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Suppose that there is an ensemble with 60% of the states prepared in

$$|a\rangle=\sqrt{\frac{2}{5}}|+\rangle-\sqrt{\frac{3}{5}}|-\rangle$$ and 40% in:

$$|b\rangle=\sqrt{\frac{5}{8}}|+\rangle+\sqrt{\frac{3}{8}}|-\rangle$$

I've considered two ways to calculate this. For the first, one could replace the $|+\rangle$ and $|-\rangle$ and calculate the density matrix as usual in $0,1$ base. Another possibility would be to calculate the density matrix in $\{|+\rangle,|-\rangle\}$ basis and then to convert it into a corresponding $|0\rangle$ $|1\rangle$ density matrix. My question, how would you proceed here?

But I'm not sure if both ways come to the same result?

The density matrix in $\{|+\rangle,|-\rangle\}$ Base is: $$\rho_a=\frac{2}{3}|+\rangle\langle+|-\frac{\sqrt{6}}{5}|+\rangle\langle-|-\frac{\sqrt{6}}{5}|-\rangle\langle+|+\frac{3}{5}|-\rangle\langle-|$$ $$\rho_b=\frac{5}{8}|+\rangle\langle+|+\frac{\sqrt{15}}{8}|+\rangle\langle-|+\frac{\sqrt{15}}{8}|-\rangle\langle+|+\frac{3}{8}|-\rangle\langle-|$$ $$\rho=\frac{2}{5}p_b+\frac{3}{5}p_a=\begin{pmatrix}\frac{49}{100}&\frac{5\sqrt{15}-12\sqrt{6}}{100}\\\frac{5\sqrt{15}-12\sqrt{6}}{100}&\frac{41}{100}\end{pmatrix}$$

Now the question is, can one translate this density matrix $p$ from $\{|+\rangle,|-\rangle\}$ into the $\{|0\rangle, |1\rangle\}$ basis?

or should one have started like this:

$$|a\rangle=\sqrt{\frac{2}{5}}|+\rangle-\sqrt{\frac{3}{5}}|-\rangle=\sqrt{\frac{2}{5}}\left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right)-\sqrt{\frac{3}{5}}\left(\frac{|0\rangle-|1\rangle}{\sqrt{2}}\right)$$

Then calculate the same for $|b\rangle$ and then the density matrix


Then I would be interested, for this question, I did not want to open a new topic. Suppose we have the following density matrix:

$$\rho=\begin{pmatrix}\frac{2}{5}&\frac{-i}{8}\\\frac{i}{8}&\frac{3}{5}\end{pmatrix}$$

What is the probability that the system is in state 0? I would say 2/5 so 40%, but according to the solution that is not true. So my question is why is not that true?

I have found these tasks out of interest and try to work on them. But I wanted to get some advice from those present ...

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  • $\begingroup$ The first occurrence of your matrix $p$ does not have trace 1, so you must have a typo/error somewhere. $\endgroup$ – DaftWullie Jun 26 at 13:47
  • $\begingroup$ In fact, I had overlooked that, then I'll calculate it again. Thank you! That was very fast of you! $\endgroup$ – P_Gate Jun 26 at 13:56
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Let's call the two states $|a_1\rangle$ and $|a_2\rangle$ to distinguish them. What you're saying is that the state has been prepared (to the best of our knowledge) as $$ \rho=\frac{3}{5}|a_1\rangle\langle a_1|+\frac{2}{5}|a_2\rangle\langle a_2|. $$ The probability that you expect it to give you an answer $|0\rangle$ if you project in the standard basis is $\langle 0|\rho|0\rangle$. You can perform that calculation in whatever basis you want to!

I would probably start by calculating $$ \langle 0|a_1\rangle=\frac{1}{\sqrt{2}}\sqrt{\frac{2}{5}}-\frac{1}{\sqrt{2}}\sqrt{\frac{3}{5}}=\frac{\sqrt{2}-\sqrt{3}}{\sqrt{10}} $$ and $$ \langle 0|a_2\rangle=\frac{1}{\sqrt{2}}\sqrt{\frac{5}{8}}-\frac{1}{\sqrt{2}}\sqrt{\frac{3}{8}}=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{16}} $$ Hence, the overall calculation is $$ \frac{3}{5}\left(\frac{\sqrt{2}-\sqrt{3}}{\sqrt{10}}\right)^2+\frac{2}{5}\left(\frac{\sqrt{5}-\sqrt{3}}{\sqrt{16}}\right)^2 $$ If my hastily done maths is correct, this is the same as $$ \frac{1}{2}-\frac{12\sqrt{6}+5\sqrt{15}}{100}. $$


Second question (really ought to be separate):

Assuming the matrix $p$ that you give (might be better to use the standard notation, $\rho$, for a density matrix) is written in the standard basis, then the probability of finding it in the 0 state when measured in the standard basis is indeed $\frac{2}{5}$. But that is not quite what your wording says (I don't know where the imprecision comes in). What you ask what is the probability that the system is in the state $|0\rangle$. This is different because measuring changes the state of the system. You measure it, and you find it to be $|0\rangle$ with a certain probability. That's not the same as it being that state with a certain probability prior to the measurement. However the literal question probably isn't really answerable.

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  • $\begingroup$ Referring to the second question from me. I can quote from the original edition, "Consider the density matrix $\rho$. Show that the probability of finding the system in the state $|0\rangle$ is 0.66." $\endgroup$ – P_Gate Jun 26 at 13:58
  • $\begingroup$ @P_Gate Where does this come from? $\endgroup$ – DaftWullie Jun 26 at 14:33
  • $\begingroup$ I am currently reading "Quantum Computing Explained" by McMahon, there are some interesting tasks, including these ones. Maybe you know that book too. $\endgroup$ – P_Gate Jun 26 at 14:35
  • $\begingroup$ @P_Gate I didn't, but have just gone and found a copy. It must be an error. $\endgroup$ – DaftWullie Jun 26 at 14:48
  • $\begingroup$ Ok, thats good! You as an expert, what do you think of the book, what you have seen so far? $\endgroup$ – P_Gate Jun 26 at 15:58

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