2
$\begingroup$

I'm confused about the state of a system after a measurement. Say we have a particle $v$ in the state: $ |\psi\rangle= \sqrt{1/4} \ |0\rangle + \sqrt{3/4} \ |1\rangle $.

From my understanding, if one were to measure the state of $v$, one would get the result $|0\rangle$ with probability $|\sqrt{1/4}|^2=1/4$, and similarly, $|1\rangle$ with probability $3/4$.

However, I've also learned that a measurement is always done by an observable (a unitary operator), e.g. $Z=|0\rangle \langle 0|-|1\rangle \langle 1|$, and that the outcome of the measurement is an eigenvalue of this operator, and that the state we get after the measurement is always dependent on the observable we use, and similarly for the probability of getting that state.

Now, by inspection, I noticed that when I measure $Z$, I do get the state $|0\rangle$ with probability $1/4$, and $|1\rangle$ with probability $3/4$, as expected. But I don't get these results when I measure the Pauli operator $X$, for example.

Does that mean that the claim in my second paragraph always assumes a measurement of $Z$?

$\endgroup$
  • 4
    $\begingroup$ Observables are Hermitian operators, not (necessarily) unitary ones. $\endgroup$ – tparker May 22 at 12:16
2
$\begingroup$

$X$ basis measurement means that after the measurement we will have either $|+\rangle$ or $|-\rangle$ state (the eigenbasis vectors for the $X$ operator). To see the corresponding probabilities we should rewrite the $|\psi\rangle$ state in $X$ basis:

$$|\psi\rangle = \frac{1}{\sqrt{4}}|0\rangle + \frac{\sqrt{3}}{\sqrt{4}}|1\rangle = \frac{1}{2}\frac{1}{\sqrt{2}}(|+\rangle + |-\rangle) + \frac{\sqrt{3}}{2}\frac{1}{\sqrt{2}}(|+\rangle - |-\rangle) = \\ = \left(\frac{1 + \sqrt{3}}{\sqrt{2^3}}\right)|+\rangle + \left(\frac{1 - \sqrt{3}}{\sqrt{2^3}}\right)|-\rangle$$

where $|+\rangle = \frac{1}{\sqrt{2}}(|0\rangle + |1\rangle)$ and $|-\rangle = \frac{1}{\sqrt{2}}(|0\rangle - |1\rangle)$. The probability of measuring $|+\rangle$ is equal to $\left| \frac{1 + \sqrt{3}}{\sqrt{2^3}}\right|^2$ and the probability of measuring $|-\rangle$ is equal to $\left| \frac{1 - \sqrt{3}}{\sqrt{2^3}}\right|^2$. In general case:

$$|\psi \rangle = \alpha |0\rangle + \beta |1\rangle = \frac{\alpha + \beta}{\sqrt{2}} |+\rangle + \frac{\alpha - \beta}{\sqrt{2}} |-\rangle$$

The statement in the second paragraph of the question assumes that the measurement is done in $Z$ basis because $|0\rangle$ and $|1\rangle$ are eigenbasis vectors of $Z$ operator. If we will do a measurement in $X$ basis we will never have $|0\rangle$ or $|1\rangle$ state after the measurement.

| improve this answer | |
$\endgroup$
2
$\begingroup$

Yes, the claim in your second paragraph implicitly assumes a measurement of $Z$.

The states $|0\rangle$ and $|1\rangle$ are not well-defined unless you specify which basis you're referring to (e.g. the $Z$-basis). To make this completely clear, it's better to write out the states as $|Z = 0\rangle$and $|Z = 1\rangle$ whenever there's any ambiguity as to the basis.

| improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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