Tuesday, March 14, 2017

hilbert space - How is conditional probability handled in quantum mechanics?


In ordinary probability theory the conditional probability/likelihood is defined in terms of the joint and marginal likelihoods. Specifically, if the joint probability of two variables is $\mathcal{L}(x,y)$, the marginal likelihood of $x$ is $\mathcal{L}(x) \equiv \int \mathcal{L}(x,y) \operatorname{d}y$, and the conditional likelihood is then defined by $$\mathcal{L}(y|x) = \frac{\mathcal{L}(x,y)}{\mathcal{L}(x)}.$$


In quantum mechanics, the probabilities/likelihoods are defined in terms of probability amplitudes by the absolute square of the amplitudes. Is there an analogous relation for a conditional probability amplitude? Concretely, say we have a joint wave function of the variables $x$ and $y$, $\Psi(x,y)$. Can we define a conditional amplitude by the following relations: \begin{align} \psi(x) & \equiv \sqrt{ \int |\Psi(x,y)|^2 \operatorname{d}y} \\ \psi_c(y|x) & \equiv \frac{\Psi(x,y)}{\psi(x)}. \end{align}


Admittedly, there is a phase function ambiguity that is, strictly, allowable. That is, transforming the marginal and conditional amplitudes by $\psi(x) \rightarrow \psi(x) \mathrm{e}^{i \alpha(x)}$ and $\psi_c(x) \rightarrow \psi_c(x) \mathrm{e}^{-i \alpha(x)}$ for any function $\alpha(x)$ does not change the predicted probabilities.


I guess the question is less whether such quantities can be defined, because I obviously just defined them. I'm more curious about whether they've been used.



Answer




Yes, conditional probabilities are studied in quantum mechanics, especially from a quantum informational perspective. The trickiest/interesting aspect of it is that, because of the nature of QM, it becomes crucial to also take the measurement choices into consideration when dealing with conditionals. Two areas that come to mind where this sort of thing is studied at length are nonlocality protocols and quantum discord.



If you don't play with measurement bases, then QM doesn't offer anything new. We can define marginals and conditional probabilities exactly as we do classically. Given any bipartite state, once we fixed a choice of measurement bases, we are left with a probability distribution $p(a,b)$ giving the probability of finding the two measurement outcomes $a$ and $b$. We can write this as something like $$p(a,b)=|\langle a,b|\Psi\rangle|^2,$$ for a suitable choice of computational basis. You can then define the marginals and conditionals as $$p(a)= \sum_b |\langle a,b|\Psi\rangle|^2, \qquad p(b|a)=p(a,b)/p(a).$$ Nothing new happens here.



Conditional states


Say Alice and Bob share a (pure) quantum state $|\Psi\rangle$. This is by definition a bipartite state, so something of the form $$|\Psi\rangle=\sum_{ij} c_{ij}|i\rangle\otimes|j\rangle$$ for some complex coefficients $c_{ij}$ and choice of bases for Alice and Bob (say Alice is represented by the kets on the left and Bob by those on the right).


You can then talk of the conditional state on Bob's side, conditioned on what Alice did on her side. This is the major difference between QM and standard probability theory: observations cannot (in general) be described by simply saying that outcomes $(a,b)$ will happen with some probability $p(a,b)$; the measurement choices of Alice and Bob need to be taken into consideration.


If Alice doesn't do anything on her side, or equivalently if she doesn't say anything to Bob about what she did, then Bob's state is effectively given by $\mathrm{Tr}_A[\mathbb P_\Psi]$, where I'm using the shorthand notation $\mathbb P_\phi\equiv|\phi\rangle\!\langle\phi|$. Explicitly, this means that Bob's state (from his point of view) is $$\rho^B \equiv \mathrm{Tr}_A[\mathbb P_\Psi]= \sum_{jj'}\Big(\sum_i c_{ij}\bar c_{ij'}\Big)|j\rangle\!\langle j'|.$$ This means that $\rho^B$ captures the maximum amount of knowledge that Bob can ever hope to gain about his state, short of Alice telling him something about what she did/found on hers. If the initial state was not pure, say we started with some $\rho_\Psi$ rather than $|\Psi\rangle$, we would get a similar result, modulo $\mathbb P_\Psi$ being replaced with $\rho_\Psi$.


On the other hand, say Alice chooses to perform some measurement $x$ on her state, and observes a measurement result $a$. This then tells us something about Bob's state, which will now be some $\rho_{a|x}$ that we can write as $$\rho_{a|x} = p(a|x)^{-1}\mathbb P[(\Pi_{a|x}\otimes I)|\Psi\rangle], $$ where $\Pi_{a|x}$ projects onto the measurement/outcome combination found by Alice, and $p(a|x)$ is a normalisation constant to ensure that we get a state (which in this case is also equal to the probability of Alice finding the outcome $a$ when measuring $x$, hence the notation).


Studying these types of situations leads to all sorts of possible communication protocols, depending on the specifics of the type of knowledge that is shared etc. See for example this review of quantum steering (which is also where I got part of the notation used here, see chapter II).



Bell nonlocality


Bell scenarios are another prominent case study of conditional probabilities in QM. Here, you think of Alice and Bob as having access to black boxes, the inner workings of which are unknown to them. They can only push buttons that sit on these boxes and observe what happens when they do. If they do this systematically and compare their results, as you might know, they can find some rather interesting results, such as types of correlations that cannot be explained classically.


Discord


Another interesting example that highlights how conditional probabilities can be tricky in quantum mechanics is quantum discord.


The quantum discord of a bipartite state is defined as the difference between two expressions for the mutual information. Classically, given random variables $A$ and $B$, you can write their mutual information in the following two equivalent ways: $$I(A:B)=H(A)+H(B)-H(A,B), \\ J(A:B)=H(A)-H(A|B).$$ As it happens, if you consider the quantum versions of these two quantities, you instead can get different results. This happens because $J(A:B)$ actually involves conditional probabilities, which in QM only make sense in a given choice of measurement basis.


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