1
$\begingroup$

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!

$\endgroup$
1
$\begingroup$

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:

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

$\endgroup$
1
$\begingroup$

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 $1/2$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.