# 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: $$\frac{1}{\sqrt{2}}(\left|00\right>-\left|11\right>)$$ 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!

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:

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

Or equivalently:

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

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 $$1/2$$.