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In statevector formalism suppose two particle $|\psi\rangle =c_1|A\rangle +c_2|B\rangle $ where $|A\rangle =a_1|u_1\rangle +a_2|u_2\rangle , |B\rangle =b_1|u_1\rangle +b_2|u_1\rangle $, but $|\psi\rangle =c_1|A\rangle +c_2|B\neq (c_1a_1+c_2b_1)|u_1\rangle +(c_1a_2+c_2b_2)|u_2\rangle $.

I'm kind of confused by this expression, because if they are orthonormal states($|u_1\rangle$ ,$|u_2\rangle$ ) , why can't we use superposition principle and add them together? Why can't we expand this expressoin($|\psi\rangle $) into $|u_1\rangle $ and $|u_2\rangle $ by the linearlity of $|u_1\rangle $ and $|u_2\rangle $ ?

Notice: Suppose numbers are all real for simplicity $(c_1a_1+c_2b_1)^2+(c_1a_2+c_2b_2)^2=(c_1a_1)^2+(c_2b_1)^2+2c_1a_1c_2b_1+(c_1a_2)^2+(c_2b_2)^2+2c_1a_2c_2b_2=c_1^2+c_2^2+2c_1a_1c_2b_1+2c_1a_2c_2b_2=1+2c_1a_1c_2b_1+2c_1a_2c_2b_2>1$.

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  • $\begingroup$ suppose two particle? I'm confused, did you try to describe 2 particles? $\endgroup$ Commented Nov 2, 2018 at 17:27
  • $\begingroup$ @OfekGillon Yes. $|\psi>$ is a suposition of particle $|A>$ and $|B>$ normalized by $c_1,c_2$. Particle $|A>$ is normalized with respect to state vector $|u_1>,|u_2>$ by $a_1,a_2$. Particle $|B>$ is normalized with respect to state vector $|u_1>,|u_2>$ by $b_1,b_2$. $\endgroup$ Commented Nov 2, 2018 at 17:28
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    $\begingroup$ Why do you normalize two particles? Why do you add them? When having two particles you should use a tensor product $\endgroup$ Commented Nov 2, 2018 at 17:29
  • $\begingroup$ @OfekGillon It's $|c_1|^2$ portion/probability of particle $A$ and $|c_2|^2$ for particle $B$. The measurement $|A><A|+|B><B|$ obtain either particle $A$ or $B$, yet, if one add them together, the secondary result for measurement $|u_1><u_1|+|u_2><u_2|$ is differed. (Acturally, if you add them together, the sate is no longer normalized) $\endgroup$ Commented Nov 2, 2018 at 17:43
  • $\begingroup$ What does it mean to have a $|c_1|^2$ probability of a particle? Can you please give an example? $\endgroup$ Commented Nov 2, 2018 at 17:47

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Why can't we expand $|\psi\rangle$ into $|u_1\rangle$ and $|u_2\rangle$ by the linearity?

Of course we can. But if you do this, you will not get the expression that you wrote. $$ c_1 ( a_1 |u_1\rangle + a_2 |u_2\rangle) + c_2 (b_1 |u_1\rangle + b_2|u_2\rangle) = (c_1 a_1 + c_2 b_1) |u_1\rangle + (c_1 a_2 + c_2 b_2) |u_2\rangle . $$


Edit: The above was an answer to v4 of the question, let me also answer v7.

Note first that you claim $2c_1 a_1 c_2 b_1 + 2c_1 a_2 c_2 b_2 > 0$, which is not true in general.

You write $|\psi\rangle = c_1|A\rangle + c_2 |B\rangle$ and assume $c_1^2 + c_2^2 = 1$. The condition "$c_1^2 + c_2^2 = 1$" is the condition for normalization if, and only if, $|A\rangle$ and $|B\rangle$ are orthogonal, so you should additionally assume $$\langle A | B \rangle = a_1 b_2 + a_2 b_1 = 0 .$$ Then your calculation shows that $(c_1 a_1 + c_2 b_1)^2 + (c_1 a_2 + c_2 b_2)^2 = 1$ as it should be.

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  • $\begingroup$ Sorry, that's an typo but you see, suppose numbers are all real for simplicity $(c_1a_1+c_2b_1)^2+(c_1a_2+c_2b_2)^2=(c_1a_1)^2+(c_2b_1)^2+2c_1a_1c_2b_1+(c_1a_2)^2+(c_2b_2)^2+2c_1a_2c_2b_2=c_1^2+c_2^2+2c_1a_1c_2b_1+2c_1a_2c_2b_2=1+2c_1a_1c_2b_1+2c_1a_2c_2b_2>1$. $\endgroup$ Commented Nov 3, 2018 at 1:35
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    $\begingroup$ @user9976437 I edited my post $\endgroup$
    – Noiralef
    Commented Nov 3, 2018 at 8:31
  • $\begingroup$ But the amswer is $\sqrt{c_1^2a_1^2+c_2^2b_1^2}|u_1>+\sqrt{c_1^2a_2^2+c_2^2b_2^2}|u_2>$, I thought AaronStevens posted a page about the states being non linear space but I don't konw where it went. $\endgroup$ Commented Nov 3, 2018 at 16:55
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    $\begingroup$ @user9976437 The information in my answer was true in general, but I was overlooking the specifics of your problem that this answer handles correctly. Therefore I removed the answer. This is the correct answer to your question. Also the solution you state in the comment above is the same solution expressed in this answer. Let me know if you can't figure out why. – $\endgroup$ Commented Nov 4, 2018 at 1:46

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