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We know that if we consider a Bell state for example $$ |\Phi^+\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}} $$ Then we can write this state in some other orthonormal basis in the same form. Like: $$ |\Phi^+\rangle = \frac{|00\rangle + |11\rangle}{\sqrt{2}} = \frac{|aa\rangle + |a^{\perp}a^{\perp}\rangle}{\sqrt{2}} $$

Can we have same result for GHZ states with $n$ qubits. In particular can we write $$ |GHZ_3\rangle = \frac{|000\rangle + |111\rangle}{\sqrt{2}} = \frac{|aaa\rangle + |a^{\perp}a^{\perp}a^{\perp}\rangle}{\sqrt{2}} $$

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The statement was already shown to not hold exactly as stated. Still, one could hope for it to hold by somewhat relaxing the notation. In other words, could it be possible that you can write the GHZ as $$|a,b,c\rangle + |a^\perp,b^\perp,c^\perp\rangle$$ for some choice of states $|a\rangle,|b\rangle,|c\rangle$?

For the purpose, consider a bipartite pure state in Schmidt form: $$|\Psi\rangle=\sum_k \sqrt{p_k} (|u_k\rangle\otimes|v_k\rangle).$$ Is it possible to write this using different choices of local bases? The answer can be readily seen to be negative, unless some Schmidt coefficients are equal. More precisely, the question is equivalent to asking about the different possible singular value decompositions of the matrix $\Psi$ such that $\operatorname{vec}(\Psi)\equiv|\Psi\rangle$. As discussed e.g. here, you can use different choices of local bases iff the matrix $\Psi^\dagger\Psi$ is degenerate (equivalently, iff some Schmidt coefficients are equal).

In particular, if $|\Psi\rangle=\sum_k |u_k,v_k\rangle$, then $\Psi=UV^T$ with $U,V$ unitaries whose columns are the vectors $|u_k\rangle,|v_k\rangle$, respectively. Thus $\Psi = W(VU^T \bar W)^T$ for any unitary $W$, which means we can equivalently write the Schmidt decomposition as $$\sum_k |u_k,v_k\rangle = \sum_k |w_k, w_k'\rangle$$ for any orthonormal basis $\{|w_k\rangle\}_k$ and with $|w_k'\rangle$ the columns of the matrix $VU^T \bar W$. A typical example of this is $$\sum_k |k,k\rangle = \sum_k |w_k, \bar w_k\rangle$$ where $|\bar w_k\rangle$ is the vector whose coefficients are the complex conjugates of those of $|w_k\rangle$.

All this to say, that if an entangled state has non-degenerate Schmidt coefficients, then there is a unique way to decompose it in such a way, and vice versa. This is relevant to the example at hand, because it allows us to write $$|000\rangle+|111\rangle = |u\rangle\otimes(\bar u_0 |00\rangle + \bar u_1 |11\rangle) + |v\rangle\otimes(\bar v_0 |00\rangle + \bar v_1 |11\rangle)$$ for any pair of orthonormal single-qubit vectors $|u\rangle,|v\rangle$. But notice how we now got non-product states on the last two qubits. The only way to get product states attached to each state of the first qubit is to have $u_0 u_1=v_0 v_1=0$. In other words, there is no other way to get a Schmidt decomposition for the GHZ state between first qubit and the rest, that is also a Schmidt decomposition with respect to different bipartitions. Or equivalently, the GHZ state only "looks like a GHZ state" with respect to a unique choice of basis.

Incidentally, this argument doesn't need to only use three-qubit GHZ states. You can use the same argument with $|0^n\rangle+|1^n\rangle$ and reach the same conclusion.

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No, you can't. Almost any choice you make for $|a\rangle$ will demonstrate this to you.

For example, if I let $|a\rangle=(|0\rangle+|1\rangle)/\sqrt{2}$, then I have $$ |aaa\rangle+|a^\perp a^\perp a^\perp\rangle=\frac{1}{4}\sum_{x\in\{0,1\}^3}|x\rangle(1+(-1)^{|x|})=\frac12\sum_{x: |x|\equiv 0\text{ mod }2}|x\rangle. $$ This is clearly not the same as $(|000\rangle+|111\rangle)/\sqrt{2}$

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  • $\begingroup$ The question is specifically talking about orthonormal bases, the notation suggests that $|a^\perp\rangle$ is orthogonal to $|a\rangle$, so I must disagree with this answer $\endgroup$ Mar 22, 2022 at 13:40
  • $\begingroup$ @QuantumMechanic yes, so I took $|a\rangle=|+\rangle$ and $|a^\perp\rangle=|-\rangle$, an orthonormal basis. $\endgroup$
    – DaftWullie
    Mar 22, 2022 at 13:42
  • $\begingroup$ Why would you not consider $|+++\rangle+|---\rangle$ to be a GHZ state?? You can use it to violate the same GHZ (Bell-type) inequalities.... oh, the question asked about strict equality, not about whether $|+++\rangle+|---\rangle$ is also a GHZ state. But then your answer should be that this is not even possible for Bell states, only for the singlet state! $\endgroup$ Mar 22, 2022 at 15:10
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    $\begingroup$ @QuantumMechanic Well, I was letting that slide because for $|\Phi^+\rangle$ you still get equality provided $|a\rangle$ and $|a^\perp\rangle$ both have real coefficients. But yes, for it to be universally true it only applies to the singlet state. $\endgroup$
    – DaftWullie
    Mar 22, 2022 at 15:40
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    $\begingroup$ @QuantumMechanic I think for Bell state it is true. I don't know the actual proof but you can represent any such $|a \rangle$ and $|a^{\perp}\rangle$ in terms of $0$ and $1$ to get the result. I am not sure about the GHZ state. But there must be some relation. $\endgroup$
    – IamKnull
    Mar 22, 2022 at 15:40
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Let's prove that it is impossible for a pair of orthonormal states $|a\rangle$ and $|a^\perp\rangle$ to exactly equal the GHZ state in the computational basis $$|0\rangle^{\otimes n}+|1\rangle^{\otimes n}=|a\rangle^{\otimes n}+|a^\perp\rangle^{\otimes n}$$ unless $n= 2$. Incidentally, the state $|a\rangle^{\otimes n}+|a^\perp\rangle^{\otimes n}$ will have the same properties as a GHZ state (it is a GHZ state) and can be brought to the same form as the canonical GHZ state $|0\rangle^{\otimes n}+|1\rangle^{\otimes n}$ via local unitaries.

Writing the most general qubit state as $$|a\rangle=c|0\rangle+e^{i\phi}s|1\rangle$$ with $c^2+s^2=1$ and all variables being real, we construct the most general state orthogonal to this one as $$|a^\perp\rangle=s|0\rangle-e^{i\phi}c|1\rangle;$$ there is no other normalized pure state orthogonal to $|a\rangle$ because we are in dimension $2$ and global phases are irrelevant.

We can now inspect the coefficients in the linear expansion of our candidate state $|a\rangle^{\otimes n}+|a^\perp\rangle^{\otimes n}$ in terms of $n$-component bit strings like $|0010\cdots\rangle=|0\rangle\otimes|0\rangle\otimes|1\rangle\otimes |0\rangle\otimes \cdots$. The coefficient of $|0\rangle^{\otimes n}$ will be $c^n+s^n$ and that of $|1\rangle^{\otimes n}$ will be $e^{in\phi}[s^n+(-1)^n c^n]$. For those to be equal, we need $e^{in\phi}$ to be real, so $\phi$ must be an integer multiple of $\pi/n$. Our first constraint is thus $$c^n+s^n=e^{in\phi}[s^n+(-1)^n c^n],$$ which has the following possibilities:

  1. $e^{in\phi}=1$, $n$ even, then satisfied for all $c,s$.
  2. $e^{in\phi}=1$, $n$ odd, only satisfied when $c=0$, which is the original GHZ state in the computational basis.
  3. $e^{in\phi}=-1$, $n$ even, only satisfied when $c^n+s^n=0$. But this cannot be the case when $n$ even because both $c^n$ and $s^n$ cannot be negative and one must be nonzero, so no solutions.
  4. $e^{in\phi}=-1$, $n$ odd, only satisfied when $s=0$, which is the original GHZ state in the computational basis.

So we immediately realize that we require $e^{in\phi}=1$ and $n$ even to be able to find an equivalent GHZ state in a different basis. This nicely tells us that we cannot have another GHZ state with $n=1$ equal to $|0\rangle+|1\rangle$. We then take the next coefficient, which must be zero, and look at the constraints it imposes; for example, the expansion coefficient of $|0\cdots 01\rangle$ must vanish, yielding $$n(c^{n-1}e^{i\phi}s- s^{n-1}c e^{i\phi})=0\quad\Rightarrow \quad c^n=s^n\quad\mathrm{or}\quad n=2.$$ So the only possible times when our candidate state will be equivalent to the original GHZ state in the computational basis is when (1a) $n=2$ and the states are constrained to have real coefficients with $e^{i\phi}=\pm 1$ or (1b) $e^{in\phi}=1$, $n$ even, and $|c|=|s|=1/\sqrt{2}$.

Next, we inspect the expansion coefficients of some generic state in the computational basis with $k$ $0$s and $n-k$ $1$s, yielding $$\binom{n}{k}(c^k s^{n-k}e^{i\phi(n-k)}+ (-1)^{n-k}s^k c^{n-k} e^{i\phi(n-k)})=0,$$ where without loss of generality we may now take $k\leq n-k$ and $k>0$. This then yields either

  1. $s=c$ and $(-1)^{n-k}$ odd.
  2. $s=-c$ and $(-1)^{n-k}$ even.

Since these relations must hold for all integers $k$, the sign of $(-1)^{n-k}$ will alternate, and we must simultaneously have $s=c$ and $s=-c$, which cannot hold when at least one of $c$ and $s$ must be nonzero. This concludes the proof.


What if we consider adding a global phase to $|a^\perp\rangle$ to define it as $$|a^\perp\rangle=e^{i\Phi}\left(s|0\rangle-e^{i\phi}c|1\rangle\right)?$$ The condition on the vanishing coefficients then becomes $$\binom{n}{k}(c^k s^{n-k}e^{i\phi(n-k)}+ e^{i\Phi n}(-1)^{n-k}s^k c^{n-k} e^{i\phi(n-k)})=0$$ for $k\neq 0,n$. This leads to the conditions $$s^{n-2k}=-(-1)^{n-k}e^{i\Phi n}c^{n-2k}$$ for all $k>n-k$. The only way for this to hold is with $e^{i\Phi n}=\pm1$, which is the phase that sends $|a^\perp\rangle^{\otimes n}\to\pm |a^\perp\rangle^{\otimes n}.$ We still find the necessity $|c|=|s|=1/\sqrt{2}$, the sign of the $c=\pm s$ changes with each integer $k$, so we again find that this condition cannot be satisfied unless $n=2$.

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  • $\begingroup$ " and global phases are irrelevant." Actually, they are relevant because once you make your GHZ superposition, they are no longer global, but relative phases (between the $a$ and $a^\perp$ terms). $\endgroup$
    – DaftWullie
    Mar 23, 2022 at 7:16
  • $\begingroup$ @DaftWullie fair... this branches the proof into even more things to verify! $\endgroup$ Mar 23, 2022 at 13:01
  • $\begingroup$ @DaftWullie actually it didn't need any more branches for the proof, luckily $\endgroup$ Mar 23, 2022 at 15:26

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