I also came across something, which is not completely understandable for me, so I ask here.
Given is a qubit in an entangled state, this is: $$ \frac{1}{\sqrt{2}}(\left|00\right>-\left|11\right>) $$ Now it is said that one could also apply the Hadamard transformation to the first bit and measure the first bit of the pair. That would mean so far: $$ \frac{1}{\sqrt{2}}(\left|H(0)0\right>-\left|H(1)1\right>) $$ My question is, where does the measurement start now?
Do I have to measure H (0) and H (1) and if so how does that work? For a single qubit, I know what a measurement looks like, say, we have $$ \frac{1}{\sqrt{2}}(\left|0\right>-\left|1\right>) $$ Then my qubit is 50% in state 0 and 50% in state 1
But how is that applied to the entangled particle?
Alternatively, I've multiplied this step out $$ \frac{1}{\sqrt{2}}(\left|H(0)0\right>-\left|H(1)1\right>) $$ $$ \frac{1}{2}(\left|00\right>-\left|01\right>+\left|10\right>+\left|11\right>) $$
Now my second question, what is the first bit of the entangled pair here? and how is the probability that the first bit is in $ \left|0\right>$ calculated?
I hope you can help me a bit. Thank you!