# In a state $\alpha \vert HL \rangle + \beta \vert VR \rangle$ can a particle be measured in the $\{|L\rangle,|R\rangle\}$ basis?

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

where:

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?

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.
• @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 ? Jul 29, 2022 at 9:28
• 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? Jul 29, 2022 at 9:31
• @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? Jul 29, 2022 at 12:43