In his lecture notes Scott Aaronson states:

Now, consider the $2$-qubit pure state $\frac{|00\rangle + |01\rangle + |10\rangle}{\sqrt{3}}$. We'll give the first qubit to Alice and the second to Bob. How does Bob calculate his density matrix? By picking some orthogonal basis for Alice's side. You can rewrite the state as $\sqrt{\frac{2}{3}}|0+\rangle + \sqrt{\frac{1}{3}}|10\rangle$, which lets you calculate Bob's density matrix as: $$ \frac{2}{3}|+\rangle\langle+| + \frac{1}{3}|0\rangle\langle0| = \frac{2}{3} \begin{bmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & \frac{1}{2} \end{bmatrix} + \frac{1}{3} \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} = \begin{bmatrix} \frac{2}{3} & \frac{1}{3} \\ \frac{1}{3} & \frac{1}{3} \end{bmatrix} $$

In general, if you have a bipartite pure state, it'll look like $$ \sum_{i,j = 1}^N \alpha_{ij} |i\rangle|j\rangle = |\psi\rangle $$ And you can get Bob's local density matrix $$ (\rho_\text{Bob})_{j, j'} = \sum_i \alpha_{ij} \alpha_{ij'}^* $$

I have two problems with this formulation.

  1. I'm not sure how this generalizes to states with more qubits.
  2. These indices are making my head explode.

Is there another way to go about this, perhaps in terms of outer products as Aaronson hints at with his example, which doesn't have these shortcomings? An example computation would also be very welcome.


1 Answer 1


For calculating a local density matrix or in other words taking a partial trace has a neat trick (IMHO):

The Trick:

  1. Write the state given in density matrix form, if not already. Which means if $\vert \psi \rangle$ is the state, the density matrix is $\vert \psi \rangle \langle \psi \vert$.
  2. Turn all the outer-products into inner-products for the dimensions you don't want.


  1. $$\vert \psi \rangle =\frac{|00\rangle + |01\rangle + |10\rangle}{\sqrt{3}}$$ $$\vert \psi \rangle \langle \psi \vert = \big(\frac{|00\rangle + |01\rangle + |10\rangle}{\sqrt{3}}\big)\big( \frac{\langle00| + \langle01| + \langle10\vert}{\sqrt{3}} \big) $$ $$=\big( \frac{\vert00\rangle\langle00\vert+\vert00\rangle\langle01\vert+\vert00\rangle\langle10\vert+\vert01\rangle\langle00\vert+\vert01\rangle\langle01\vert+\vert01\rangle\langle10\vert+\vert10\rangle\langle00\vert+\vert10\rangle\langle01\vert+\vert10\rangle\langle10\vert}{3}\big)$$
  2. Since you want the state of 2nd system, you would turn all the outer-products for 1st system to inner-prodcuts(This implies all the terms of form, $\langle0\vert1\rangle=\langle1\vert0\rangle=0$ and $\langle0\vert0\rangle=\langle1\vert1\rangle=1$, $$\rho_2 = \big( \frac{1*\vert0\rangle\langle0\vert+1*\vert0\rangle\langle1\vert+0*\vert0\rangle\langle0\vert+1*\vert1\rangle\langle0\vert+1*\vert1\rangle\langle1\vert+0*\vert1\rangle\langle0\vert+0*\vert0\rangle\langle0\vert+0*\vert0\rangle\langle1\vert+1*\vert0\rangle\langle0\vert}{3}\big)$$ $$\rho_2 = \big( \frac{\vert0\rangle\langle0\vert+\vert0\rangle\langle1\vert+\vert1\rangle\langle0\vert+\vert1\rangle\langle1\vert+\vert0\rangle\langle0\vert}{3}\big)$$ $$\rho_2 = \big( \frac{2\vert0\rangle\langle0\vert+\vert0\rangle\langle1\vert+\vert1\rangle\langle0\vert+\vert1\rangle\langle1\vert}{3}\big)$$ Which is the state in the example you quote. Now, this will become second nature when you become little experienced and you wouldnt even be doing all this algebra!

Example on 3-qubit system:

  1. I deliberately choose an state with 2 terms just to show the proof of principle, you can work out a complicated state if you wish so!, $$\vert \psi \rangle = \sqrt{\frac{2}{3}}\vert 011 \rangle + \sqrt{\frac{1}{3}}\vert100\rangle$$ $$ \vert \psi \rangle \langle \psi \vert = \frac{2}{3}\vert011\rangle\langle011\vert +\frac{1}{3}\vert100\rangle\langle100\vert +\frac{\sqrt{2}}{3}\vert011\rangle\langle100\vert +\frac{\sqrt{2}}{3}\vert100\rangle\langle011\vert$$
  2. Lets say you want the state of 2nd qubit, turn all the outer-products for 1 and 3 qubit to inner-products. $$\rho_2=\frac{2}{3}\langle0\vert0\rangle\vert1\rangle\langle1\vert\langle1\vert1\rangle +\frac{1}{3}\langle1\vert1\rangle\vert0\rangle\langle0\vert\langle0\vert0\rangle +\frac{\sqrt{2}}{3}\langle0\vert1\rangle\vert0\rangle\langle0\vert\langle1\vert0\rangle +\frac{\sqrt{2}}{3}\langle1\vert0\rangle\vert0\rangle\langle0\vert\langle0\vert1\rangle$$ $$\rho_2 = \frac{2}{3}\vert0\rangle\langle0\vert + \frac{1}{3}\vert1\rangle\langle1\vert$$

Explaination of trick: Trace on a single qubit state(in computational basis), $\sigma$ is given by sum of diagonal elements, $$Tr(\rho) = \sum_{i|i\in\{0,1\}}\langle i \vert\sigma\vert i\rangle.$$ Similarly for 2-qubit state $\rho$, $$Tr(\rho) = \sum_{i,j |i,j\in\{0,1\} }\langle ij\vert\rho\vert ij\rangle.$$

Now, given this 2-qubit state if you want to know the local state of 1st qubit you would take a partial trace over 2nd qubit: $$Tr_2(\rho) = \sum_{j|j\in\{0,1\}}\langle j \vert\rho\vert j\rangle.$$ In above you can see that if there was an element of form $\vert0\rangle\langle1\vert,\vert1\rangle\langle0\vert$ for 2nd qubit, they will go to zero while the terms of form $\vert0\rangle\langle0\vert,\vert1\rangle\langle1\vert$ goes to 1, and hence the trick! and if you want to know the local state of 2nd qubit you would take a partial trace over 1st qubit: $$Tr_1(\rho) = \sum_{i|i\in\{0,1\}}\langle i \vert\rho\vert i\rangle.$$


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