What is the best way to notate probability in QM?

Question

In a post I was reading upon, I saw the following: $$\text{probability} = \int_a^b\Psi^*\Psi\,\mathrm{d}x\quad\biggl(= \int_a^b\Psi^2\,\mathrm{d}x,\text{ if }\Psi\text{ is a real function}\biggr)$$ Is that not equivalent to: $$\int_a^b\langle\Psi|\Psi\rangle\,\mathrm{d}x ?$$ (I’m just starting to learn QM and I want to clear out the differences in notations and forms to minimise misconceptions while I’m learning)

Answer 1

No, that would not be correct. The definition of the inner product $\langle\Psi|\Psi\rangle$ already includes the integation. $$\langle\Psi|\Psi\rangle:=\int\Psi^*(x)\Psi(x)dx$$

The product of a state $|\Psi\rangle$ with itself is defined to be $1$ (the so-called normalization condition), Then we can interpret $\Psi^*(x)\Psi(x)$ as the probability density, and the probability of the particle to be anywhere is the integral, i.e. $1$.

Answer 2

You probably should have a look at the general description of the postulates of Quantum Mechanics. I am going to suggest you look up the Quantum Mechanics textbook by Cohen-Tannoudji. These postulates can be stated as follows:

1. To describe a quantum system there is a Hilbert Space $\mathcal{H}$ whose elements represent the possible states of the system.

2. For each physical quantity associated to the system there is one hermitian operator defined on $\mathcal{H}$. We call these operators observables.

3. The only possible values to be measured of an observable are the elements of its spectrum.

4. If $A$ is an observable with discrete and non-degenerate spectrum $\sigma(A)=\{a_i : i \in \Bbb N\}$, then the probability of measuring $a_i$ on the state $|\psi\rangle$ is $P(a_i)=|\langle \varphi_i|\psi\rangle|^2$, where $|\varphi_i\rangle$ is the eigenstate corresponding to the eigenvalue $a_i$. Analogously, if $A$ is an observable with continuous spectrum $\sigma(A)$ then the probability density for the values of $A$ on the state $|\psi\rangle$ is $\rho(x)=|\langle x|\psi \rangle|^2$ where $x\in \sigma(A)$ and $|x\rangle$ is the generalized eigenvector corresponding to $x$.

5. When one performs a measurement of the observable $A$ the state colapses to the eigenstate corresponding to the eigenvalue measured.

6. The time evolution is governed by the requirement that the observable corresponding to the energy is the generator of time translations. That is, the time evolution equation is $i\hbar \frac{d|\psi\rangle}{dt}=H|\psi\rangle$.

In the above list, postulate 4 is the one that answers your question.

Answer 3

$\langle\Psi|\Psi\rangle$ is just a number, equal to the squared norm of the ket $\lvert\Psi\rangle$. Since kets are generally assumed to be normalised, this is just equal to 1. Thus, the expression you gave would simplify to $$\int_a^b\langle\Psi|\Psi\rangle\,\mathrm{d}x=\langle\Psi|\Psi\rangle\int_a^b\mathrm{d}x=b-a.$$ Clearly, this is not a probability.

The usual way to introduce the Born Rule is to say that the probability of measuring some observable to be in some set is equal to the integral of the squared norm of the wavefunction, expressed in the basis of that observable, over the set of interest. In other words, for a system in the state $\lvert\alpha\rangle$, the probability of measuring the position to be in the interval $[a,b]$ is equal to $$\int_a^b\mathrm{d}x\,\psi_\alpha^*(x)\psi_\alpha(x),$$ where $\psi_\alpha(x)$ is the position space wavefunction corresponding to $\lvert\alpha\rangle$.$^*$ While this is an accurate way to state the Born Rule, it's not very enlightening about what's really going on here, so we instead look at another way of stating it. An alternative statement of the Born Rule is that the probability of measuring some observable to be in some set is equal to the expectation value of the projection operator onto the subspace corresponding to the set of interest. For the case of position in the interval $[a,b]$, this subspace is spanned by the set of position eigenkets with eigenvalues in that interval, so the projection operator is given by $$\Lambda_{[a,b]}=\int_a^b\mathrm{d}x\,\lvert x\rangle\langle x\rvert.$$ With the projection operator defined above, we can then say that the probability of measuring the position in the interval $[a,b]$ is equal to $$\langle\Lambda_{[a,b]}\rangle=\int_a^b\mathrm{d}x\,\langle\alpha| x\rangle\langle x|\alpha\rangle.$$

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$^*$Note that I have used slightly different notational conventions from you here, since I think it can be confusing and misleading to use the same symbol for kets and wavefunctions.

Answer 4

If you want to write the integral from $[a,b]$ in terms of Dirac notation you can use a projection operator. An example of a projection operator is $$P=\pmatrix{1&0\\0&0}$$ which projects all vectors down to the x-axis. If you define $|1\rangle=\pmatrix{1\\0}$ then you can also write $P=|1\rangle\langle 1|$. A more general projection operator can be written as $$\sum_n|n\rangle\langle n|$$ where all the $|n\rangle$ form an orthonormal basis: $\langle m|n\rangle=\delta_{m,n}$. For example $P=|1\rangle\langle 1|+|3\rangle\langle 3|$. An important property of projection operators is that $P^2=P$.

Finally we can define a projection operator for a continuous basis (which is what we want here) $$P_{[a,b]}=\int_a^b\mathrm dx\,|x\rangle\langle x|$$ Now we can write \begin{align}\text{probability}&=\langle \psi|P_{[a,b]}|\psi\rangle\\ &=\int_{a}^b\mathrm dx\,\langle \psi|x\rangle\langle x|\psi\rangle\\ &=\int_{a}^b\mathrm dx\,\psi^*(x)\psi(x) \end{align} Note that this notation is almost never used in practice and that you could just as easily say "the probability that the particle is in $[a,b]$ after a measurement is given by..."

Finally note the notation $\langle\psi|\psi\rangle$ is equal to \begin{align} \langle\psi|\psi\rangle&=\langle\psi\left(\int_{-\infty}^\infty\mathrm dx\,|x\rangle\langle x|\right)|\psi\rangle\\ &=\int_{-\infty}^\infty\mathrm dx\,\langle \psi|x\rangle\langle x|\psi\rangle\\ &=\int_{-\infty}^\infty\mathrm dx\,\psi^*(x)\psi(x) \end{align} We can do this because $\int_{-\infty}^\infty\mathrm dx\,|x\rangle\langle x|$ is equal to the identity operator ; applying this to any ket does nothing, i.e. $|\psi\rangle=|\psi\rangle$.