Hadamard transformation and measurement of one qubit

Question

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!

Answer 1

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:

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

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 $|\psi>_{pre-measurement}$, then trace out the 2nd qubit. Then apply bra-ket to your reduced density matrix:

${p_1(m)}$ = $<m|_{1}$ $\rho_{1} |m>_{1}$

Or equivalently:

${p_1(m)}$ = $\sum_{n}|<m|_1<n|_2 \sum_{ij} c_{ij} |i>_1|j>_2|^2$

Answer 2

After applying $H$ to the first qubit you have this state:

$$\frac{1}{2}(\left|00\right>-\left|01\right>+\left|10\right>+\left|11\right>)$$

You square the absolute values of the amplitudes to obtain probabilities of each of possible outcomes. In this case it is $1/4$ for each of the outcomes. Half of them has $\left|0\right>$ for first qubit, so its sum probability is $1/2$.