c) Compute $$\text{Prob}(\uparrow_\hat{n}\uparrow_\hat{m}) \equiv \text{tr}(\pmb{E}_A(\hat{n})\pmb{E}_B(\hat{n})\pmb{p}(\lambda)), \tag{4.164}$$

where $\pmb{E}_A(\hat{n})$ is the projection of Alice's qubit onto $\left|\uparrow_{\hat{n}}\right\rangle$ and $\pmb{E}_B(\hat{m})$ is the projection of Bob's qubit onto $\left|\uparrow_{\hat{m}}\right\rangle$.

So what they are actually asking is, you have an entangled system of two qubits $\rho$ and we measure the first qubit on some (random) axis n and then the second on some other (random) axis $m$. What is the probability of both spins being "spin-up" on their respective axes? The system in the actual exercise is a bit hard, but to show my problem, let's take an easy example:

$$\rho = \left|\uparrow \downarrow \right\rangle _{AB} \left\langle \uparrow \downarrow\right|_{AB}$$

Now without actually using the formula, let's assume we measure in the z-direction twice. We will measure for example a spin-up for qubit A, so looking at the system we know that the second qubit should be spin-down, because this density matrix comes from the state $|\psi \rangle_{AB} = \left|\uparrow \downarrow\right\rangle $. Now if we look at the given formula, $E_B(z)$ measures the spin of qubit B, which can be for example -1, then the second operator measures $E_A(z)$, as we said, is +1 in the given state (because A & B have opposite spins). We know that the expectation value of an operator is $\langle E_A E_B \rangle$, which is the same as taking the trace $\text{Tr}(E_A E_B \rho)$, which is the described formula above. We will not explicitly calculate this with the trace formula, because we have the corresponding state here, so we know that the probability is

$$\text{Tr}(E_A E_B \rho)= \langle E_A E_B \rangle = \langle \psi | E_A E_B | \psi \rangle = \langle \psi | (+1)(-1) | \psi \rangle = -1$$

So we have a "probability" of minus 1. How can this be a probability? In the original example, I have also calculated this and we get an expression in the function of the angle between the two axes, which can also be negative. This seems quite logical for two qubits A and B which are entangled in a way they will almost always have opposite spin direction, for example. So my question is: how can this probability be negative?


If $\boldsymbol{E}_A(\hat{n})$ is the projection on $\left|\uparrow_{\hat{n}}\right\rangle$, then $$ \boldsymbol{E}_A(\hat{n}) = \left|\uparrow_{\hat{n}}\right\rangle\left\langle \uparrow_{\hat{n}}\right|, $$ Similarly, $$ \boldsymbol{E}_B(\hat{m}) = \left|\uparrow_{\hat{m}}\right\rangle \left\langle \uparrow_{\hat{m}}\right|. $$

The corresponding projection on the composite system $AB$ is the tensor product of projections, that is $$ \boldsymbol{E}_{AB}(\hat{n},\hat{m}) = \boldsymbol{E}_A(\hat{n}) \otimes \boldsymbol{E}_B(\hat{m}) = \left|\uparrow_{\hat{n}}\right\rangle \left\langle \uparrow_{\hat{n}}\right| \otimes \left|\uparrow_{\hat{m}}\right\rangle \left\langle \uparrow_{\hat{m}}\right| = \left|\uparrow_{\hat{n}}\uparrow_{\hat{m}}\right\rangle \left\langle \uparrow_{\hat{n}}\uparrow_{\hat{m}}\right| . $$ Now you can take the usual matrix product of $\boldsymbol{E}_{AB}(\hat{n},\hat{m})$ and $\rho$ because they have the same size 4 x 4.
The trace of the product of two non-negative (positive semi-definite) hermitian matrices always non-negative. So the answer must be non-negative.

  • $\begingroup$ This helps a lot, but as I do get an answer that is not non-negative. How do I explicitly calculate this. Let's work in the xz-plane, a projection operator for the axis $\hat{n}= (cos(\theta), sin(\theta))$ then will be a matrix with rows $((cos(\theta), sin(\theta))$ and $(sin(\theta), -cos(\theta))$ right $(E_n = \sigma \cdot \hat{n})$? Or is this where I go wrong $\endgroup$ – CFRedDemon Apr 19 '19 at 8:30
  • $\begingroup$ Okay, I found my mistake I think, the projection operator is defined as $E(\hat{n}) = 1/2 ( 1 + \hat{n} \cdot \sigma) $. if I add the extra term in front of it, my answer is indeed a non-negative answer... Wow, feeling pretty dumb right now $\endgroup$ – CFRedDemon Apr 19 '19 at 8:43
  • $\begingroup$ Yes, that is the correct formula. $\endgroup$ – Danylo Y Apr 19 '19 at 8:44

$\langle \psi \vert E_A E_B \vert \psi \rangle$ is just the expectation value of the composite operator $E_AE_B$ which, of course, can be negative. This is not the same as the probability of measuring $\left\vert\uparrow \downarrow \right\rangle_{AB}$ which in your case of $\rho = \left\vert\uparrow \downarrow \right\rangle_{AB} \left\langle \uparrow \downarrow \right\vert_{AB}$ is 1.

The two values are related by:

$\langle \alpha \rangle = \sum_{i} h_i \phi_i $ where $\langle \alpha \rangle$ is the expectation value of $\alpha$, the $h_i$'s are the eigenvalues of the state in question and the $\phi_i$'s their respective probability amplitudes squared (read probabilities).

An easy way to remember the difference is with the basic Schrodinger equation

$$ \hat{H} \vert \psi\rangle = E \vert \psi \rangle$$

where then the expectation value $\langle \hat{H} \rangle = \langle \psi \vert E \vert \psi \rangle = E$ which is an energy and can be any real number you like, positive or negative. This, again, is not the probability of measuring state $\vert\psi\rangle$.

  • $\begingroup$ The expectation value of an operator is defined as $Tr($ operator $ \rho)$ right, which can be negative! So the probability here is defined as $Tr( E_A E_B \rho)$ so how can this not be negative ( as probability cannot be negative). The system I am talking about is $\rho = (1 − \lambda)|\psi ^-\rangle \langle \psi ^- | + \frac{1}{4} \lambda I$ with $\psi^-$ a Bell state. So if we compute the probability here that is stated in the exercise, there should be a positive answer? My answer for the probability $\rho$ is a function of the angle between the chosen axes, which can be negative... $\endgroup$ – CFRedDemon Apr 18 '19 at 21:58
  • $\begingroup$ $E_A$ and $E_B$ are projections, and $E_AE_B$ is in fact a tensor product of projections, which is a positive operator. Hence that expectation can't be negative. Also in the context of the question that trace $Tr(E_AE_B\rho)$ is indeed the probability of event, not the expectation. $\endgroup$ – Danylo Y Apr 19 '19 at 7:52

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.