# Dirac's definition of probability in quantum mechanics

I'm currently reading "The principles of quantum mechanics" by Dirac, and I'm having some trouble understanding some of his assumptions, because in the quantum mechanics course I'm following at university we are taking a different approach (a more modern one I guess).

I got to the point where Dirac explains why he decides to define the average value of the observable $$\xi$$ in the state $$|x\rangle$$ as $$\langle x|\xi|x\rangle$$. What I don't get is the following part, where he writes that the probability $$P_a$$ of $$\xi$$ having the value $$a$$ is given by the average value of the Kronecker delta $$\delta_{\xi a}$$:

$$P_a=\langle x|\delta_{\xi a}|x\rangle$$

I don't understand if he's deciding to define the probability of $$\xi$$ having the value $$a$$ in this way, or if he's deducing this formula.

In the following similar question: Why does the probability of obtaining a value of a measurement follow from Dirac's general assumption? By Symmetry writes:

The Projection Operator $$\delta_{\varepsilon\, a}$$ is $$1$$ for some particular $$|a\rangle$$ and is $$0$$ on any orthogonal state. The definition of the expectation in classical probability theory for this operator is given by $$$$\left\langle \delta_{\varepsilon\, a}\right\rangle = 1\Pr(|x\rangle = |a\rangle) + 0\Pr(|x\rangle \perp |a\rangle) = \Pr(|x\rangle = |a\rangle)$$$$

So, from this answer it seems to me that he's saying that $$P_a=\langle x|\delta_{\xi a}|x\rangle$$ by definition, but from Dirac's book I have the impression that he's deducing it from previous assumptions.

So my question is: is the formula for the probability of $$\xi$$ having value $$a$$ in the state $$|x\rangle$$ just a definition or can it be deduced by previous assumptions in Dirac's book? If he's deducing it, how is he doing it?

• What is your actual question? At the top of your post you wrote "What I don't get is the following part..." and then you provide a long quotation from Dirac... Are you just generally asking someone to explain the quote? At the bottom of your post you cited to another answer and then said that answer is unsatisfying and then wrote "is it actually just a definition?" Is what just a definition? Neither of these are a very clear questions. Can you edit your post to include a clearer question? This will help you get a clearer answer.
– hft
Commented Nov 13, 2023 at 23:21
• That being said, I will try to give an answer, but I'm not totally sure I'm answering the right question...
– hft
Commented Nov 13, 2023 at 23:22
• it is possible to base probability theory on the concept of expectation, see this quite readable book on the subject amazon.com/dp/0387989552; this approach is similar to that, instead of Lebesgue measure, one uses the (Daniell) integral as the fundamental concept, and was much preferred by Nagy and Riesz. Commented Nov 13, 2023 at 23:23
• Judging from "but he's deducing it." and "So how can the probability coincide with the mean of such function?" I think @hft is getting it correct that the OP is wondering why Dirac considered his choice of postulate to be superior to simply postulating that measurements get eigenvalues per probability. Dirac is proving that his choice of postulates is able to get the latter! It is not a definition! Commented Nov 14, 2023 at 3:18
• @Fede Compare the answer by hft and the paragraph you quoted from Dirac; you should see that "expectation value of Kronecker delta function = probability of observing eigenvalue", and both of them wrote down a proof. Commented Nov 14, 2023 at 9:20

I'm currently reading "The principles of quantum mechanics" by Dirac...

...What I don't get is the following part, where he writes:

In the general case we cannot speak of an observable having a value for a particular state, but we can speak of its having an average value for the state. We can go further and speak of the probability of its having any specified value for the state, meaning the probability of this specified value being obtained when one makes a measurement of the observable. This probability can be obtained from the general assumption in the following way. Let the observable be $$\xi$$ and let the state correspond to the normalized ket $$|x\rangle$$. Then the general assumption tells us, not only that the average value of $$\xi$$ is $$\langle x|\xi|x\gt$$, but also that the average value of any function of $$\xi$$, $$f(\xi)$$ say, is $$\lt x|f(\xi)|x\rangle$$. Take $$f(\xi)$$ to be that function of $$\xi$$ which is equal to unity when $$\xi =a$$, $$a$$ being some real number, and zero otherwise. This function of $$\xi$$ has a meaning according to our general theory of functions of an observable, and it may be denoted by $$\delta_{\xi a}$$ in conformity with the general notation of the symbol $$\delta$$ with two suffixes given on p. 62 {equation (17)). The average value of this function of $$\xi$$ is just the probability, $$P_a$$ say, of $$\xi$$ having the value $$a$$. Thus $$P_a=\langle x|\delta_{\xi a}|x\rangle$$. If $$a$$ is not an eigenvalue of $$\xi$$, $$\delta_{\xi a}$$ multiplied into any eigenket of $$\xi$$ is zero, and hence $$\delta_{\xi a}$$ and $$P_a$$. This agrees with a conclusion of §10, that any result of a measurement of an observable must be one of its eigenvalues.

Based on his Section 16 at Equation 16, Dirac is using the $$\delta$$ symbol with two suffixes (like $$\delta_{ab}$$ or $$\delta_{\xi a}$$) to denote a Kronecker delta, not a Dirac delta.

Next, note that Dirac denotes "the normalized ket" by the symbol $$|x\rangle$$. He seem to sometimes use $$|x\rangle$$ in his text for kets with a continuous label, but he also refers to this ket as "the normalized ket," so in this case he may be using the notation differently.

Anyways, instead of calling the normalized ket "$$|x\rangle$$," I will call it "$$|\Psi\rangle$$."

And, instead of calling the observable "$$\xi$$," I will call it "$$\hat A$$."

Let $$\hat A$$ have eigenvectors $$\{|a\rangle\}$$.

Let the expansion of $$|\Psi\rangle$$ in terms of the $$|a\rangle$$ be: $$|\Psi\rangle = \sum_{a}\Psi_a |a\rangle\;.$$

The expectation value of $$\hat A$$ in the state $$|\Psi\rangle$$ is: $$\langle\Psi|\hat A|\Psi\rangle = \sum_{a,b}\Psi_a\Psi_b^*\langle b|\hat A|a\rangle = \sum_{a}|\Psi_a|^2 a$$

The expectation value of $$f(\hat A)$$ in the state $$|\Psi\rangle$$ is: $$\langle\Psi|f(\hat A)|\Psi\rangle =\sum_{a,b}\Psi_a\Psi_b^*\langle b|f(\hat A)|a\rangle = \sum_{a}|\Psi_a|^2 f(a)$$

The expectation value of a function: $$f_{a_0}(a) \equiv \delta_{aa_0} = \left\{ \begin{matrix} \;1 \qquad a=a_0\\ 0 \qquad \text{else}\end{matrix}\right.$$ is $$\sum_a |\Psi_a|^2 \delta_{aa_0} = |\Psi_{a_0}|^2\;,$$ which, as usual, is the probability to measure $$a_0$$ when the state is $$|\Psi\rangle$$.

If the $$a$$ are continuous, $$|\Psi(a)|^2$$ is a probability density rather than a probability.

Update:

To spell out the above explanation again, maybe a bit more clearly:

1. Dirac introduces the expectation value of $$\hat A$$, which is written as $$\langle \Psi|\hat A|\Psi\rangle$$. He does not perform the expansion in terms of eigenstates of $$\hat A$$, but we did so above and we see that $$\langle \Psi|\hat A|\Psi\rangle = \sum_{a}|\Psi_a|^2 a$$.
2. [Dirac may have noted it previously, but I note here that $$0\le|\Psi_a|^2\le 1$$ and $$\sum_a|\Psi_a|^2=1$$, which are also properties that we desire of a set of probability values.]
3. Dirac further notes that the expectation value of any function $$f(\hat A)$$ is $$\langle \Psi|f(\hat A)|\Psi\rangle$$, which, again, he does not expand, but we expanded to show that it is $$\sum_a|\Psi_a|^2 f(a)$$.
4. Dirac considers a special function $$f_{a_0}(a)$$ that is $$0$$ for all the $$a$$ values except for one of them $$a_0$$ (where it is $$1$$) and he asserts (correctly, see below) that the expectation value of such a function is the probability that $$\hat A$$ takes on the value $$a_0$$. He does not show this explicitly, but we have shown explicitly above that $$\langle \Psi|f_{a_0}(\hat A)|\Psi\rangle = |\Psi_{a_0}|^2$$.

Regarding why the expectation value of a function $$f_{a_0}(a)$$ that is zero for all $$a$$ other than $$a_0$$ and is $$1$$ at $$a_0$$ is the probability of $$a_0$$:

Consider the general definition of an expectation value (not specific to quantum mechanics, but completely general) for a given discrete set of probability values $$P(a)$$: $$\langle f \rangle = \sum_a P(a) f(a)\;.$$ Clearly $$\langle f_{a_0}\rangle = P(a_0)\;.$$

• Thank you, this helps me understand why Dirac's definition of probability coincides with the one they teach us at university. However, up to the point in the book I quoted, Dirac had never defined the probability of $\xi$ having value $a$ in the state $|x\rangle$ as the square of the coefficient of the eigenvector corresponding to $a$ in the expansion of $|x\rangle$, therefore my question was more about how Dirac himself could write that formula in that point of his book.
– Fede
Commented Nov 14, 2023 at 9:28
• I added an update. I'm not sure it addresses your exact question, since I'm not entirely sure what "that formula" refers to in your comment. But, maybe the update helps?
– hft
Commented Nov 14, 2023 at 16:53
• thank you for your patience, now you really clarified every detail
– Fede
Commented Nov 15, 2023 at 20:58
• @Fede Glad to hear it!
– hft
Commented Nov 15, 2023 at 21:07

Once one identifies $$\langle x | A |x \rangle$$ with the average value of $$A$$, and the states $$|x\rangle$$ are normalized, it is just the frequentist definition of probability that forces to identify the expectation value of $$\langle x | \delta_{\xi, a} |x \rangle$$ with the probability $$P_a$$.

Dirac's approach was to introduce the average value and, from that, to deduce the probability.

This has nothing to do with quantum mechanics. Imagine you're tossing a fair die. You can assign any numbers to the six outcomes. If you number them $$1,2,3,4,5,6$$ then the expected value of a roll is $$3.5$$. If you number them $$0,1,0,0,0,0$$ then the expected value of a roll is $$1/6$$, which is not coincidentally also the probability of rolling two pips. If the die is loaded and that probability isn't $$1/6$$, numbering the outcomes in that way still gives you the correct probability.