# Probabilities in Quantum Mechanics: Measurement Outcomes or More?

In all treatments of quantum mechanics, the probabilistic nature of the theory enters via the Born rule for the statistical properties of the measurement outcomes of some observable. In short, this says that for an observable $$\Omega \in \mathcal{L(H)}$$ on the Hilbert space $$\mathcal{H}$$, the probability of measuring the eigenvalue $$\omega$$ in a normalized state $$|\psi \rangle \in \mathcal H$$ is given by:

$$\mathrm{Pr}(\Omega = \omega) \equiv \Vert P_\omega \vert \psi \rangle \Vert^2$$

with $$P_\omega$$ being the projector onto the eigenspace of $$\Omega$$ associated with the eigenvalue $$\omega$$.

Now as far as I know, as far as pure states are concerned, this is the only way in which probabilities enter quantum mechanics; speaking of probabilities without there being a measurement on an observable doesn't really make sense. There is one problem however, in many cases, I've seen people use the notion of a quantity

$$\vert \langle \phi \vert \psi(t) \rangle \vert^2$$

as some kind of probability as well, without there being any mention of a measurement/observable. For example, for a spin $$1/2$$ particle, if $$|\psi(0) \rangle = \vert \left\downarrow \right \rangle$$ and $$|\phi \rangle = \vert \left\uparrow \right \rangle$$, I've seen the quantity

$$p(t) = \vert \langle \uparrow \vert \psi(t) \rangle \vert^2$$

being called the "spin flip probability". Now I understand that $$p(t)$$ satisfies all axioms necessary for it to be a valid probability, and I also intuitively understand that this quantity measures the closeness of the states $$\vert \psi(t) \rangle$$ and $$\vert \left\uparrow \right \rangle$$, with them being exactly the same (up to a phase) when $$p(t) = 1$$. But still, I keep trying to find a way to connect this probabilistic statement with the Born rule, without any success.

As another example of this discrepancy, consider the standard formulation of time-dependent perturbation theory, in which the transition probability from the energy eigenstate $$\vert n \rangle$$ to $$\vert m \rangle$$ is defined as:

$$p_{n\rightarrow m}(t) \equiv \vert \langle m \vert U(t) \vert n \rangle \vert ^2$$ where $$U$$ is the propagator. This can then be used to calculate a transition rate, etc.

Again, I don't see any mention of there being a measurement taking place at all. The only way I can make sense of it is by thinking of it as a shorthand for saying:

"If the system started in the state $$\vert n \rangle$$ and evolved for time $$t$$, what would be the probability of measuring the system's energy and getting $$E_m$$ as the result".

But this interpretation is not without problems either, since it breaks down if the $$m^\mathrm{th}$$ eigenstate is degenerate. Then there is no way I can think of to connect this to the Born rule.

Am I missing something fundamental that connects these probabilistic interpretations in a satisfactory way, or should I just interpret these second "probabilities" as more of a measure for the "closeness" of the two states $$\vert \phi \rangle$$ and $$\vert \psi(t) \rangle$$ ?

• Yes, probability always stands for the probability of seeing some result when you do a meeasurement, for example in your first case the spin flip probability is the probability of measuring $|\uparrow \rangle$ if you do a spin measurement. – knzhou Jul 19 at 12:23
• @knzhou I have mentioned that in my second example as one interpretation. But what about "transition probabilities" from one energy eigenstate to the other in the context of time-dependent perturbation theory? If either eigenstate in question is associated with a degenerate eigenvalue, you can't make that association with the Born rule anymore. See my second example. – Sahand Tabatabaei Jul 19 at 12:26
• Sure you can, you just have to measure something other than energy. – knzhou Jul 19 at 12:29
• That's true. You could always define some Hermitian operator for which the probabilities of the measurement outcomes correspond to the $\vert \langle n \vert U \vert m \rangle \vert^2$ transition probabilities. But I don't feel like this actually addresses my issue. The only thing that tells you is that that quantity is the probability of measuring a specific value for some (perhaps) artificial and abstract observable you've defined on your Hilbert space. Why would you call it the transition probability then? Shouldn't there be a connection with the physical intuition of a "transition". – Sahand Tabatabaei Jul 19 at 12:35
• When talking about probability measurement, it makes me think about higher dimensions searchers fifth or 10th where probability is relative to space and time but is hard to comprehend – C. Jordan Jul 19 at 14:01


What you should realise is that e.g. $$P_\uparrow \equiv \vert\!\uparrow\,\rangle\langle\,\uparrow\!\vert$$ is the projector onto the spin-up eigenspace. Thus

$$\mathrm{Pr}\left(S_z=\frac{1}{2}\right) = \Vert P_\uparrow \vert\,\Psi\,\rangle \Vert^2 = \underbrace{\langle\,\uparrow\!\vert\!\uparrow\,\rangle}_{=1}\vert\langle\,\uparrow\!\vert\Psi\,\rangle\vert^2$$

and both expressions represent the same quantity.

If the subspace is degenerate, you may write the projector onto that subspace in an orthonormal(!) basis as

$$P_\omega = \sum_k \ket{\omega,k}\bra{\omega,k}$$ where the sum runs over a basis spanning the $$\omega$$-subspace. Then the Born rule yields

$$\mathrm{Pr}(\Omega=\omega) = \Vert P_\omega \ket{\Psi}\Vert^2 = \sum_k \vert \braket{\omega,k}{\Psi}\vert^2$$ i.e the total probability to measure $$\omega$$ is found by summing over the degeneracy (in terms of probability theory: you marginalize).

Maybe as an example consider the hydrogen atom with energy & angular momentum basis $$\ket{n,l,m}$$. Each energy eigenstate $$H\ket{n,l,m}=E_n\ket{n,l,m}$$ is $$n^2$$ fold degenerate. If the system begins e.g. in the state $$\ket{000}$$, the probability of finding it at a later time $$t$$ in a state with principal quantum number $$n$$ is

$$\sum_{l=0,...,n-1}\sum_{m=-l,...,l} \vert \braket{n,l,m}{U(t)\vert 000}\vert^2$$ What angular momentum the system might have is irrelevant for the measurement of energy at hand.

If you'd ask the question: What is the probability to find the electron in the p-orbital of the second shell, that would be

$$\sum_{m=-1,0,1} \vert \braket{2,1,m}{U(t)\vert 000}\vert^2$$ since in that case the observable at hand is $$H\otimes L^2$$.

What is up with $$p_{n\rightarrow m}(t) \equiv \vert \langle m \vert U(t) \vert n \rangle \vert ^2\;\;?$$ This is the probability of finding the system in state $$\ket{m}$$. If the corresponding energy eigenvalue is non-degenerate than this is - as you well understand - the probability of finding the system at energy $$E_m$$ when making a measurement after time $$t$$. But if it is degenerate, how could you possibly tell in which of the degenerate states the system is in by measuring energy alone? You cannot. You need (at least) another observable that you can measure simultaneously to tell the degenerate states apart. In conclusion, you are right. The expression above is not the measurement probability of energy $$E_m$$. To obtain this, one needs to sum over the subspace.

It's not in contradiction to the Born rule though. Any outcome of a (projective) measurement corresponds to an orthogonal projector $$P$$. Whether that projector is of rank one and projects onto a single state, or whether it's projecting onto a degenerate subspace. So, asking "if the system is in state $$\ket{n}$$ is a perfectly valid measurement. One you may not be able to carry out in practice, but that's fine. And so is asking "if the system has energy E_n". The Born rule applies to all those situations equally and the probability is always given as $$\Vert P\ket{\Psi}\Vert^2$$

• Yes, I have mentioned that in the context of my second example of the discrepancy. As soon as you're discussing an observable with a degenerate spectrum, your point breaks down. See the second example in my question to see what I mean. – Sahand Tabatabaei Jul 19 at 12:23
• @SahandTabatabaei I have expanded my answer. Although I am not convinced I fully understand the root of your problem which seems to be of semantic rather than physical nature? – Nephente Jul 19 at 13:19
• Exactly! So the quantity $\vert \langle n \vert U \vert m \rangle \vert^2$ (with no summation over the degenerate eigenspace) is only commensurate with the Born rule if the states are non-degenerate as you say. I completely understand everything as long as the spectrum is non-degenerate. That's exactly what my problem is. I've seen this quantity being called the "transition probability from state $m$ to state $n$" with no mention of any assumptions on degeneracy. See for example the chapter on time-dependent perturbation theory in Sakurai and Griffiths (and tons of other books). – Sahand Tabatabaei Jul 19 at 13:27
• @SahandTabatabaei You are perfectly right. I've revised the section. – Nephente Jul 19 at 14:54
• @SahandTabatabaei Asking what state the system is in, is a valid measurement! Any rank one projector is a perfectly fine observable. – Nephente Jul 19 at 17:36