I'm attempting to clarify my understanding in the example presented in Section 2.2 of 1-way Quantum Finite Automata: Strengths Weaknesses and Generalizations (this alternative link may also be useful). This example provides a very stripped down example of a 1-QFA with the following transition rules:

$V_a|q_0\rangle = \frac{1}{2}|q_0\rangle + \frac{1}{2}|q_1\rangle + \frac{1}{\sqrt 2}|q_{rej}\rangle$,

$V_a|q_1\rangle = \frac{1}{2}|q_0\rangle + \frac{1}{2}|q_1\rangle - \frac{1}{\sqrt 2}|q_{rej}\rangle$,

$V_{\$}|q_0\rangle = |q_{rej}\rangle$,

$V_{\$}|q_1\rangle = |q_{acc}\rangle$

For instance, if I'm in $q_0$ and I process an $a$ as input, I apply the first rule. My understanding is that I would have a $||\frac{1}{2}||^2 = \frac{1}{4}$ chance of staying in state $|q_0\rangle$, a $||\frac{1}{2}||^2 = \frac{1}{4}$ chance of progressing to state $|q_1\rangle$ and a $||\frac{1}{\sqrt 2}||^2 = \frac{1}{2}$ chance of ending the computation and rejecting the string.

I would imagine the automata for this to look like the following image

enter image description here

I'm not entirely sure if that is correct however. The probabilities mentioned in the paper for acceptance of the string $aa$ is $\frac{1}{4}$ while the probability of rejection is $\frac{3}{4}$. Just wondering if someone could either point out a flaw or validate what I have conceptually in mind for the example.


Reworked automata model to more accurately reflect the probabilities: enter image description here

  • 1
    $\begingroup$ I suggest you look up "quantum superposition". It looks like you are interpreting it purely probabilisticly, which ignores the possibility of interference, which is central in quantum computing. $\endgroup$
    – funkstar
    Mar 30, 2012 at 19:52
  • $\begingroup$ Well, I'm considering the state to be in a superposition upon each input. Given the input to the automata, it progresses to the next state based on the collapse of the superposition from measurement. This collapsed value obtained from the previous superposition serves as the weight for transition. For instance in step one, the probability to transition from $q_0$ to $q_{rej}$ is $||\frac{1}{\sqrt 2}||^2$. Each step of the computation induces a measurement. $\endgroup$ Mar 31, 2012 at 5:12
  • 1
    $\begingroup$ Note that the measurement is partial - it doesn't give you an exact state unless it is a final state. That way, you can do a (partial) measurement after applying a unitary transform corresponding to some symbol, and if it doesn't collapse to a final state then it will still be in a proper superposition, which leaves the possibility of interference open. $\endgroup$
    – funkstar
    Mar 31, 2012 at 7:25
  • $\begingroup$ So just to be sure I understand: 1.)-- Read first $a$: Reject string with probability $\frac{1}{2}$, otherwise state can be either in $q_0$ or $q_1$ and the superposition collapses to $\frac{1}{2}|q_0\rangle + \frac{1}{2}|q_1\rangle$. 2.)-- For the second $a$ I'm a little unsure: Since the state could be in either $|q_0\rangle$ or $|q_1\rangle$, both rules are applied to the current superposition I would assume as: $\endgroup$ Mar 31, 2012 at 19:16
  • $\begingroup$ $(\frac{1}{2}\langle q_0| + \frac{1}{2}\langle q_1|) (\frac{1}{2}|q_0\rangle + \frac{1}{2}|q_1\rangle + \frac{1}{\sqrt 2}|q_{rej}\rangle) ((\frac{1}{2}|q_0\rangle + \frac{1}{2}|q_1\rangle - \frac{1}{\sqrt 2}|q_{rej}\rangle))$ $ = (\frac{1}{4}\langle q_0 | q_0 \rangle + \frac{1}{4} \langle q_1 | q_1 \rangle) (\frac{1}{2}|q_0\rangle + \frac{1}{2}|q_1\rangle - \frac{1}{\sqrt 2}|q_{rej}\rangle)$ $ = \frac{1}{8}|q_0\rangle + \frac{1}{8}|q_1\rangle $ $\endgroup$ Mar 31, 2012 at 19:17

3 Answers 3


For sure, the automata performs a measurement after reading each symbol "a" and applying an associated unitary $V_a$. Yet, it is not really meaningful to compute the amplitudes of the state to be measured over $|q_{0}⟩$ and $|q_{1}⟩$ and to put them on the diagram, for the automata does neot measure the projectors $|q_{0}⟩\langle q_{0}|$, and $|q_{1}⟩\langle q_{1}|$. In other words, this numbers do not represent probabilities, they do not correspond to outcomes of the measurement to be performed. Therefore, labeling the arrows of the diagram with this numbers would give a potentially-misleading illustration of how the measurement process is working.

Although I might say things you already know, let us elaborate on the topic a bit to clarify the meaning of the transition diagrams you depicted. It is important to highlight that after reading a symbol $a$ and applying $V_a$, the automata does not perform a measurement in the standard computational basis: $$\{|q_0⟩,|q_1⟩,|q_{acc}⟩,|q_{rej}⟩\}$$ Instead, the automata measures this complete set of orthogonal projectors:

  • $P_{acc}=|q_{acc}⟩\langle q_{acc}|$
  • $P_{rej}=|q_{rej}⟩\langle q_{rej}|$
  • $P_{non}= |q_{0}⟩\langle q_{0}|+|q_{1}⟩\langle q_{1}|$

In other words, the measurement has three possible outcomes: (acc) the automata measures an accepting state and halts; (rej) the automata measures a rejecting state and halts; (non) the automate measures something else, does not halt, and reads next symbol (non states for non-halting).

Now, this is a problem I see in your diagrams: if you had a state $(|q_{0}⟩+|q_{1}⟩)/2$ before some measurement, and you happened to obtain the outcome (non), the state would remain invariant after the measurement (just apply $P_{non}$ and check). Therefore, depicting the probability of transition to either $|q_{0}⟩$ or $|q_{1}⟩$ creates confusion.

Taking all said into account it is easy to follow the calculation given in your main reference. In order to illustrate everything said, and, for completeness, I will quote it with here with some minor comments (though I added some modifications, I do not know whether this type of quoting is acceptable; if its is not, please, let me know or edit the answer yourself):

The automaton starts in $|q_0\rangle$.

  1. Read "a". $V_a$ is applied, giving $\frac{1}{2} |q_0\rangle+\frac{1}{2} |q_1\rangle+ \frac{1}{\sqrt{2}} |q_{rej}\rangle$. This is observed. Two outcomes are possible. With probability $(1/\sqrt{2})^2=1/2$, a rejecting state is observed. Then, the superposition collapses to $|q_{rej}\rangle$, the word is rejected and the computation terminates. Otherwise (with probability $1/2$), a non-halting state is observed and the superposition collapses to $\frac{1}{2} |q_0\rangle+\frac{1}{2} |q_1\rangle$. In this case, the computation continues.

  2. Read "a" again. A simple computation (we leave the details out) shows that $\frac{1}{2} |q_0\rangle+\frac{1}{2} |q_1\rangle$ is mapped to itself by $V_a$. After that, a non-halting state is observed. (There are no accepting or rejecting states in this superposition.)

  3. Read final symbol \$. The transformation $V_{\$}$ corresponding to the right endmarker $\$$ is done. It maps the superposition to $\frac{1}{2} |q_{rej}\rangle+\frac{1}{2} |q_{acc}\rangle$. This is observed. With probability $(1/2)^2=1/4$, the rejecting state $q_{rej}$ is observed. With probability $1/4$, the accepting state $q_{acc}$ is observed.

The total probability of accepting is $1/4$, the probability of rejecting is $1/2+1/4=3/4$.

  • 1
    $\begingroup$ Ah terrific answer, thank you so much for taking the time to write it out in more explicit terms. Also, on a side note, is there any obvious way or perhaps previous reference that depicts a visual representation in a more accurate and non-misleading way? It seems that perhaps laying out the evolution in some type of a tree-like structure would allow a better illustration of the branching that occurs throughout the execution of the automata. Thanks again for your help. $\endgroup$ Apr 7, 2012 at 21:54
  • 1
    $\begingroup$ @VincentRusso: you can't really describe it in a tree-based manner. The very point is that there can be destructive interference between amplitudes on the various finite automaton 'states'; this being the principal difference between stochastic and quantum computation. The graphical depiction of the automaton isn't really misleading if you take quite seriously that it is describing amplitudes for vectors, rather than probabilistic transitions. Of course, for both quantum or stochastic automata, the model is actually about linear transformations, so the picture is mostly beside the point. $\endgroup$ Apr 8, 2012 at 14:41
  • $\begingroup$ At the beginning of chapter 6 of Kaye, Laflamme, Mosca's there is a nice discussion about the differences between classical-probabilistic and quantum computation; the authors illustrate the text with state-diagrams. They actually discuss that these diagrams are not entirely adequate to describe quantum-interference --as Niel de Beaudrap has pointed out and nicely explained--. I personally recommend this reference for further reading. $\endgroup$ Apr 9, 2012 at 18:47

Juan Bermejo Vega has given an accurate summary of what is said in the original paper. I will give you a higher-level description.

I will recommend, in your case, to avoid thinking of the amplitudes as probabilities altogether. They are related to probabilities, but this is not especially helpful to think about for these finite automata. I suspect that things will be much clearer if you think of this as a slightly abstract recipe for transforming complex-valued vectors instead.

What vectors are we transforming? Well: suppose you have a finite automaton with n states. The automaton represents a (yes, probabilistic) model for transforming vectors, which at any time step may give rise to an accepting or rejecting decision.

  • Define n axes (coefficients of the vectors being transformed). Label each axis with one of the labels of the states. In particular, there is an accept axis and a reject axis. Call this vector space $\mathcal H$. The vectors we will be transforming are unit vectors in $\mathcal H$, and the only probabilities we will ever consider have to do with the coefficients of the vectors along the accept or reject axes.
  • We have a collection of transformations $V_q : \mathcal H \to \mathcal H$ for each state q, such that $V_q^\dagger V_q = I$ (i.e. they are unitary). We also have an initial transformation $V_c$ (neglected in the example you describe; we have $V_c = I$ in that case) and a final transformation $V_\$$ which are also unitary.
  • We start with a vector $V_c\,\mathbf{\hat e}_1$, where $V_c$ is the initial transformation and $\mathbf{\hat e}_1$ is the unit vector in the direction of the axis of the first state ($q_0$ in your example). We then perform the following transformations for each letter:

    1. If the next letter to be read is $a$, we perform the transformation $V_a$ on the vector.
    2. The resulting vector $\mathbf v$ might have coefficients $u_A$ along the accept axis, and $u_R$ along the reject axis. If so, we immediately accept with probability $|u_A|^2$ and reject with probability $|u_R|^2$.
    3. If some of the other coefficients are non-zero, there is a probability that the computation continues; if it does so, we multiply the vector by a factor of $(1 - |u_A|^2 - |u_R|^2)^{-1/2}$, and set the accept and reject coefficients to zero.
    4. Proceed for the next letter.
  • We perform these transformations for each letter of the word, and also for a terminating symbol \$ which we append to the end.

The only time probabilities come into play is with the accepting and rejecting axis. While this model of computation is obviously inspired by finite automata, it is simple not useful to interpret any of the other coefficients of the vector, at any point in time (even at the end!), as being probabilities.

Note that the accept and reject probabilities at each timestep are conditional probabilities, i.e. they depend on the computation not having yet halted; if you want to compute the total probability of acceptance/rejection, you can most easily do this by dispensing with the "renormalization" step in the evolution (the part where we multiply by the inverse-square-root value; but still setting the accept/reject coefficients to zero if the computation continues), and simply sum all of the probabilistic contributions to accepting/rejecting that arise along the way.

  • 1
    $\begingroup$ Niel, thank you again for your comment as well. It was very helpful to get a more intuitive picture of how this computation takes place. Thank you very much for taking the time to explain, it's much clearer to me now. $\endgroup$ Apr 8, 2012 at 18:26

You are incorrectly assuming that the state collapses fully (to a computational basis state) after reading each symbol. The measurement at each stage is only partial.

  • $\begingroup$ I was under the impression that since this is utilizing the "measure many" model that a measurement is performed on each step of the computation, which collapses the superposition to a definite probabilistic value. $\endgroup$ Mar 31, 2012 at 4:52
  • $\begingroup$ @Vincent Russo You are right that the state is measured for each symbol that is read. $\endgroup$
    – funkstar
    Mar 31, 2012 at 6:10
  • $\begingroup$ I think that a QFA where the state of the automaton is measured after each step would be completely equivalent to a Markov Chain. So what would be the motivation to study them? $\endgroup$
    – Shitikanth
    Apr 2, 2012 at 9:05
  • 3
    $\begingroup$ I only wanted to show why his "reworked automaton model" does not accurately describe the QFA system. This is, in fact, because the automaton does not fully collapse in the computational basis at each step. $\endgroup$
    – Shitikanth
    Apr 4, 2012 at 14:58
  • 2
    $\begingroup$ In any case, I think that the question has already been answered sufficiently clearly. Let's not get lost in technicalities. $\endgroup$
    – Shitikanth
    Apr 4, 2012 at 15:01

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.