What is the probability that measurement finds it in the $|0\rangle$ state?

Question

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

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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 ...

Answer 1

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 $$ \frac{1}{2}-\frac{12\sqrt{6}+5\sqrt{15}}{100}. $$

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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.