Commented Feb 25, 2020 at 17:50

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