Probability of measuring state $|+\rangle$ and state $|-\rangle$ given a state and a basis I am given a basis $|+\rangle = \frac{1}{\sqrt 2}(|0\rangle + |1\rangle)$ and $|-\rangle = \frac{1}{\sqrt 2}(|0\rangle - |1\rangle)$ and i am given a three qubit state $|\phi\rangle = \frac{1}{\sqrt 3}|1\rangle |0\rangle |1\rangle + \frac{2}{\sqrt 3}|0\rangle |1\rangle |0\rangle $
What is the probability of measuring state $|+\rangle|+\rangle|+\rangle$ and what is the probability of measuring state $|-\rangle|-\rangle|-\rangle$?
I know how to express the state $|0\rangle $ and $|1\rangle $ in the basis above, that would be $|0\rangle = \frac{1}{\sqrt 2}(|+\rangle + |-\rangle)$ and $|1\rangle = \frac{1}{\sqrt 2}(|+\rangle - |-\rangle)$
but when attempting to do $|\left(\langle+|\langle+|\langle+|\right)|\phi\rangle|^2$ I can't really get anywhere.
How do I solve this?
 A: So I don't think your state is normalized and probably what was meant was $$|\phi\rangle = \sqrt{\frac13}|101\rangle + \sqrt{\frac23}|010\rangle.$$(But as written you can also normalize it by multiplying by $\sqrt{3/5}$ and converting the denominator to 5.)
There are several ways to do this problem. Possibly the easiest is to rewrite the state that you are dotting with, $$|{+}{+}{+}\rangle = \sqrt{\frac18} \big(|000\rangle + |001\rangle + \dots + |111\rangle\big),$$ to reveal that $$\langle{+}{+}{+}|\phi\rangle = \sqrt{\frac18}\sqrt{\frac13} + \sqrt{\frac18}\sqrt{\frac23},$$ while $|{-}{-}{-}\rangle,$ having negative signs in front of every term with an odd number of ones, instead generates
$$\langle{-}{-}{-}|\phi\rangle = \sqrt{\frac18}\sqrt{\frac13} - \sqrt{\frac18}\sqrt{\frac23},$$and the probabilities are of course the squares of these amplitudes. So, uh, $\frac18 \pm \frac1{12}\sqrt{2}$ if I am doing the quadratic equation right in my head?
The other way is indeed as the other answer says, to convert the above into the appropriate basis, so each $|1\rangle$ becomes a $|+\rangle - |-\rangle$ and each $|0\rangle$ becomes a $|+\rangle + |-\rangle$ and so we have$$\begin{align}
|\phi\rangle =& \sqrt{\frac1{24}}\big(|{+}{+}{+}\rangle - |{+}{+}{-}\rangle +
|{+}{-}{+}\rangle-|{+}{-}{-}\rangle \\
&\hphantom{\frac1{24}}~~-|{-}{+}{+}\rangle + |{-}{+}{-}\rangle - |{-}{-}{+}\rangle + |{-}{-}{-}\rangle\big)\\
&+\sqrt{\frac1{12}}\big(|{+}{+}{+}\rangle + |{+}{+}{-}\rangle -
|{+}{-}{+}\rangle-|{+}{-}{-}\rangle \\
&\hphantom{+\frac1{12}}~~~~+|{-}{+}{+}\rangle + |{-}{+}{-}\rangle - |{-}{-}{+}\rangle - |{-}{-}{-}\rangle\big),\\
\end{align}$$
from which you can read not just those amplitudes but diverse other ones that you might be interested in in the Hadamard basis. Note that there is really no “hard work” here in either case, it is all mostly just keeping track of signs of terms. This is one reason why the Hadamard basis is so nice to think about problems with.
A: Step 1: Notice that $\vert 0\rangle = \frac{1}{\sqrt{2}}(\vert +\rangle + \vert - \rangle)$ and $\vert 1\rangle = \frac{1}{\sqrt{2}}(\vert +\rangle - \vert - \rangle)$.
Step 2: Expand $\vert\phi\rangle = \frac{1}{\sqrt 3}\vert 1\rangle \vert0\rangle \vert1\rangle + \frac{\sqrt{2}}{\sqrt 3}\vert0\rangle \vert1\rangle \vert0\rangle$ using the substitution in Step 1.
Step 3: The squares of the coefficients in your expansion in Step 2 should add to 1. The squares of the coefficients of the $\vert +\rangle\vert +\rangle\vert +\rangle$ and $\vert -\rangle\vert -\rangle\vert -\rangle$ states are the probabilities you desire.
The statement below is incorrect. $\vert 0\rangle$ and $\vert 1\rangle$ are as written above, not as you have written them.

I know how to express the state $|\phi\rangle $ in the basis above,
that would be $|0\rangle = \frac{1}{\sqrt 2}(|+\rangle + |-\rangle)$
and $|1\rangle = \frac{1}{\sqrt 2}(|+\rangle - |-\rangle)$

