Last semester, 14.04 — Intermediate Microeconomic Theory covered choice theory under uncertainty; at the same time, I was taking 8.05 — Quantum Physics II, where we had talked about 2-state systems and the formalism of quantum states being complex vectors in a Hilbert space, and as choice under uncertainty discusses how consumers make choices based on states of the world, I thought it would be cool to extend it to quantum states, but I wasn't sure how to do that at that time. Now that 14.03 — Microeconomic Theory and Public Policy is talking about choice under uncertainty as well this semester, and now that I have had some time for the stuff about 2-state systems to simmer in my head, I think I have a slightly better idea of how to think about extended the states of the world to include the superpositions as allowed by quantum mechanics.
One of the simplest examples would be the 2-state system. In the class I am taking now, a typical 2-state system would be a fair coin which can either take on values of heads or tails (each with probability $\frac{1}{2}$). What might this look like in quantum mechanics? We could replace the coin with the spin $S_z$ of an electron. A state of the electron corresponding to being measured as $|\uparrow_z \rangle$ or $|\downarrow_z \rangle$ each with equal probability could be labeled as $|\psi \rangle = \frac{1}{\sqrt{2}} \left(|\uparrow_z \rangle + |\downarrow_z \rangle\right)$. This indeed gives equal probabilities of being measured as spin-up or spin-down in the $z$-direction. The problem is that this state is exactly $|\psi \rangle = |\uparrow_x \rangle$, so the probability of the state being measured in the $x$-direction as spin-up is unity. This leads to issues in trying to reconcile the interpretation of payoffs for different states of the world; this particular state would pay $w(|\uparrow_z \rangle)$ or $w(|\downarrow_z \rangle)$ with equal probabilities until $S_z$ is measured, but would pay $w(|\uparrow_x \rangle)$ with certainty as $S_x$ has essentially already been measured, collapsing a previously unknown state into this one. So there seems to be an issue with trying to stuff the interpretation of probabilistic measurement of a state of the world into the idea of superposing quantum states, as certain measurable states of the world do not commute ($[S_i, S_j] = i\hbar \epsilon_{ijk} S_k$ for instance). So what can be done now?
Recall that each state of the world $j$ has an associated probability $\mathbb{P}_j$. Yet, once a state is measured, those probabilities are meaningless, because a state becomes observed or not observed; it cannot be "observed in fraction $\mathbb{P}_j$" or anything like that. Taking the frequentist view, these probabilities are only meaningful in the large ensemble limit; after a large number of observations, the fraction of those observations corresponding to the state $j$ converges to $\mathbb{P}_j$. This is exactly the same rationale behind the density matrix, which describes the state of an ensemble of systems which are usually not all prepared in the same quantum state, but in the large number limit, the fraction of those prepared in the state $|\psi_j \rangle$ is $\mathbb{P}_j$ such that $\sum_j \mathbb{P}_j = 1$. It is then defined as \[ \rho = \sum_j \mathbb{P}_j |\psi_j \rangle \langle \psi_j | . \] Note that in general $\langle \psi_j | \psi_{j'} \rangle \neq \delta_{jj'}$ because the states do not in general need to be orthogonal. Furthermore, different state preparations can have the same density matrix, in which case the "different" state preparations are actually physically indistinguishable if a quantum mechanical measurement is made.
So what does this have to do with the states of the world? If the example of a fair coin flip is again translated into measuring each eigenvalue of $S_z$ with equal probability, then the density matrix becomes $\rho = \frac{1}{2} \left(|\uparrow_z \rangle \langle \uparrow_z | + |\downarrow_z \rangle \langle \downarrow_z |\right)$. Now it is much more plausible to define a wealth $w(|\psi_j \rangle)$ dependent on the state of the world.
For example, let us now consider an unfair "coin" represented by the density matrix $\rho = \frac{1}{3} \left(|\uparrow_z \rangle \langle \uparrow_z | + 2|\downarrow_z \rangle \langle \downarrow_z |\right)$. What is the probability of measuring this system in the state $|\uparrow_x \rangle$? That would be $\langle \uparrow_x | \rho | \uparrow_x \rangle = \frac{1}{2}$. Similarly, $\langle \downarrow_x | \rho | \downarrow_x \rangle = \frac{1}{2}$. It is interesting to note that equal probabilities are achieved for spin-up and spin-down in the $x$-direction (and $y$-direction as well) for this system; the difference is that now there are off-diagonal terms in the density matrix describing the degree of interference between the states of $S_x$ in the two populations of spins.
What can we do with this? Payoffs now need to be defined over every plausible noncommuting observable in the Hilbert space (because commuting observables yield the same state, and the payoff is really defined for the state). Thankfully this is fairly simple for spins, as the noncommuting observables in question are the components of $\vec{S}$.
Let us return to the previous unfair "coin". Classically (so it would be a coin), the result of the coin flip would be the end of the story. Let us suppose that the consumer had a risk-neutral utility $V(w) = w$. Let us also suppose that there was a game which payed off 20 if heads and 0 if tails and cost 10 to enter. The expected payoff, accounting for the cost of entering, would be $\mathbb{E}(w) = -\frac{10}{3}$, so the consumer would prefer to not play.
Classically that would be the whole story. Quantum mechanically, though, there is more to the story. If the owner of the game decided to measure $S_z$, then the result would be the same as the classical result. If the owner decided instead to measure $S_x$ or $S_y$, then $\mathbb{E}(w) = 0$ so the consumer would be indifferent between playing and not playing, which is certainly a different outcome from preferring not to play. In fact, if the consumer does not know for sure what the owner decides to measure, then probabilities could be assigned to the measurement choice itself, and these could then be used to find the expectation value of payoff or utility over all possible choices, where each of those choices will have an expected payoff or utility as well.
One thing to look further at is how to extend this to include continuous ensembles. The example in class was how a state of the world might be the outdoor temperature measured in some range. The quantum mechanical equivalent might be having a continuum of systems, each infinitesimal one having a position inside a given range of allowed positions; the only issue with this is that matter is not continuous, so I don't think it is possible to have a density matrix in the form of $\rho = \int |\psi(s) \rangle \langle \psi(s)| \, ds$ for some continuous index $s$. However, that can certainly be further investigated, and maybe I'll do that next time. The point is that states of the world in analyzing expected utility can easily be generalized to include quantum states through density matrices, and the question of which observable to measure in a noncommuting set brings out some interesting behavior not seen in classical states.
One of the simplest examples would be the 2-state system. In the class I am taking now, a typical 2-state system would be a fair coin which can either take on values of heads or tails (each with probability $\frac{1}{2}$). What might this look like in quantum mechanics? We could replace the coin with the spin $S_z$ of an electron. A state of the electron corresponding to being measured as $|\uparrow_z \rangle$ or $|\downarrow_z \rangle$ each with equal probability could be labeled as $|\psi \rangle = \frac{1}{\sqrt{2}} \left(|\uparrow_z \rangle + |\downarrow_z \rangle\right)$. This indeed gives equal probabilities of being measured as spin-up or spin-down in the $z$-direction. The problem is that this state is exactly $|\psi \rangle = |\uparrow_x \rangle$, so the probability of the state being measured in the $x$-direction as spin-up is unity. This leads to issues in trying to reconcile the interpretation of payoffs for different states of the world; this particular state would pay $w(|\uparrow_z \rangle)$ or $w(|\downarrow_z \rangle)$ with equal probabilities until $S_z$ is measured, but would pay $w(|\uparrow_x \rangle)$ with certainty as $S_x$ has essentially already been measured, collapsing a previously unknown state into this one. So there seems to be an issue with trying to stuff the interpretation of probabilistic measurement of a state of the world into the idea of superposing quantum states, as certain measurable states of the world do not commute ($[S_i, S_j] = i\hbar \epsilon_{ijk} S_k$ for instance). So what can be done now?
Recall that each state of the world $j$ has an associated probability $\mathbb{P}_j$. Yet, once a state is measured, those probabilities are meaningless, because a state becomes observed or not observed; it cannot be "observed in fraction $\mathbb{P}_j$" or anything like that. Taking the frequentist view, these probabilities are only meaningful in the large ensemble limit; after a large number of observations, the fraction of those observations corresponding to the state $j$ converges to $\mathbb{P}_j$. This is exactly the same rationale behind the density matrix, which describes the state of an ensemble of systems which are usually not all prepared in the same quantum state, but in the large number limit, the fraction of those prepared in the state $|\psi_j \rangle$ is $\mathbb{P}_j$ such that $\sum_j \mathbb{P}_j = 1$. It is then defined as \[ \rho = \sum_j \mathbb{P}_j |\psi_j \rangle \langle \psi_j | . \] Note that in general $\langle \psi_j | \psi_{j'} \rangle \neq \delta_{jj'}$ because the states do not in general need to be orthogonal. Furthermore, different state preparations can have the same density matrix, in which case the "different" state preparations are actually physically indistinguishable if a quantum mechanical measurement is made.
So what does this have to do with the states of the world? If the example of a fair coin flip is again translated into measuring each eigenvalue of $S_z$ with equal probability, then the density matrix becomes $\rho = \frac{1}{2} \left(|\uparrow_z \rangle \langle \uparrow_z | + |\downarrow_z \rangle \langle \downarrow_z |\right)$. Now it is much more plausible to define a wealth $w(|\psi_j \rangle)$ dependent on the state of the world.
For example, let us now consider an unfair "coin" represented by the density matrix $\rho = \frac{1}{3} \left(|\uparrow_z \rangle \langle \uparrow_z | + 2|\downarrow_z \rangle \langle \downarrow_z |\right)$. What is the probability of measuring this system in the state $|\uparrow_x \rangle$? That would be $\langle \uparrow_x | \rho | \uparrow_x \rangle = \frac{1}{2}$. Similarly, $\langle \downarrow_x | \rho | \downarrow_x \rangle = \frac{1}{2}$. It is interesting to note that equal probabilities are achieved for spin-up and spin-down in the $x$-direction (and $y$-direction as well) for this system; the difference is that now there are off-diagonal terms in the density matrix describing the degree of interference between the states of $S_x$ in the two populations of spins.
What can we do with this? Payoffs now need to be defined over every plausible noncommuting observable in the Hilbert space (because commuting observables yield the same state, and the payoff is really defined for the state). Thankfully this is fairly simple for spins, as the noncommuting observables in question are the components of $\vec{S}$.
Let us return to the previous unfair "coin". Classically (so it would be a coin), the result of the coin flip would be the end of the story. Let us suppose that the consumer had a risk-neutral utility $V(w) = w$. Let us also suppose that there was a game which payed off 20 if heads and 0 if tails and cost 10 to enter. The expected payoff, accounting for the cost of entering, would be $\mathbb{E}(w) = -\frac{10}{3}$, so the consumer would prefer to not play.
Classically that would be the whole story. Quantum mechanically, though, there is more to the story. If the owner of the game decided to measure $S_z$, then the result would be the same as the classical result. If the owner decided instead to measure $S_x$ or $S_y$, then $\mathbb{E}(w) = 0$ so the consumer would be indifferent between playing and not playing, which is certainly a different outcome from preferring not to play. In fact, if the consumer does not know for sure what the owner decides to measure, then probabilities could be assigned to the measurement choice itself, and these could then be used to find the expectation value of payoff or utility over all possible choices, where each of those choices will have an expected payoff or utility as well.
One thing to look further at is how to extend this to include continuous ensembles. The example in class was how a state of the world might be the outdoor temperature measured in some range. The quantum mechanical equivalent might be having a continuum of systems, each infinitesimal one having a position inside a given range of allowed positions; the only issue with this is that matter is not continuous, so I don't think it is possible to have a density matrix in the form of $\rho = \int |\psi(s) \rangle \langle \psi(s)| \, ds$ for some continuous index $s$. However, that can certainly be further investigated, and maybe I'll do that next time. The point is that states of the world in analyzing expected utility can easily be generalized to include quantum states through density matrices, and the question of which observable to measure in a noncommuting set brings out some interesting behavior not seen in classical states.
