Let $| \psi \rangle = \frac{1}{\sqrt{2}}|00\rangle + \frac{1}{2}|01\rangle + \frac{\sqrt{3}}{4} |10\rangle + \frac{1}{4}|11\rangle$ be a state vector describing a closed quantum mechanical system.

Note the convention is $| ab \rangle$ with its self - adjoint being $\langle ab|$

To determine the marginal probability for $| a \rangle $ with measurement along computational basis $\hat{z}$ where a = 0 (this means: I look only at states for which Alice's qubit shows $0$ or $|0\rangle$):

$p(|a\rangle_{\hat{z}} | |\psi \rangle ) = |\frac{1}{\sqrt{2}}|^{2} + |\frac{1}{2}|^{2} = \frac{3}{4}$

To arrive at the same value to the above in the language of density matrix, the state vector for a tensor product state can be cast into a density matrix like

$\delta_{AB} = | \psi \rangle \langle \psi |$.

At this point I am stuck but I do have an expression describing the probability (a marginal probability) $P(|a\rangle_{\hat{z}} | \delta_{AB})$ for which a measurement on Alice's qubit in the system $\delta_{AB}$ yields an outcome $|a\rangle \rightarrow |0\rangle$ along measurement basis $\hat{z}$:

$P(|a\rangle_{\hat{z}} | \delta_{AB}) = Tr[\frac{1}{2}(I + (-1)^{a}\vec{a}\cdot\hat{z})] = \frac{1}{2}[1 + (-1)^{a}\vec{a}\cdot\hat{z}]$ where $I$ is an identity matrix.

  • $\begingroup$ You need to calculate $\mathrm{Tr} [\rho E_a ]$, where $E_a$ is the measurement operator. For $\rho = \sum_i p_i | \psi_i \rangle \langle \psi_i |$, it equals $\sum_i p_i \langle \psi_i |E_a | \psi_i \rangle$. In your case, $E_a$ is just a projector and there is only one state in mixture, thus you'll get $\langle \psi |E_a | \psi \rangle$ $\endgroup$
    – EvgeniyZh
    Commented Sep 17, 2023 at 15:56

2 Answers 2

  • Using pure states and kets, as you described, you compute the marginal probabilities by computing the squared norm of the projection on the states you're interested in. In this case, this means projecting on $|0\rangle$ on the first qubit, i.e. applying the operator $\langle 0|\otimes I$ to your state. This leaves you with $$\frac{1}{\sqrt2}|0\rangle+\frac12|1\rangle,$$ the norm of which is $3/4$, as you wrote.

  • Given a (not-necessarily-bipartite) density matrix $\rho$, you can compute the probability of getting an outcome $|a\rangle$ (when measuring the state in a projective basis containing $|a\rangle$) by computing $$\langle \mathbb{P}_a,\rho\rangle\equiv \operatorname{tr}(\mathbb{P}_a\rho)\equiv \langle a|\rho|a\rangle,$$ where I used the shorthand notation $\mathbb{P}_a\equiv |a\rangle\!\langle a|$. Note that these are all equivalent notations to write the same thing; the last one being the most standard. The inner product in expressions like $\langle \mathbb{P}_a,\rho\rangle$ is understood as the Hilbert-Schmidt inner product between operators (which is defined via the trace as in the identity immediately after it).

  • Given a bipartite density matrix, you do almost the same thing, except instead of outright projecting with some $\mathbb{P}_a$, you have to account for the fact that you're not being picky about the measurement outcome on the second party. This effectively amounts to using as projection $\mathbb{P}_a\otimes\sum_b \mathbb{P}_b$, which is like saying "I want the first party to give outcome $a$, but any outcome $b$ is fine for the second party". This operator is identical to $\mathbb{P}_a\otimes I$ with $I$ the identity. In conclusion, you compute the probability as: $$\langle \mathbb{P}_a\otimes I,\rho\rangle = \operatorname{tr}[(\mathbb{P}_a\otimes I)\rho] = \langle a\lvert \operatorname{tr}_2(\rho) |a\rangle.$$ In the last identity I exploited the properties of trace and partial trace to get a simplified (or at least, different) expression. You could also immediately get to $\langle a\lvert \operatorname{tr}_2(\rho) |a\rangle$ observing that you're neglecting what happens on the second party, thus you're effectively working with the reduced state $\operatorname{tr}_2(\rho)$, and you can then use the standard rule to obtain the probability of getting the outcome $a$.


Let's first see how projective measurements work in 'density operator language'.

If you perform a measurement on your density operator $\rho$, corresponding to a set of projective measurement operators $\{P_k\}$, then you will get outcome $k$ with probability $$p_k = \text{Tr} ( P_k \cdot \rho)\,,$$ and your state after measurement, say $\rho^{\prime}$, will be: $$\rho^{\prime} = \frac{P_k \cdot \rho \cdot P_k^\dagger}{p_k}\,.$$

For your question, your density operator is$^1 $

$$\rho_{AB} = | \psi \rangle \langle \psi |\,.$$

Now, let's construct the measurement operators $\{P_k\}$. You are measuring Alice's qubit in $Z$ basis and doing nothing to Bob's qubit. Hence, you will have two possible outcomes of getting state $| 0 \rangle $ or $| 1 \rangle$ after performing $Z$-measurement on Alice's qubit. Therefore, we will have two measurement operators:

$$P_0 = | 0 \rangle \langle 0 |_A \otimes I_B\,,$$ $$P_1 = | 1 \rangle \langle 1 |_A \otimes I_B\,.$$

Now, the probability of getting outcome 0, i.e. measuring Alice's qubit in state $| 0 \rangle $, will be:

$$p_0 = \text{Tr}(P_0 \cdot \rho_{AB})\,.$$

Similarly, the probability of finding Alice's qubit in state $| 1 \rangle $, will be: $$p_1 = \text{Tr}(P_1 \cdot \rho_{AB})\,.$$

Performing explicit calculations will give you:

$$p_0 = \frac{3}{4} = p(|0\rangle_A \, | \, \rho_{AB})\,,$$ $$p_1 = \frac{1}{4} = p(|1\rangle_A \, | \, \rho_{AB})\,.$$

$^1$: It is common practice to use $\rho$ to denote a density matrix and we use $\delta$ alot in quantum computing as Kronecker delta; therefore I avoided using $\delta_{AB}$ as the density matrix you have specified and instead used $\rho_{AB}$.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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