Given a pair of entangled particles: $\alpha \vert HL \rangle + \beta \vert VR \rangle$


  1. $\alpha^2 + \beta^2=1$
  2. $H$ and $V$ are orthonormal vectors.
  3. $L$ and $R$ are orthogonal vectors (but not necessarily orthonormal).

Is it possible to measure one of the particles in $L$ and $R$ basis using a polarizer (without any normalization etc.)? Or does quantum mechanics forbids it? And if yes, does the probability of it being $L$ or $R$ given by $\alpha^2$ and $\beta^2$ respectively?


1 Answer 1


The overall state must have unit norm, i.e. for $|\psi\rangle=\alpha |HL\rangle+\beta|VR\rangle$ we should have $\langle\psi|\psi\rangle=1$. If I understood the question, by your assumption $\langle H |V\rangle=0$, and $\langle L |R\rangle=0$, hence $$\langle\psi|\psi\rangle=|\alpha|^2\langle H|H\rangle \langle L|L\rangle +|\beta|^2\langle V|V\rangle \langle R|R\rangle=1$$

Assuming further that $|H\rangle$ and $|V\rangle$ have unit norm and $|\beta|^2=1-|\alpha|^2$ this reduces to $$|\alpha|^2\langle L|L\rangle +(1-|\alpha|^2)\langle R|R\rangle=1$$ Although for states $|L\rangle$ and $|R\rangle$ of unit norm this constraint is satisfied, they do not need to have unit norm, but only subject to satisfy the constraint.

With all that said, defining normalized versions of $|R\rangle$, $|L\rangle$ and adjusting the coefficients $\alpha,\beta$ correspondingly would probably make the analysis much simpler.


If the vectors $|L\rangle, |R\rangle$ are not normilized, then coefficients $|\alpha|^2, |\beta|^2$ in decomposition $|\psi\rangle=\alpha |L\rangle+\beta |L\rangle$ do not correspond to measurement probabilities. Consider a simple example $|L\rangle = \frac1{\sqrt{2}}|0\rangle, |R\rangle = \sqrt{\frac{3}{2}}|1\rangle$ and $$|\psi\rangle=\frac{1}{\sqrt{2}}|L\rangle+\frac{1}{\sqrt{2}}|R\rangle=\frac12|0\rangle+\frac{\sqrt{3}}{2}|1\rangle$$ The actual measurement probabilities are $\frac14,\frac34$. It is not clear to me though why go into the trouble of working with non-normilized states at all in this context.

  • $\begingroup$ Hi, great analysis Nikita. Could you extend your reply to address the OP's question about measurement when states are not normalized and how this relates to completeness relation of projective measurements? $\endgroup$
    – MonteNero
    Jul 28, 2022 at 22:01
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    $\begingroup$ @MonteNero done. $\endgroup$ Jul 29, 2022 at 8:16
  • $\begingroup$ @NikitaNemkov Thank you for the answer. understood quiet a bit. But I am still unclear about the central Q. As you mentioned in (addition) the $|L\rangle = \frac1{\sqrt{2}}|0\rangle, |R\rangle = \sqrt{\frac{3}{2}}|1\rangle$. Thus, they are not normalized. But the probability we calculated is of $|0\rangle, |1\rangle$ by inserting the values and effectively normalizing. Is the calculation of probability and measurement only possible for $|0\rangle, |1\rangle$ but not for $|L\rangle, |R\rangle$. That was the original query and I am still unclear about it ? $\endgroup$ Jul 29, 2022 at 9:28
  • $\begingroup$ I mean $|L\rangle, |R\rangle$ are vectors too even if not normalized. How can we measure their probability if at all possible and not $|0\rangle, |1\rangle$'s? $\endgroup$ Jul 29, 2022 at 9:31
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    $\begingroup$ @TheoryQuest1 For your original state $|\psi\rangle=\alpha |HL\rangle+\beta |VR\rangle$ the probability to find the first particle in state $|H\rangle$ is not equal to $|\alpha|^2$, but to $|\alpha|^2 \langle L|L\rangle$. If your first particle is measured in state $|H\rangle$ the second is in (normalized) $|L\rangle$ with certainty. Unnormalized vectors $|L\rangle, |R\rangle$ may be useful to keep in intermediate computations, but they are not valid states. And if you only work with normalized states, there is no ambiguity, right? $\endgroup$ Jul 29, 2022 at 12:43

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