Measurement of a superposition and Born's rule

[mbond]

Let be a superposition $|\psi\rangle=\sum_j a_j|j\rangle$ with one amplitude $a_x$ much greater than the others, where $x$ is not known. For example, $|\psi\rangle$ may result from the quantum Fourier transform of a periodic wave function with an unknown period. I expect a measurement of $|\psi\rangle$ to project it on $|x\rangle$ because the probability $|a_x|^2$ is very high.

But how do I do that? I would like to use Born's rule, but what projector would project on $|x\rangle$?

I appreciate your help.

 

[vanhees71]

Given the system being prepared in the state represented by $|\psi \rangle$ the probability to find the value $x$ when measuring the observable related with the operator
$$\hat{J}=\sum_j j |j \rangle \langle j|,$$
is
$$P(x)=|\langle x|\psi \rangle|^2=|a_x|^2.$$
The projector to the corresponding eigenstate is
$$\hat{P}_x=|x \rangle \langle x|.$$
I hope I understood the question right. It's a bit vague ;-).

 

[mbond]

The point is that I don't know $x$...
I am looking for a projector that would return $x$ because it has the highest amplitude $|a_x|^2$.

 

[vanhees71]

Then I don't understand the question.

 

[Demystifier]

I am looking for a projector that would return $x$ because it has the highest amplitude $|a_x|^2$.

There is no such projector. The only eigenvalues of any projector are 0 and 1, so a projector cannot return the number $x$.

But you can measure the observable $J$ defined by @vanhees71. If you do that, it is almost certain that the result of measurement will be $x$.

 

[vanhees71]

A projector never returns a number but another state:
$$\hat{P}_x |\psi \rangle =|x \rangle \langle x|\psi \rangle.$$

 

[vanhees71]

I think, given the confusion of the OP, one should be concise:

An operator is applied to a Hilbert-space vector, giving another vector, not a number. An operator has eigenvalues (numbers) and eigenvectors. A projection operator $\hat{P}$ has eigenvalues 0 and 1, because it obeys $\hat{P}^2=\hat{P}$. From this it also follows that there's one eigenvector (moduloa an arbitrary phase factor) with eigenvalue 1, i.e., $\hat{P}=|\psi \rangle \langle \psi|$ with some normalized $|\psi \rangle$ (which is the one eigenvector with eigenvalue 1).

 

[mbond]

Let me rephrase the question another way:
I have a wave function resulting from a quantum Fourier transform. This wave function is a superposition with one of the amplitudes very high (because it corresponds to the period). How do I recover the period?

 

[Morbert]

Let be a superposition $|\psi\rangle=\sum_j a_j|j\rangle$ with one amplitude $a_x$ much greater than the others, where $x$ is not known. For example, $|\psi\rangle$ may result from the quantum Fourier transform of a periodic wave function with an unknown period. I expect a measurement of $|\psi\rangle$ to project it on $|x\rangle$ because the probability $|a_x|^2$ is very high.

But how do I do that? I would like to use Born's rule, but what projector would project on $|x\rangle$?

I appreciate your help.

Measure $\sum_j a_j|j\rangle\langle j|$ a large number of times and see which outcome occurs the most. That is likely your $a_x$.

Or if you want to keep it theoretical, $p(a_j) = |\langle j|\psi\rangle|^2$ so compute for each $j$ and see which $p(a_j)$ is largest.

 

[mbond]

A wave function can't be measured more than once; it's collapsed after the 1st measurement.

 

[Demystifier]

A wave function can't be measured more than once; it's collapsed after the 1st measurement.

In general, a single measurement cannot determine the wave function, unless you you have some knowledge about the wave function before measurement.

 

[Demystifier]

Let me rephrase the question another way:
I have a wave function resulting from a quantum Fourier transform. This wave function is a superposition with one of the amplitudes very high (because it corresponds to the period). How do I recover the period?

This period refers to periodicity either in time or in space. If it is in time, then your wave function is approximately an energy eigenstate. If it is in space, then your wave function is approximately a momentum eigenstate. Hence you need to measure either the energy or the momentum, the value that you obtain is probably the value that you need. There is a probability that you will get a wrong value, but this probability is small. Since your wave function is not an exact eigenstate, there is no way to get the needed value with 100% certainty.

 

[mbond]

In quantum computing, there are algorithms such as the quantum Fourier transform or Grover's algorithm that ends up with a superposition $\sum_j a_j|j\rangle$ with one amplitude $a_x\sim 1$ and the other amplitudes $\sim 0$. I am looking for a way to recover the $x$...

 

[DrClaude]

Slightly off-topic suggestion: when discussing QM, stop using $x$ for anything else than position.

 

[martinbn]

It is clear by the answers that your question is not clear. You need to clarify it. You probably have been thinking about it for a while, and it is all obvious to you, but no one here has. So you need to write down all the details of what is given and what you are looking for. It seems, but that is just a guess, that you a looking for something along the lines of Simon's algorithm, or Shor's.

 

[Nugatory]

In quantum computing, there are algorithms such as the quantum Fourier transform or Grover's algorithm that ends up with a superposition $\sum_j a_j|j\rangle$ with one amplitude $a_x\sim 1$ and the other amplitudes $\sim 0$. I am looking for a way to recover the $x$...

.
That does not exist. The best we can do, and what is done in quantum computing, is to measure $x$. It is probable, and approaches certainly as one of the $a_i$ approaches unity and the others approach zero, that the result will be the eigenvalue corresponding to that $a_i$.

 

[Morbert]

A wave function can't be measured more than once; it's collapsed after the 1st measurement.

Sorry, I mean measure the observable $\sum_j \lambda_j |j\rangle\langle j|$ a large number of times, each on a different member of your ensemble. If you know the preparation has the form $\sum_{j} a_j|j\rangle$ with one amplitude approximately 1 and the others 0, then measuring $\sum_j \lambda_j |j\rangle\langle j|$ a large number of times will identify it.

 

[mbond]

OK, I have $|\psi\rangle=\displaystyle\sum_{j=0}^{2^n-1}a_j|j\rangle$ with a peak at $j=r$, i.e., $|a_r|^2\approx 1$ and $|a_{j\neq r}|^2\approx 0$. I take the observable $\hat{O}=\displaystyle\sum_{k=0}^{2^n-1}k|k\rangle\langle k|$ and I make the measurement

$\langle\psi|\hat{O}|\psi\rangle=r|a_r|^2+\displaystyle\sum_{j\neq r}j|a_j|^2\approx r$

and the period I am looking for in the quantum Fourier transform $|\psi\rangle$ is $P\approx\displaystyle \frac{2^n}{r}$.

Sorry for not being clear, and many thanks for the help.

 

[vanhees71]

Sorry, I mean measure the observable $\sum_j \lambda_j |j\rangle\langle j|$ a large number of times, each on a different member of your ensemble. If you know the preparation has the form $\sum_{j} a_j|j\rangle$ with one amplitude approximately 1 and the others 0, then measuring $\sum_j \lambda_j |j\rangle\langle j|$ a large number of times will identify it.

You cannot figure out the state by just measuring the observable. All you get are the probabilities, i.e., the $|a_j|^2$. You get no information about the phases, i.e., the complex numbers $a_j$.

There's a thorough discussion about, how to empirically find out the quantum state ("state tomography") in Ballentine, Quantum Mechanics.

 

[Demystifier]

I make the measurement
$\langle\psi|\hat{O}|\psi\rangle$

You cannot get the average value with a single measurement.

 

[Demystifier]

Slightly off-topic suggestion: when discussing QM, stop using $x$ for anything else than position.

Actually, in the quantum computing community it is quite common to use $x$ for other observables or states.

 

[Demystifier]

In quantum computing, there are algorithms such as the quantum Fourier transform or Grover's algorithm

But you said that you want to obtain the result with a single measurement. A quantum algorithm is not a single measurement. Moreover, most quantum algorithms still cannot be realized with actually existing quantum computer hardware.

 

[Morbert]

You cannot figure out the state by just measuring the observable. All you get are the probabilities, i.e., the $|a_j|^2$. You get no information about the phases, i.e., the complex numbers $a_j$.

There's a thorough discussion about, how to empirically find out the quantum state ("state tomography") in Ballentine, Quantum Mechanics.

I interpreted his question to be asking to identify which $j$ had the largest amplitude: " If you know the preparation has the form $\sum_j a_j|j\rangle$with one amplitude approximately 1 and the others 0"