4
$\begingroup$

Consider a ket-space spanned by the eigenkets of an observable $A$ and let $B$ be an additional observable on the same ket-space. We can build a filter that only lets an eigenvalue $a$ of $A$ through and afterwards subject the outgoing beam to a second filter that selects an eigenvalue $b$ of $B$.

If $|\alpha\rangle$ is an arbitrary ket and we're interested in computing the probability of a particle whose state is described by $|\alpha\rangle$ to survive the ensemble described above, we can firstly compute the probability of measuring $a$ in $|\alpha\rangle$ which is simply given by: $$ P_a = |\langle a | \alpha \rangle | ^2 $$ If the particle comes through the first filter, its state is now described by the ket $|a\rangle$. Just like before, we can calculate the probability of the particle to pass through the second filter: $$ P_b = |\langle b | a \rangle | ^2 $$ I'm not sure if my question is completely trivial, but I'm having trouble understanding why the probability for the initial state $|\alpha\rangle$ to survive the whole ensemble is then given by the product: $$ P_{tot} = P_bP_a = |\langle b | a \rangle \langle a | \alpha \rangle | ^2 $$ Do we have to postulate this? This result is well known for the computation of the probability of two independent events in mathematics, but I don't know if I can just directly translate such results to the probabilities of quantum mechanics.

$\endgroup$
2
  • $\begingroup$ Have you contemplated two light polarizers at an angle? $\endgroup$ Commented Jan 4, 2023 at 16:13
  • $\begingroup$ @CosmasZachos I tried to understand this using that example but the difference between both situations is that the amplitude of a beam of light gets reduced when passing through a filter (namely by the scalar product of the direction of the polarisation of the beam and the polarisation direction of the filter). The outgoing light then has an amplitude that depends on the polarisation of the beam prior to the polaroid filter. In the QM-case however, the state gets projected into a normalised eigenstate and the information of the initial state is lost. So the formula can't be directly translated $\endgroup$ Commented Jan 4, 2023 at 16:59

3 Answers 3

5
$\begingroup$

$$ P_a = |\langle a | \alpha \rangle | ^2 $$

$$ P_b = |\langle b | a \rangle | ^2 $$

I'm having trouble understanding why the probability for the initial state $|\alpha\rangle$ to survive the whole ensemble is then given by the product: $$ P_{tot} = P_bP_a = |\langle b | a \rangle \langle a | \alpha \rangle | ^2 $$

Do we have to postulate this?

No. The meaning of probability remains unchanged regardless of if we are dealing with classical or quantum mechanics. Quantum mechanics just gives us a new (well, new in the 1920s) way of calculating the probabilities of certain physical processes as squared amplitudes.


"The concept of probability is not altered in quantum mechanics. When we say the probability of a certain outcome of an experiment is p, we mean the conventional thing, i.e., that if the experiment is repeated many times, one expects that the fraction of those which give the outcome in question is roughly p."

(Feynman and Hibbs, "Quantum Mechanics and Path Integrals," Section 1-1, page 2.)


In your case, the initial state is $|\alpha\rangle$ and the first event $A$ is the event that the measurement yields $a$, which we assume is the eigenvalue of $|a\rangle$ only. Therefore the probability of the first event is $$ P_1 = |\langle a|\alpha\rangle|^2\;, $$

The state is now $|a\rangle$ (per the specification of the problem that there is only one state with that eigenvalue). Therefore, the probability of the subsequent event $b$ measurement (where again we assume only the states $|b\rangle$ yields this eigenvalue) is $$ P_2 = |\langle b|a\rangle|^2 $$ and therefore, the total probability of measuring $a$ followed by measuring $b$ is: $$ P_{1,2} = |\langle b|a\rangle|^2|\langle a|\alpha\rangle|^2 $$

This is not the same as the probability of measuring b first then a: $$ P_{2,1} = |\langle a|b\rangle|^2|\langle b|\alpha\rangle|^2 $$ unless $|\langle b|\alpha\rangle|^2=|\langle a|\alpha\rangle|^2$.

$\endgroup$
12
  • 1
    $\begingroup$ But for non-commuting observables it might not make sense to define the joint probability distribution, tho. See for example section 9.6 of Ballentine. $\endgroup$ Commented Jan 4, 2023 at 19:55
  • $\begingroup$ @TobiasFünke Okay, I believe you, but is your caveat relevant for OP's physical case: "a filter that only lets an eigenvalue a of A through and afterwards... a second filter that selects an eigenvalue b of B." $\endgroup$
    – hft
    Commented Jan 4, 2023 at 20:30
  • 1
    $\begingroup$ @TobiasFünke See also this: "The concept of probability is not altered in quantum mechanics. When we say the probability of a certain outcome of an experiment is $p$, we mean the conventional thing, i.e., that if the experiment is repeated many times, one expects that the fraction of those which give the outcome in question is roughly $p$." (Feynman and Hibbs, page 2.) $\endgroup$
    – hft
    Commented Jan 4, 2023 at 23:17
  • 2
    $\begingroup$ @hft I think Tobias's comment is relevant here. If $A$ and $B$ are incompatible, then the probability of measuring an eigenvalue $a$ followed by $b$ is different from the probability of measuring an eigenvalue $b$ followed by $a$ $\endgroup$
    – Ryder Rude
    Commented Jan 5, 2023 at 1:55
  • 1
    $\begingroup$ @justaphase you're right the logic events and logical and are not the same as the physical/temporal/causal events. The order matters. I updated my answer. $\endgroup$
    – hft
    Commented Jan 5, 2023 at 20:23
4
$\begingroup$

Below I've derived this rule for the joint probability of two consecutive measurements that holds in any axiomatic formulation of quantum mechanics without further assumption. This derivation applies whether or not the two observables commute, and does not assume conditional probabilities, etc.

The way I represent measurements follows from the Stinespring Dilation Theorem, which says that all quantum channels can be represented via isometries (which are lenght-preserving maps between Hilbert spaces of different size; see below for further details).

For measurements, the Hilbert space is "dilated" to include degrees of freedom that codify the measurement outcome. By Choi's Theorem, this is equivalent to the Kraus representation. Moreover, I can embed isometries in unitaries, which leads to a unitary picture of measurements for qubits and more generally. That unitary picture of measurement is equivalent to the Many-Worlds Interpretation (MWI).

Suppose I have two operators $A$ and $B$ with spectral decompositions $$ A \, = \, \sum\limits_{a=0}^{N^{\,}_A-1} \, A^{\,}_a \, \mathbb{P}^{\,}_a ~~,~~B \, = \, \sum\limits_{b=0}^{N^{\,}_B-1} \, B^{\,}_b \, \widetilde{\mathbb{P}}^{\,}_b ~,~~$$ where $\{A^{\,}_a\}$ are the $N^{\,}_A$ unique eigenvalues of $A$ (and thus the possible outcomes upon measurement), and $\{B^{\,}_b\}$ are the $N^{\,}_B$ unique eigenvalues of $B$. The projector $\mathbb{P}^{\,}_a$ projects onto the $a$th eigenspace of $A$, so that $A \mathbb{P}^{\,}_a = A^{\,}_a \mathbb{P}^{\,}_a$; likewise the projector $\widetilde{\mathbb{P}}^{\,}_b$ projects onto the $b$th eigenspace of $B$, so that $B \widetilde{\mathbb{P}}^{\,}_b= B^{\,}_b \widetilde{\mathbb{P}}^{\,}_b$. The projectors are orthogonal, complete, and idempotent: $$ \sum\limits_{a=0}^{N^{\,}_A-1} \, \mathbb{P}^{\,}_a \, = \, \mathbb{1}~~~~\text{and}~~~~\mathbb{P}^{\,}_a \, \mathbb{P}^{\,}_{a'} \, = \, \delta^{\,}_{a,a'} \mathbb{P}^{\,}_a~,~~$$ and likewise for the projectors for $B$.

We encode measurement channels by representing the degrees of freedom corresponding to the measurement apparati. In this case, we have two measurement devices: One for each of $A$ and $B$, with $N^{\,}_A$ and $N^{\,}_B$ internal states, respectively. It will help to define the cyclic shift operators $$ X^k_A \, = \, \sum\limits_{q=0}^{N^{\,}_A-1} \, \left| q+k \, \text{mod}\, N^{\,}_A \middle\rangle \hspace{-0.4mm} \middle\langle q \right| ~~,~~~\widetilde{X}^k_B \, = \, \sum\limits_{q=0}^{N^{\,}_B-1} \, \left| q+k \, \text{mod}\, N^{\,}_B \middle\rangle \hspace{-0.4mm} \middle\langle q \right| ~,~~$$ which are both unitary operators that reduce to the Pauli $X$ matrix for $N^{\,}_{A,B}=2$, and are also known as "Weyl $X$" operators or "shift" operators in clock models.

The unitary channel that captures measurement of $A$ (and $B$) requires that we identify a "default state" of the apparatus, which we take to be $\left| 0 \right\rangle$ without loss of generality (this is the case in experiments as far as I'm aware). With this choice, the operator that captures projective measurement of the operator $A$ is the channel $$ \mathsf{M}^{\vphantom{\dagger}}_{[A]} \, = \, \sum\limits_{a=0}^{N^{\,}_A-1} \, \mathbb{P}^{\,}_a \otimes X^{a}_A \, ,~~$$ while the operator that captures measurement of $B$ is $$ \mathsf{M}^{\vphantom{\dagger}}_{[B]} \, = \, \sum\limits_{b=0}^{N^{\,}_B-1} \, \widetilde{\mathbb{P}}^{\,}_b \otimes \widetilde{X}^{b}_B \, ,~~$$ and I'll note that various useful relations can be found in this paper.

Now, suppose that we measure $A$ only. Starting in the state $\left| \alpha \right\rangle$ of the physical system (and the default state $\left|0 \right\rangle$ of the measurement apparatus), we write $$\left| \Psi (0)\right\rangle \, \equiv \, \left| \alpha \right\rangle \otimes \left| 0 \right\rangle \, ,~~$$ and we then have $$ \left| \Psi (t)\right\rangle \, = \, \mathsf{M}^{\vphantom{\dagger}}_{[A]} \, \left| \Psi (0)\right\rangle \, = \, \sum\limits_{a=0}^{N^{\,}_A-1} \, \mathbb{P}^{\,}_a \left| \alpha \right\rangle \otimes X^{a}_A \,\left| 0\right\rangle \, = \, \sum\limits_{a=0}^{N^{\,}_A-1} \, \mathbb{P}^{\,}_a \left| \alpha \right\rangle \otimes \left| a\right\rangle \, , ~$$ where the state $\mathbb{P}^{\,}_a \left| \alpha \right\rangle$ of the physical system is just an unnormalized verison of the collapsed (Copenhagen) wavefunction corresponding to observing outcome $a$. In the case where the $a$th eigenvalue $A$ is nondegenerate, we have $\mathbb{P}^{\,}_a = \left| a \middle\rangle \hspace{-0.4mm}\middle\langle a \right|$ so that $\mathbb{P}^{\,}_a \left| \alpha \right\rangle = \left\langle a \middle| \alpha \right\rangle \, \left| a \right\rangle$.

The operator $\mathsf{M}^{\,}_{[A]}$ is unitary acting on $\left| \Psi (0) \right\rangle = \left| \alpha \right\rangle \otimes \left| 0 \right\rangle$. The isometry $\mathsf{V}^{\,}_{[A]}$ corresponding to the measurement of $A$ is similar, but we do not include the ancilla state (for the apparatus) from the outset. Instead, we have: $$ \mathsf{V}^{\vphantom{\dagger}}_{[A]} \, : \, \left| \alpha \right\rangle \to \left| \alpha'\right\rangle \, = \, \sum\limits_{a=0}^{N^{\,}_A-1} \, P^{\,}_a \left| \alpha \right\rangle \otimes \left| a \right\rangle \, ,$$ where there is no apparatus state prior to the isometry. Thus, the isometry dilates the Hilbert space by introducing the state of the apparatus, and the Stinespring theorem tells us this is always a way to represent a quantum channel, including measurements. To see that this mapping is isometric (meaning that it is length preserving), we compute $$ \left\langle \alpha' \middle| \alpha' \right\rangle \, = \, \left\langle \alpha \middle| \mathsf{V}^{\dagger}_{[A]} \, \mathsf{V}^{\vphantom{\dagger}}_{[A]} \middle| \alpha \right\rangle \, = \, \sum\limits_{a,a'=0}^{N^{\,}_A-1} \, \left\langle \alpha \middle| P^{\,}_{a'} \, P^{\,}_a \middle| \alpha \right\rangle \times \left\langle a' \middle| a \right\rangle $$ $$ = \sum\limits_{a=0}^{N^{\,}_A-1} \, \left\langle \alpha \middle| P^{\,}_a \middle| \alpha \right\rangle \, = \, \left\langle \alpha\middle| \alpha \right\rangle \, = \, 1 \, , ~$$ where I used the fact that (i) the basis states for the apparatus are orthonormal (i.e., $ \left\langle a' \middle| a \right\rangle = \delta^{\,}_{a,a'}$), (ii) the projectors are idempotent (i.e., $P^2_a = P^{\,}_a$), and (iii) the projectors are complete (i.e., $\sum_a \, P^{\,}_a = \mathbb{1}$ is the identity). The unitary version $\mathsf{M}$ also preserves length, and hence the state $\left| \Psi (t) \right\rangle = \left| \alpha' \right\rangle$ is properly normalized. Note that an isometry from a Hilbert space to itself is always unitary.

Importantly, the probability of recovering outcome $a$ upon measuring $A$ is given by the expectation value (in the post-measurement state $\left| \Psi (t)\right\rangle$) of $\mathbb{1}^{\,}_{\rm phys} \otimes \left| a \middle\rangle \hspace{-0.4mm} \middle\langle a \right|^{\,}_A$, which is just the projector onto the state $a$ of the measurement apparatus! In other words, $$ p^{\vphantom{\prime}}_a \, \equiv \, \left\langle \mathbb{1}^{\,}_{\rm phys} \otimes \left| a \middle\rangle \hspace{-0.4mm} \middle\langle a \right|^{\,}_A \right\rangle^{\vphantom{\dagger}}_{\Psi(t)} \, = \, \sum\limits_{n,n'=0}^{N^{\,}_A} \, \left\langle \alpha \middle| \mathbb{P}^{\,}_n \mathbb{1} \mathbb{P}^{\,}_{n'} \middle| \alpha \right\rangle \otimes \left\langle 0 \middle| X^{-n}_A \, \mathbb{P}^{\,}_a \, X^{n'}_A \middle| 0 \right\rangle \, = \, \left\langle \alpha \middle| \mathbb{P}^{\,}_a \middle| \alpha \right\rangle \, , ~~$$ which is exactly the same as what the Copenhagen interpretation tells us! If the eigenvalue $A^{\,}_a$ is nondegenerate (i.e., there is exactly one eigenstate $\left| a \right\rangle$ with eigenvalue $A^{\,}_a$ under $A$), then we have $$p^{\,}_a \, \to \, \left| \left\langle a \middle| \alpha \right\rangle \right|^2 \, , ~$$ as expected.

Now, suppose we measure $A$ and then we measure $B$. The combination of the two channels acts as $$ \left| \Psi (t)\right\rangle \, = \, \mathsf{M}^{\vphantom{\dagger}}_{[B]} \, \mathsf{M}^{\vphantom{\dagger}}_{[A]} \, \left| \Psi (0)\right\rangle \, = \, \sum\limits_{b=0}^{N^{\,}_B-1}\, \sum\limits_{a=0}^{N^{\,}_A-1} \, \widetilde{\mathbb{P}}^{\,}_b \, \mathbb{P}^{\,}_a \, \left| \alpha \right\rangle \otimes X^{a}_A \,\left| 0\right\rangle \otimes \widetilde{X}^{b}_B \,\left| 0\right\rangle \,, ~$$ and now we can compute the joint probability to recover outcome $A^{\,}_a$ upon measuring $A$ and outcome $B^{\,}_b$ upon measuring $B$, starting from the state $\left| \alpha \right\rangle$ of the physical system: $$ p^{\vphantom{\prime}}_{b|a} \, \equiv \, \left\langle \mathbb{1}^{\,}_{\rm phys} \otimes \left| a \middle\rangle \hspace{-0.4mm} \middle\langle a \right|^{\,}_A\otimes \left| b \middle\rangle \hspace{-0.4mm} \middle\langle b \right|^{\,}_B \right\rangle^{\vphantom{\dagger}}_{\Psi(t)} \, = \, \sum\limits_{m,m'=0}^{N^{\,}_B} \, \sum\limits_{n,n'=0}^{N^{\,}_A} \, \left\langle \alpha \middle| \mathbb{P}^{\,}_{n} \, \widetilde{\mathbb{P}}^{\,}_m \,\mathbb{1} \, \widetilde{\mathbb{P}}^{\,}_{m'} \, \mathbb{P}^{\,}_{n'} \middle| \alpha \right\rangle \times \left\langle 0 \middle| X^{-n}_A \middle| a \middle\rangle \hspace{-0.4mm} \middle\langle a \middle| X^{n'}_A \middle| 0 \right\rangle \times \left\langle 0 \middle| \widetilde{X}^{-m}_B \middle| b \middle\rangle \hspace{-0.4mm} \middle\langle b \middle| \widetilde{X}^{m'}_B \middle| 0 \right\rangle \, ,~~ $$ which simplifies to $$p^{\vphantom{\prime}}_{b|a} \, = \, \left\langle \alpha \middle| \mathbb{P}^{\,}_a \, \widetilde{\mathbb{P}}^{\,}_b \, \mathbb{P}^{\,}_a \middle| \alpha \right\rangle \, \to \, \left| \left\langle b \middle| a \right\rangle \right|^2 \, \left| \left\langle a \middle| \alpha \right\rangle \right|^2 \, , ~~$$ where the latter expression holds when $A$ and $B$ are both nondegenerate. This is precisely what you wrote! If the order of measurements is reversed, simply swap $a \leftrightarrow b$ above, which gives $p^{\,}_{a|b}$!

I think it is much more straightforward to understand and derive the various relations for expectation values and joint/conditional probabilities in this Stinespring/MWI picture of measurement. It also accurately captures what happens in experiments. Let me know if anything is unclear or if additional details at any step would help!

$\endgroup$
2
  • $\begingroup$ What are isometries? $\endgroup$ Commented Jan 4, 2023 at 23:25
  • 3
    $\begingroup$ An isometry is a length-preserving map between vector spaces of different sizes, I'll update the answer to explain! $\endgroup$ Commented Jan 4, 2023 at 23:26
3
$\begingroup$

The issue is whether the probabilities are independent. I believe that this is an assumption of the notion of "projective measurements" which are approximations to what actually happens in an experiment. You really need a Hilbert space ${\mathcal H}_A\otimes {\mathcal H}_B \otimes {\mathcal H}_{\rm system}$ where $A$ and $B$ are the detectors measuring the system. You then ask what is the amplitude for the system to evolve from $|0\rangle\otimes |0\rangle\otimes |\alpha \rangle$ an end up in $|a\rangle\otimes |b\rangle\otimes |{\rm something}\rangle$. Whether the square of this is the product $|\langle a|b\rangle|^2| \langle b|\alpha\rangle|^2$ depends on the physics of the situation, but experiments are usually designed so that it works. Mott shows how the projective measuremnts are approximations to reality in hos famous paper on cloud chamber tracks.

$\endgroup$
4
  • $\begingroup$ What exactly do you mean here with independent (in your first sentence)? It might not make sense to define the joint probability (as in the classical case) if both observables do not commute. $\endgroup$ Commented Jan 4, 2023 at 21:54
  • $\begingroup$ I don't think commuting is relevent. As @Cosmas Zachos said consider light going through polarizers. You multiply the amplitudes so the probabilities mutiply as well. The projection operators commute, but the photon spin opertors which are being measured do not. $\endgroup$
    – mike stone
    Commented Jan 4, 2023 at 22:13
  • $\begingroup$ But what should independent probabilities mean here? OP is computing, as far as I can see, the probability that we measure $a$ in the state $\psi$ and then measure $b$ in the state $|a\rangle$. $\endgroup$ Commented Jan 4, 2023 at 22:32
  • 1
    $\begingroup$ He measures $|\psi \rangle$ and finds $|a\rangle$ with amplitude $\langle a |\psi\rangle$ so that $|\psi\rangle \to |a\rangle \langle a |\psi\rangle$ and then measures this state and finds $|b\rangle$ with overall amplitude $\langle b|a\rangle \langle a |\psi\rangle$. So overall $|\psi\rangle \to |b\rangle \langle b|a\rangle \langle a |\psi\rangle$. The probility of this is $|\langle b|a\rangle \langle a |\psi\rangle|^2=|\langle b|a\rangle|^2| \langle a |\psi\rangle|^2 $ $\endgroup$
    – mike stone
    Commented Jan 4, 2023 at 23:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.