I understand that an measurement operator always returns its eigenvalues, with a probability. I also understand that it is a postulate of QM, so why can't prove it Why is the measured value of some observable $A$, always an eigenvalue of the corresponding operator?

However, I would like to see that an operator in a random state returning its eigenvalue in action. So I turned to the Hadamard gate, do an spectral decomposition, hoping that by doing a projection measurement to the eigenbases, I can somehow get its eigenvalues back. But I cannot seem to do it. Below is how I worked

We know that $H^\dagger H=1 $, so it satisfies the completeness requirement

Eigenvectors and value:

  • $[1+\sqrt2, 1], \lambda=1$
  • $[1-\sqrt2, 1], \lambda=-1$

Spectral decomposition: $$H=\begin{bmatrix} \frac{1+\sqrt{2}}{\sqrt{4+2\sqrt{2}}} & \frac{1-\sqrt{2}}{\sqrt{4-2\sqrt{2}}}\\\\ \frac{1}{\sqrt{4+2\sqrt{2}}} & \frac{1}{\sqrt{4-2\sqrt{2}}} \end{bmatrix} \begin{bmatrix} 1 & 0\\\\ 0 & -1 \end{bmatrix} \begin{bmatrix} \frac{1+\sqrt{2}}{\sqrt{4+2\sqrt{2}}} & \frac{1-\sqrt{2}}{\sqrt{4-2\sqrt{2}}}\\\\ \frac{1}{\sqrt{4+2\sqrt{2}}} & \frac{1}{\sqrt{4-2\sqrt{2}}} \end{bmatrix} ^T$$

Now when measuring state $[\alpha, \beta]$ there are two possible outcomes, 1 and -1

$p(1)=\begin{bmatrix} \alpha& \beta \end{bmatrix}\begin{bmatrix} \frac{1+\sqrt{2}}{\sqrt{4+2\sqrt{2}}}\\\\ \frac{1}{\sqrt{4+2\sqrt{2}}} \end{bmatrix}\begin{bmatrix} \frac{1+\sqrt{2}}{\sqrt{4+2\sqrt{2}}}& \frac{1}{\sqrt{4+2\sqrt{2}}} \end{bmatrix} \begin{bmatrix} \alpha\\\\ \beta \end{bmatrix}$

And now I don't know what to do next? All I can get is a probability, but not the eigenvector

  • 1
    $\begingroup$ You're almost there! The row vector and the final column vector multiply together to give a scalar. The first column vector stays the same - that is the new state (up to normalization) - and it is the eigenstate associated with eigenvalue 1. $\endgroup$ Jan 8 at 1:39
  • $\begingroup$ I see. To me it is not clear what "outcome of the measurement" is. Is it the probability that $\phi$ magically takes the eigenvalues (clarify that magic is the crux of my question), or the state after the measurement. $\endgroup$
    – Minh Triet
    Jan 8 at 21:13
  • $\begingroup$ I think that "clarifying that magic" is actually one of the biggest open problems in theoretical physics :p $\endgroup$
    – Plop
    Jan 9 at 13:36

2 Answers 2


I also understand that it is a postulate of QM, so why can't prove it?

You appear to be confused between the philosophical distinction between a postulate and a proposition. A postulate is asserted without proof simply because there must be certain things that we must begin from. Descartes called them clear and distinct ideas. What he means by this is that they must be as clear as possible.

A proposition is a question about the field written in the language peculiar to that field. It is proven when it can be derived from the postulates.

We cannot prove the postulates of QM because tautologically they are asserted without proof. What we can demonstraye is that they are useful postulates yo take by showing that the consequences derived from them hold experimentally. One might say this is proof.

More formally, we can derive the postulates if we have a deeper theory. At present, such a thoery is out of reach.


The theory of quantum mechanics, at its minimum, does not describe the process of measurement. It only says that before a measure, the system is in a state described by a vector in Hilbert space, that the measure itself is modelled by an orthonormal basis in this Hilbert space together a list of numbers (one number for each vector in the basis, so you can think of the whole as a hermitian operator, but you don't have to), and that after the measure has been performed, the system is in a state described by some vector of the basis, and the result of the measure is the corresponding number.

Quantum mechanics does not tell you anything but probabilities. So, when you say

I don't know what to do next? All I can get is a probability, but not the eigenvector

then you got everything right. Quantum mechanics doesn't tell you anything more than that.

There are some theories that intend to describe what happens in the measurement process, but this is something else (and I am not competent to tell you much about it).

  • $\begingroup$ By calculation, I managed to got (1) the state$\rvert A \rangle$ after measurement, which is naturally in the eigenspace of the measurement operator, and (2) a probability that the system end up in that state. Are you saying that just because this state $\rvert A \rangle$ correspond to a eigenvalue $\lambda_A$, I should accept that the result of the measurement is $\lambda_A$? $\endgroup$
    – Minh Triet
    Jan 9 at 17:14
  • $\begingroup$ I didn’t really look at your computations. Unless the starting state is already an eigenstate, at least two probabilities for distinct eigenvalues will be nonzero, so you can’t tell what the state of the vector will be after the measurement. $\endgroup$
    – Plop
    Jan 10 at 0:46

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