# Finding amplitude of probability

The mathematical structure of quantum mechanics, follows almost inevitably from the concept of a probability amplitude. For James Binney and David Skinner:

"With every value in the spectrum of a given measurement there will be a quantum amplitude that we will find this value if we make the relevant measurement. Quantum mechanics is the science of how to calculate such amplitudes given the results of a sufficient number of prior measurements."

I would like to know how the expression of probability amplitude is determinated!

Would quantum amplitudes be determined from the statistic experiment from the system to study? We have to repeat the same experiment (system to study is prepared in the same state) to evaluate the probabilities to obtain the measurements outcomes and then deduce, from this probabilities, the amplitudes of probabilities ?

One does not measure probability amplitudes (they are complex). One measures relative probabilities of possible outcomes. These outcomes (they are eigenvalues $\lambda_k$ of the operators you measure) are tied to eigenvectors $\vert\psi_k\rangle$ of the operators. The probability of getting $\lambda_k$ is the mod square of the amplitude $\langle\psi_k\vert \phi\rangle$ with $\vert\phi\rangle$ the initial state. One then compares the theoretical probability $\vert\langle\psi_k\vert\phi\rangle\vert^2$ with the measured probability.

Thus, if $$\vert\phi\rangle=\alpha \vert \psi_1\rangle +\beta\vert\psi_2\rangle \tag{1}$$ with energies $E_1$ and $E_2$ for states $\vert \psi_1\rangle$ and $\vert \psi_2\rangle$ respectively, then the prob. of getting energy $E_1$ is $\vert\alpha\vert^2$ and the prob. of getting $E_2$ is $\vert\beta\vert^2$, with $$\vert\alpha\vert^2 +\vert\beta\vert^2=1\, .$$ $N$ repeated measurements of energy of a set of identically prepared states $\vert\phi\rangle$ should yield $E_1$ approximately $N \vert\alpha\vert^2$ times and $E_2$ approximately $N\vert\beta\vert^2$ times.

In the limit of very large $N$, this should be accurate. Of course the situation can be reversed, i.e. given some frequencies of various outcomes one could determine the various coefficients in the expansion of $\vert \phi\rangle$. This does not quite determine $\vert\phi\rangle$ as the phase of the coefficients cannot be determined, i.e. the set of states \begin{align} \vert\phi(\zeta,\eta)\rangle=e^{i\zeta}\alpha \vert \psi_1\rangle +e^{i\eta}\beta\vert\psi_2\rangle \end{align} for arbitrary $\zeta,\eta$ (they are in general distinct states as they differ in general by a relative phase) will produce the same probabilities for $E_1$ and $E_2$.

As an example, if measuring the spin projection along $\hat z$ yields $+$ half the time, and $-$ half the time, then one cannot distinguish between the states $\frac{1}{\sqrt{2}}\left(\vert +\rangle -\vert -\rangle\right)$ and $\frac{1}{\sqrt{2}}\left(\vert +\rangle +i \vert -\rangle\right)$.

It is sometimes possible to completely reconstruct $\vert\phi\rangle$ using a procedure called quantum tomography. In general, it requires more than a single operator.

• Okay, my turn to ding you. Probability densities are not complex—they have to be real. And, yes, you can measure probability densities. – njspeer Apr 15 '18 at 15:34
• @njspeer good catch. I fixed it. – ZeroTheHero Apr 15 '18 at 15:35
• Thanks for this clear answer. I know that we does not measure probability amplitude. Thus to find the probability amplitude we reverse the situation i.e. given some frequencies of various outcomes one could determine the various coefficients in the expansion of |ϕ⟩ ? in situation where we know that the probabilities are equiprobable we don't need to do the experiment just reverse the situation to find the mathematical expression of probabilities amplitudes (alpha and beta) ? – Paradigm Apr 15 '18 at 16:38
• @Paradigm yes but this will still not fix the relative phases between the coefficients, since the probabilities are insensitive to those phases. In quantum tomography, one chooses a number of distinct types of measurements and one can then reconstruction completely $\vert\phi\rangle$, including the phase of the coefficients. – ZeroTheHero Apr 15 '18 at 16:47

Would quantum amplitudes be determined from the statistic experiment from the system to study? We have to repeat the same experiment (system to study is prepared in the same state) to evaluate the probabilities to obtain the measurements outcomes and then deduce, from this probabilities, the amplitudes of probabilities?

Yes. Experimentally, the probability density function (PDF) would need to be determined by repeated measurement.

Theoretically, the PDF is the spectrum of the square of the absolute values of the eigenstate of the operator associated with the measurement in question.

• My question is about probabilty amplitude which is use to calculate PDF. Are the amplitudes of probabilities determined experimentally ? – Paradigm Apr 15 '18 at 14:25
• @Paradigm Yes, as I said, the amplitudes can be determined or varified experimentally. Unnormalized, the amplitudes of the PDF would be the number of times you measured the system in that particular state. Just think of a poll where you are trying to determine the outcome of an election. If you have 5 candidates and you poll 100 people, you might get [5,13,28,24,30]. This would be your unnormalized PDF. Normally we want normalized PDFs, so we divide each element by the sum. This yields [0.05,0.13,0.28,0.24,0.3], which are the probabilities of each outcome for a single vote (measurement). – njspeer Apr 15 '18 at 15:09
• Not is it not. For the harmonic oscillator state $\frac{1}{\sqrt{2}}\vert 0\rangle+\frac{1}{\sqrt{2}}\vert 7\rangle$, the prob. of getting $n=0$ is the same as getting $n=7$. This has nothing to do with the eigenvalues, which are $0$ and $7$ (or $\hbar\omega/2$ and $7\hbar\omega/2$ if you measure energy). – ZeroTheHero Apr 15 '18 at 15:17
• @ZeroTheHero correct. You can only construct the PDF experimentally. – njspeer Apr 15 '18 at 15:31
• I cancelled my downvote; your final edited answer is much much better than the original. – ZeroTheHero Apr 15 '18 at 15:45