Hadamard transformation and measurement of one qubit 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:
$$ \frac{1}{\sqrt{2}}(\left|00\right&gt;-\left|11\right&gt;) $$
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:
$$ \frac{1}{\sqrt{2}}(\left|H(0)0\right&gt;-\left|H(1)1\right&gt;) $$ 
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
$$ \frac{1}{\sqrt{2}}(\left|0\right&gt;-\left|1\right&gt;) $$
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
 $$ \frac{1}{\sqrt{2}}(\left|H(0)0\right&gt;-\left|H(1)1\right&gt;) $$ 
 $$ \frac{1}{2}(\left|00\right&gt;-\left|01\right&gt;+\left|10\right&gt;+\left|11\right&gt;) $$

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&gt;$ calculated?

I hope you can help me a bit. Thank you!
I'm not the best person to answer your question but I'll give it a try.

1st Question:

You do not yet make a measurement after applying the Hadamard Operator on your first qubit. Afterwards say you make a measurement on your 1st qubit, the state you'll end up with is:

$|\psi&gt;_{post-measurement}$ = $1/\sqrt{p_1(m)}$ $\sum_{j} c_{mj} |m&gt;_1 |j&gt;_2$

whereby $m$ is the measurement result of your 1st qubit (either 0 or 1) , $p_1(m)$ is the probability of getting that result on your 1st qubit, $c_{mj}$ the usual coefficient infront of your ket. 

2nd Question:

This will answer how to get ${p_1(m)}$. 
Write out the density matrix of $|\psi&gt;_{pre-measurement}$, then trace out the 2nd qubit. Then apply bra-ket to your reduced density matrix:

${p_1(m)}$ = $&lt;m|_{1}$ $\rho_{1} |m&gt;_{1}$ 

Or equivalently:

${p_1(m)}$ = $\sum_{n}|&lt;m|_1&lt;n|_2 \sum_{ij} c_{ij} |i&gt;_1|j&gt;_2|^2$