# I've read that $\langle a | b\rangle$ is a probability amplitude but $\langle a | a\rangle$ is a probability. Why the inconsistency?

I'm studying elementary quantum mechanics, and I've read that $\langle a \vert b \rangle$ is the probability amplitude of a transition from state $a$ to state $b$. Thus, $|\langle a | b \rangle|^2$ is the probability of the transition. However, $\langle a | a \rangle$ is supposed to be the probability (not the probability amplitude) of being in state $a$, or, if you will, making the transition from state $a$ to state $a$. This seems like a contradiction. What don't I understand?

There are a few misconceptions here.

The probabilistic nature of quantum mechanics is not related to what state a system is at a particular time, but rather to what state the system will be in after a measurement is performed.

For example, consider a simple spin-1/2 system. A measurement of the a particular component of the particle's spin will yield either "spin up" (denoted by $\uparrow$) or "spin down" (denoted by $\downarrow$).

One example for a possible state of the system is $$|\psi\rangle = \frac{1}{2}|\uparrow \rangle + \frac{\sqrt{3}}{2}|\downarrow\rangle$$ in which case we have that

$$\langle \uparrow | \psi \rangle = \frac{1}{2} \hspace{1 cm} \left|\langle \uparrow | \psi \rangle\right|^2 = \frac{1}{4}$$

$$\langle \downarrow | \psi \rangle = \frac{\sqrt{3}}{2} \hspace{1 cm} \left|\langle \downarrow | \psi \rangle\right|^2 = \frac{3}{4}$$

This does not mean that the particle is in state $\uparrow$ with probability $\frac{1}{4}$ and in state $\downarrow$ with probability $\frac{3}{4}$. The state of the particle at this moment is not in question - it is simply $\psi$.

However, if we measure the spin of the particle, the probability that we find the particle to be spin up or spin down are $\frac{1}{4}$ and $\frac{3}{4}$ respectively.

As to the other part of your question, probabilities are always found by taking the square modulus of a probability amplitude. If I'm reading your question correctly, the answer is that the second assertion you make is simply incorrect. Perhaps if you could give more specifics as to why you might think it to be true, then we could better resolve the issue.

Technically speaking, $$\langle a | a \rangle$$ is also a transition amplitude "from a to a", which is always 1 by convention since any eigenvector is a member of a projective space, hence a normalization factor won't break eigenvectorness.

Also, the "probability of a to remain in a" is $$\langle a | a \rangle \langle a | a \rangle$$ which is also 1. If there was some small time evolution allowed, then you would indeed compute the probability as $$\lim_{\Delta t \rightarrow 0}\langle a |e^{i\bf{H}\sf{t}}| a \rangle \langle a |e^{-i\bf{H}\sf{t}}| a \rangle = 1$$

So, is not dimensionally proper to say that $$\langle a | a \rangle$$ is by itself a probability, but people have been known to get away with it since the result is numerically the same

If $|a\rangle$ is normalised then $\langle a|a\rangle$ should be 1, if it is unnormalised and you want to relate it to probability then how you constructed $|a\rangle$ is important.

One example (and maybe more advanced then where you are at the moment) is if you were working with a particular Krauss decomposition of a CPTP map (for example if you were reading Nielsen and Chuang). A CPTP map acts on density matrices by $\varepsilon (\rho) = \sum_{j} K_{j} \rho K_{j}^{\dagger}$. This can be reinterpreted as acting on a pure state so that state $|a\rangle$ has a probability to have evolved under each $K_{j}$ with some probability $p_{j}$.

In this case starting with state |a> the $j^{\mathrm{th}}$ Krauss operator is $|t_j\rangle = K_j |a\rangle$ and the norm of $|t_j\rangle$ is the probability that the result $j$ happened, $p_{j} = \langle t_j|t_j\rangle$. This is related to the probability $|\langle b|a\rangle|^{2}$ that you mentioned in the original post by choosing your Krauss operators to be $K_{1} = P_{b}U$ ($P_{b}$ being the projection onto $b$ and $U$ some unitary evolution) and $K_{2} = (1-P_{b})U$, then $p_{1} = \langle t_{1}|t_{1}\rangle = \langle a| P_{b} P_{b} | a\rangle = \langle a| P_{b} | a\rangle = \langle a | b\rangle \langle b|a\rangle$ which is the probability of measuring $|b\rangle$ after evolution $U$.

As a side note a Krauss operators have the requirement $\sum_{j} K_{j}^{\dagger} K_{j} = 1$ in order for this to be a CPTP map.