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The pure state

$|\psi\rangle = \sum \alpha_i |i\rangle$, $\rho = \sum_{i,j}\alpha_i^*\alpha_j |j\rangle\langle i|$,

why is the form not be $\rho = \sum_{i}\alpha_i^*\alpha_i |i\rangle\langle i|$ ???

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2 Answers 2

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In short, we need to consider cross-terms when combining two summations. Recall that for a pure state $|\psi\rangle$, its density matrix representation is $\rho = |\psi\rangle \langle \psi |$.

For example, if our state was $|\psi\rangle = \frac{1}{\sqrt{2}} (|0\rangle + |1\rangle)$, then we would have $$\rho = |\psi \rangle \langle \psi | = \underbrace{\left [ \left( \frac{1}{\sqrt{2}} \right) (|0\rangle + |1\rangle) \right]}_{|\psi\rangle} \cdot \underbrace{ \left[ \left( \frac{1}{\sqrt{2}} \right)^* (\langle 0 | + \langle 1 |) \right]}_{\langle \psi|}$$ $$ = \frac{1}{2} (|0\rangle \langle 0 | + |0 \rangle \langle 1 | + |1\rangle \langle 0 | + |1\rangle \langle 1|),$$ which is the same as the form you've given on the first line of your question with $\alpha_0 = \alpha_1 = \frac{1}{\sqrt{2}}$.

More generally, for a pure state $|\psi\rangle = \sum_k \alpha_k |k\rangle$, we have

$$\rho = |\psi\rangle \langle \psi | = \left( \sum_j \alpha_j |j\rangle \right) \left( \sum_i \alpha_i^* \langle i | \right) = \sum_{i} \sum_{j} \alpha_i \alpha_j^* |i\rangle \langle j|. $$

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  • $\begingroup$ That's Right, thanks! $\endgroup$
    – KarryMa
    Jul 8, 2022 at 13:49
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Imagine that my pure state is one of two possible states $|\psi\rangle=|\pm\rangle=(|0\rangle\pm|1\rangle)/\sqrt{2}$. These are very definitely different states, and you can easily perform an experiment to distinguish them: apply the Hadamard gate and measure in the standard basis.

Now, let's write your version of the density matrix for these two states $$ \rho_+=\frac12(|0\rangle\langle 0|+|1\rangle\langle 1|)=\rho_-. $$ They are written in the same way. This would suggest that your way of writing them is missing something because I've lost any possibility of distinguishing them.

On the other hand, the standard way of converting a pure state into a density matrix is $$ \rho=|\psi\rangle\langle\psi|. $$ Remember that each time you write out $|\psi\rangle$ you have a sum, and those different sums must use different indices, $$ \rho=\left(\sum_i\alpha_i|i\rangle\right)\left(\sum_j\alpha_j^\star\langle j|\right). $$

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  • $\begingroup$ Are you sure the density matrix for these two states are same? $\endgroup$
    – KarryMa
    Jul 8, 2022 at 14:03
  • $\begingroup$ They shouldn't be, but my point is that the way you propose to write them, they are. $\endgroup$
    – DaftWullie
    Jul 8, 2022 at 14:06
  • $\begingroup$ Yes, I see, thank you! $\endgroup$
    – KarryMa
    Jul 8, 2022 at 15:03

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