# Is there any operator behind probability, in quantum mechanics?

In Quantum mechanics, the probability of finding a particle at position $x$ is given by $|\psi(x)|^2$, where $\psi$ is the wave function. Wonder what is the operator which gives this probability? Is probability the result of any operator acting on $\psi$?

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Correction - You mean $|\psi^2|$, and not $|\psi|$. Also, it is not the probability, but the probability density. – The Dark Side Jun 4 '14 at 8:49
I corrected it. Thanks @New_new_newbie – Rajesh Dachiraju Jun 4 '14 at 8:50
Note also that you should distinguish between probability and probability density. – Emilio Pisanty Jun 9 '14 at 1:35

The theory of quantum mechanics has been developed to explain observations, i.e. measurements. Without observations it is a floating mathematical construct.

One of the postulates to connect the mathematics with reality is:

To every observable there corresponds an operator which operating on the state function will give an eigenvalue. So the question becomes : is probability an observable? and then it becomes :what is an observable.

An observable in the framework of quantum mechanics is a variable of the system under consideration which a measurement can evaluate . The energy of a single photon. The momentum of a proton. The spin of an electron. We can always measure these variables on single particles with one observation, measurement. This is not possible with probability. It is an emergent value from a great number of measurements with the same boundary conditions: it is a normalized distribution, varying from 0 to 1, of the spread of the values found in the measurements.

So no, there exists no quantum mechanical operator for probability, as it is not an observable of a variable entering the quantum mechanical problem but an emergent quantity from a lot of measurements.

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A good answer; funnily everyone missed the obvious objection: probability isn't observable. – JamalS Jun 4 '14 at 12:36
I do not agree with this answer, though it is formally correct. Events are definitely observable: a particle's position is either in a given interval $[a,b]$ or it is not. These correspond directly to spectral projectors such as $\int_a^b|x\rangle\langle x|\,\text d x$. The expectation values of the projector that corresponds to an event is the probability of that event - so therefore an operator exists with that property. What else do you need? – Emilio Pisanty Jun 4 '14 at 21:59
I also don't really see the difference between probability and e.g. energy. Given a state, one cannot meaningfully interrogate any of its properties using a single realization. A single measurement lets you estimate the state's expected energy, but you need to adjust this as you perform more and more measurements. However, the same is true of the probability of an event! It either happened or not, in which case your first estimate is either 1 or 0; as you perform more realizations your estimates get better and better. I do not really see the difference between these two cases. – Emilio Pisanty Jun 4 '14 at 22:05
Finally, if you have a single realization of a state, then a measurement of energy does not give much information about the state itself (only that the outcome eigenstate had nonzero amplitude). The only sure-fire information you obtain is of the state after the measurement. But this is also true of probability: after checking, you know whether the event happened or not, and you have complete certainty on the probability of whether it happened (and indeed whether it will happen again after an infinitesimal time) or not. (Sorry for the rant =).) – Emilio Pisanty Jun 4 '14 at 22:09
@EmilioPisanty I am not saying one cannot measure probability. I am saying the measurement does not fall within the postulate of quantum mechanics : to every observable there corresponds an operator. One measurement of energy is a fixed number ( with experimental errors and will be added to create a probability distribution). One measurement counted as probability carries no information except the number 1, without even a measurement error. At the level of an individual interaction the probability operator is ? – anna v Jun 5 '14 at 3:36

The answer is negative for two distinct reasons.

(1) In QM operator means linear operator and the map $\psi \mapsto |\psi(x)|^2$ is not linear, evidently.

(2) Wavefunctions are elements of $L^2(\mathbb R)$ and these elements are defined up to zero-measure sets. I mean that, if $\psi(x) \neq \psi'(x)$ for $x\in E$ where $E$ has zero measure, then $\psi=\psi'$ as elements of $L^2(\mathbb R)$, i.e. (if $\int |\psi(x)|^2 dx =1$) pure quantum states. In particular every set of the form $\{x_0\}$ has always zero measure. Therefore, for any fixed $x_0$, the map $$L^2(\mathbb R) \ni \psi \mapsto |\psi(x_0)|^2$$ makes no sense. It is not defined at all, as a map associating states (if $\int |\psi(x)|^2 dx =1$) with numbers.

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Isn't the wavefunction twice differentiable? So that could rule out off points. – Rajesh Dachiraju Jun 4 '14 at 15:03
It is not. $-d^2/dx^2$ is not self-adjoint. Is self-adjoint if defined on a suitable Sobolev space made of non differentiable functions... – Valter Moretti Jun 4 '14 at 15:55

If one switches to the density operator formalism the situation changes. Density operators $\rho$ are positive trace class operators with trace norm $\mathrm{tr}\rho =1$. They not only describe pure states but also mixed ones and satisfy the Liouville-von Neumann equation \begin{equation*} \partial _{t}\rho (t)=-i[H,\rho (t)]:=-iL\rho (t), \end{equation*} where $H$ is the Hamiltonian of the system and $[.,.]$ the commutator. The right hand side defines the Liouville operator $L$ acting on elements of the trace class. If, in your case, $\psi (x)\in \mathcal{H}$ is square integrable with unit norm, then \begin{equation*} \rho =|\psi ><\psi |, \end{equation*} and the probability to find the particle with coordinate in $\mathcal{% M\subset H}$ is \begin{equation*} E_{\mathcal{M}}=\mathrm{tr}P_{\mathcal{M}}\rho =\mathrm{tr}P_{\mathcal{M}% }|\psi ><\psi |=\int_{\mathcal{M}}dx|\psi (x)|^{2} \end{equation*} Here $P_{\mathcal{M}}$ is the projector defined by the characteristic function $\chi _{\mathcal{M}}(x)$, $\chi _{\mathcal{M}}(x)=1$ for $x\in \mathcal{M}$ and 0 otherwise. Thus $P_{\mathcal{M}}$ is the operator associated with the probability in coordinate space. Similarly, let $\mathcal{N}$ be a set in momentum space and $Q_{\mathcal{N}}$ be projector define by $\chi _{\mathcal{N}}(p)$. Then \begin{equation*} F_{\mathcal{N}}=\mathrm{tr}Q_{\mathcal{N}}\rho =\int_{\mathcal{N}}dp|\tilde{% \psi}(p)|^{2}, \end{equation*} where $\tilde{\psi}(p)$ is the Fourier transform of $\psi (x)$, is the probability to find the particle with momentum in $\mathcal{N}$.

In more abstract approaches, such as the C$^{\ast }$-algebra formalism, states are defined as positive linear functionals over the set of observables. Density operators belong to this class but there are more general ones.

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"Probability", by itself, is a very squishy concept, and doesn't really make much sense on its own. However, if you attach a statement, as in "probability that X will happen", then you can assign an operator for it.

Let me start with a simple example, a single particle in one dimension, where you want the probability of its coordinate $x$ being between $a$ and $b$. As you know, this can be written as $$P(x\in[a,b])=\int_a^b|\psi(x)|^2\text dx.$$ To bring this into an operator formalism, you express the wavefunction as a braket between the state $|\psi⟩$ and a position eigenstate $|x⟩$, to get $$P(x\in[a,b])=\int_a^b⟨\psi|x⟩⟨x|\psi⟩\text dx.$$ Finally, you can 'factor out' the $\psi$'s to obtain the matrix element of an operator: $$P(x\in[a,b])=⟨\psi|\left(\int_a^b|x⟩⟨x|\text dx\right)|\psi⟩=⟨\psi|\hat\Pi_{[a,b]}|\psi⟩.$$

The operator $\hat\Pi_{[a,b]}$ is called the spectral projector of the position operator $\hat x$ for the set $[a,b]\in \mathbb R$. It has the property that its expectation value in a state $|\psi⟩$ is the probability of $x$ being in $[a,b]$ in that state.

This extends to any well-behaved observable $\hat Q$ and any measurable set of real numbers $A$. In fact, the 'diagonalizability' of self-adjoint operators (including the position and momentum!) is phrased, when derived rigorously in terms of a spectral theorem, exactly in those terms. If $\hat Q$ is self-adjoint (which includes hermiticity as we know it but also some additional constraints on the domains of operators) you are not guaranteed eigenstates but rather a spectral measure. This is a function $\Pi$ which takes sets of real numbers $A$ and returns the corresponding spectral projectors $\Pi(A)$, which have the property that their expectation values are the probability that $q$ is in $A$: $$P(q\in A)=⟨\psi|\hat\Pi(A)|\psi⟩.$$

If $\hat Q$ does have eigenstates, then the spectral projectors are the sum, or integral, of the individual eigenstate projectors $|q⟩⟨q|$ over the $q$'s in $A$. If $Q$ has a continuous spectrum, in fact, the individual projectors $|q⟩⟨q|$ don't make much sense on their own, and must be integrated over to give physical predictions.

This matches up with something that V. Moretti mentioned already. The probability density, $$|\psi(x)|^2=⟨\psi|\left(|x⟩⟨x|\right)|\psi⟩,$$ is not particularly well-defined because $\psi$ can change its value at single points without affecting the state. This is OK, because probability densities are integrated over to give physical results, and these pointwise changes do not affect the total integral.

However, as it is clear from its form, the probability density is also, at least formally, the expectation value of an operator, which in this case is $$|x⟩⟨x|.$$ Because of the above-mentioned problems, this operator is not actually that well-defined, and you need to be very careful with your states (and, in particular, restrict yourself to certain parts of a rigged Hilbert space formalism) for this to make sense. You can get a slightly sturdier definition as $$|x⟩⟨x|=\frac{d}{dx}\hat \Pi_{(-\infty,x]},$$ except that now you need to worry about what it means to differentiate in operator space. The takeaway message on this is that the formal manipulations, if done correctly, do work, but if you want to make them rigorous then it gets very messy very fast. So: use your physical intuition for what quantities make sense and which ones don't, follow the manipulation rules, and you'll be safe.

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No. These cancerous probabilities come in because of the probabilistic interpretation, which also brings in all kinds of famous paradoxes that infect QM. By itself, the formalism of the theory requires just the solution of a second order differential equation to calculate $\psi(x,t)$ - a process which is completely deterministic, just like solving something coming out of Newton's second law. (I'm talking about the Schr\"odinger equation solution here, Heisenberg's matrix mechanics gives identical answers.)

The formalism of the theory and this interpretation are absolutely independent issues. One can have another interpretation (e.g. Many-Worlds, besides others that you can find mentioned in the link) tacked on to the same formalism, which would give us a different way of making sense of these answers. But by itself, there is nothing in the formalism of QM that requires or necessitates probabilities.

So, your answer is - no. Probability is not the result of any operator acting on $\psi$.

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I mean, if the downvote was because of any issue with the answer, you can make that explicit. We may try to reason each other out. – The Dark Side Jun 4 '14 at 9:06
Downvotes are anonymous, and I did not downvote your answer. Although, now that I have read it, I do not think it answers the question fully, or rather goes off on tangents which are not necessarily relevant or appropriately phrased. – JamalS Jun 4 '14 at 9:19
OK. 1. I have deleted my allegation''. Sorry :) 2. What exactly do disagree with ? (We'll both learn by discussing) – The Dark Side Jun 4 '14 at 9:22
Not my downvote either, but you appear to have used the question as an excuse to discuss your own views on the interpretation of QM rather than answering what the OP actually asked. – John Rennie Jun 4 '14 at 9:36
you do need measurements though, and a postulate to interpret the measurements and connect with reality. In my opinion many worlds is just the same thing taking the mathematics to extremes.( I did not downvote either, though I do find "cancerous" distasteful ). It is reality that the mathematics has to model, and to do that postulates are necessary. – anna v Jun 4 '14 at 10:44