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Since the issue of deriving the Born rule comes up from time to time in the forum and I'm always a little mystified by some of the opinions people have. I recently fleshed out the details about the derivation I'm about to present and realized it pretty much makes the Born rule a triviality, so I thought I'd present it to see how others take it.
Also, it shows off the operator-focused picture of quantum mechanics, which I like.
As an aside, this is an "external" derivation. (For the record, I generally prefer an "internal" ones, but those seem to stir up much controversy -- especially from those who think internal treatments are impossible by definition)Recall that the main thing about the operator-centric picture of quantum mechanics is that the central notion is that there is an algebra of operators. This algebra is supposed to represent what sorts of "measurements" can be done, and possibly more general things. There are a variety of features of this algebra, but the main thing I want is the following: there is a quantity [itex]||x||[/itex] that has the meaning
One property of a "quantum state" [itex]\psi[/itex] is that you can combine a state with an operator [itex]x[/itex] to produce a complex number [itex]E_\psi(x)[/itex] to be interpreted as something like the "expected value of the operator over that state". This function has various properties, such as
If we assume our operators form a C*-algebra, the properties of [itex]E_\psi[/itex] can be summed up by saying it's a "positive linear functional of norm 1". I'm not making that assumption, but I'm going to make some similar requirements on [itex]E_\psi[/itex] to satisfy the properties I require below.Now, given that we intend to interpret [itex]E_\psi(x)[/itex] as the "expected value" of the operator x, we can also compute the higher moments [itex]m_k = E_\psi(x^k)[/itex] of the operator x over the state [itex]\psi[/itex].
Now, I'm pretty sure the properties of [itex]E_\psi[/itex] ensure that there is a actual probability distribution that has these moments [itex]m_k[/itex] about zero -- and that probability distribution is unique -- and thus we can say that this is probability distribution of the outcomes of x over the state [itex]\psi[/itex]. But I feel (ATM) it's enough to treat the case of a qubit, where we don't need anything fancy.
The operator corresponding to measuring a qubit about some orientation satisfies [itex]x^2 = x[/itex]; intuitively because the "outcomes" are 0 and 1. The properties of E ensure [itex]0 \leq E_\psi(x) \leq 1[/itex] and clearly, all of the moments [itex]E_\psi(x^k)[/itex] are equal. This corresponds to the probability distribution
Born rule derived. QEDOh wait, you were expecting me to prove the ket version; that the probability of measuring the state [itex]a |0\rangle + b |1\rangle[/itex] to be 0 or 1 is [itex]|a^2|[/itex] or [itex]|b^2|[/itex], didn't you? It's a mere triviality at this point; the very definition of the phrase
What time evolution is really is a foundational assumption of quantum mechanics. That probabilities correspond to something in the real world (and what precisely they correspond to) is also a foundational assumption and a matter of interpretation.
But the formula for computing probabilities from kets is a banality, and the specific form of time evolution in a particular way of representing things is just a matter of translation. If we had some other representation, the equation would take some other form.
Also, it shows off the operator-focused picture of quantum mechanics, which I like.
As an aside, this is an "external" derivation. (For the record, I generally prefer an "internal" ones, but those seem to stir up much controversy -- especially from those who think internal treatments are impossible by definition)Recall that the main thing about the operator-centric picture of quantum mechanics is that the central notion is that there is an algebra of operators. This algebra is supposed to represent what sorts of "measurements" can be done, and possibly more general things. There are a variety of features of this algebra, but the main thing I want is the following: there is a quantity [itex]||x||[/itex] that has the meaning
[itex]||x||[/itex] is the magnitude of the largest outcome possible of the measurement.
(really this should be phrased as a supremum, rather than in terms of a maximum)One property of a "quantum state" [itex]\psi[/itex] is that you can combine a state with an operator [itex]x[/itex] to produce a complex number [itex]E_\psi(x)[/itex] to be interpreted as something like the "expected value of the operator over that state". This function has various properties, such as
[tex]| E_\psi(x) | \leq || x ||[/tex]
which is obvious from its intended interpretation. As an aside, by listing out an adequate list of the properties that [itex]E_\psi[/itex] must satisfy, one can even go so far as to define a quantum state to be a function that satisfies those properties.If we assume our operators form a C*-algebra, the properties of [itex]E_\psi[/itex] can be summed up by saying it's a "positive linear functional of norm 1". I'm not making that assumption, but I'm going to make some similar requirements on [itex]E_\psi[/itex] to satisfy the properties I require below.Now, given that we intend to interpret [itex]E_\psi(x)[/itex] as the "expected value" of the operator x, we can also compute the higher moments [itex]m_k = E_\psi(x^k)[/itex] of the operator x over the state [itex]\psi[/itex].
Now, I'm pretty sure the properties of [itex]E_\psi[/itex] ensure that there is a actual probability distribution that has these moments [itex]m_k[/itex] about zero -- and that probability distribution is unique -- and thus we can say that this is probability distribution of the outcomes of x over the state [itex]\psi[/itex]. But I feel (ATM) it's enough to treat the case of a qubit, where we don't need anything fancy.
The operator corresponding to measuring a qubit about some orientation satisfies [itex]x^2 = x[/itex]; intuitively because the "outcomes" are 0 and 1. The properties of E ensure [itex]0 \leq E_\psi(x) \leq 1[/itex] and clearly, all of the moments [itex]E_\psi(x^k)[/itex] are equal. This corresponds to the probability distribution
- [itex]P(x = 0) = 1 - E_\psi(x)[/itex]
- [itex]P(x = 1) = E_\psi(x)[/itex]
Born rule derived. QEDOh wait, you were expecting me to prove the ket version; that the probability of measuring the state [itex]a |0\rangle + b |1\rangle[/itex] to be 0 or 1 is [itex]|a^2|[/itex] or [itex]|b^2|[/itex], didn't you? It's a mere triviality at this point; the very definition of the phrase
The ket [itex]|\psi\rangle[/itex] represents the quantum state [itex]\psi[/itex]
(in some representation of our algebra as being actual operators) is that we have an identity[tex]E_\psi(x) = \langle \psi | x | \psi \rangle[/tex]
The usual form of the Born rule is not some foundational assumption of quantum mechanics; it's just a simple consequence of what we mean by using a ket to represent a quantum state.I assert that opinions otherwise are really talking about something like "Why does time evolution take the form of the Schrödinger equation if we use the ket representation of a state?" What time evolution is really is a foundational assumption of quantum mechanics. That probabilities correspond to something in the real world (and what precisely they correspond to) is also a foundational assumption and a matter of interpretation.
But the formula for computing probabilities from kets is a banality, and the specific form of time evolution in a particular way of representing things is just a matter of translation. If we had some other representation, the equation would take some other form.
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