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I am familiar with the true (or general) notion of orthogonality, given in the Linear Algebra and Pythagoras theorem derived from the $\vec x \cdot \vec y = 0$. I have also recently got to know that true or general definition of orthogonality is that orthogonal things are mutually exclusive and spin "up" is orthogonal to spin "down".

I always believed that two elements are orthogonal when measuring one component does not give any information about the other. It is therefore paradoxical to me to hear that spin up state is orthogonal to spin down. Having x and y in opposite directions implies that if you measure $x=n$, you are sure that $y=-n$. They have perfect overlap/correlation. Despite we agree with physicians that orthogonality is opposite of "overlap", they say that up and down, which perfectly overlap, are orthogonal! Orthogonality is identified with its opposite. I cannot screw my brain around this.

The last time I heard this idea was in Susskind's Theoretical Minimum, lecture 2, where he recalls that "overlap/correlation is opposite of orthogonality and orthogonality means mutual exclusiveness so that you can clearly distinguish between orthogonal things" (no overlap between basis vectors). Why does he speak about "measurably distinct"? Which measurement in conventional space does distinguish between x and y?

I don't understand why spin down is orthogonal to spin up rather than left and right are orthogonal to spin up. Does this orthogonality have anything to do with the conventional, 90° angle, like up-left, orthogonality? What does Euclidian orthogonality of 90° have in common with mutual exclusiveness?

This question has another dimension. Susskind comes up with 3 orthogonal basises. He says that we can have up-down orthogonality, but, we can also have left-right orthogonality. I want to know how these orthogonalities are related? If spin up and spin down are orthogonal along z axis, then what is relationship between $x$ and $z$? Why there are only 3 such relationships? Here is how Susskind derives the "left-right" orthogonality relationship from the up and down relationship.

In short, I want to know what is common between all kinds of orthogonalities, what makes up exclusive with down, aren't up and down 100% correlated/overlapped, why they are not exclusive/orthogonal in the ordinary space and what is the relationship between between up-down and left-right bases?

In math you can define 3 curls, the vectors of rotation. Can you rotate a thing in all 3 planes simultaneously or 3rd rotation will be a combination of (degree of freedom is two). Also, I see that

$${left - right \over \sqrt 2} = up$$ $${left + right \over \sqrt 2} = down$$

Is this because of the same orthogonality between sine, cos and complex exponential?

There is a question asking how statistical independence is related to orthogonality. Can we say which properties are common between statistical independence and orthogonality? This can be interesting particularly in the context of QM, which is sorta statistical mechanics.

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    $\begingroup$ You misunderstand in which space these are orthogonal. Up and down spin configurations are orthogonal in the Hilbert space of states. Yes, this Hilbert space is describing the possible configurations of a vector (or spinor) in physical space, and there is also a notion of orthogonality in that space, but it is in the Hilbert space of states where up and down spin configurations are considered orthogonal. Does that help? $\endgroup$
    – josh
    Commented Jun 9, 2013 at 16:03
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    $\begingroup$ To further clarify, consider instead the quantum states of a simple massive spinless particle in a box. The physical space has no vectors to speak of, and so there is no orthogonality in their spatial configurations to worry about. But the Hilbert space of states (ground state, first excited state, etc.) still admits a notion of orthogonality. It is that sense of orthogonality that applies to the spin configurations you ask about too. $\endgroup$
    – josh
    Commented Jun 9, 2013 at 16:05
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    $\begingroup$ @josh, I want to understand of what is common for the different notions of orthogonality. To further clarify, I see no problem with vectors being orthogonal when their inner product is 0. I have problem with up being orthogonal with down. $\endgroup$
    – Val
    Commented Jun 9, 2013 at 16:05
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    $\begingroup$ Linear algebra is a common framework. A Hilbert space is an inner product space. The link is mathematical, not physical. $\endgroup$
    – josh
    Commented Jun 9, 2013 at 16:06
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    $\begingroup$ @Val, Josh has already answered your question. Up is the vector (1,0), down is (0,1). Using the normal inner product, they are orthogonal, 0*1+1*0 = 0. It's OK to represent the directions up and down by these vectors and every other direction may be represented by a similar pair of complex numbers, too. It's called the spinor representation of $Spin(3)$. In quantum physics, every two states that are clearly "mutually exclusive" are orthogonal to each other, and spin "up" and "down" is such a pair of mutually excluding possibilities. $\endgroup$
    – Luboš Motl
    Commented Jun 9, 2013 at 16:46

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The fundamental difficulty here is that

If two elements are orthogonal, it means that measuring one component does not give any information about the other.

is incorrect. Orthogonality mean exactly that the inner product between the two things is zero. If $a \cdot b = 0$ then $a$ and $b$ are said to be orthogonal.

In a Cartesian space this has the the condition that you name as a consequence, but the space of spin-states is not a Cartesian space and you can't import the same concept. And yes, that means that the $\ell_z$ basis states are not orthogonal to the $\ell_y$ basis states because their inner product is not zero.

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  • $\begingroup$ Do you mean that in 2-dimensional space, measuring one vector gives you the other through the equation $a\cdot b=0$, so that there is a perfect correlation between orthogonal vectors? Is it the same in the Cartesian space? $\endgroup$
    – Val
    Commented Jun 13, 2013 at 19:34
  • $\begingroup$ It's not completely clear to me what you are asking but you can't use the dot product to determine 'the' orthogonal vector (there are an infinite number of answers and even 2 orthogonal unit vectors, after all) but to determine if two given vectors are orthogonal. The important thing is to not think of the space occupied by angular momentum as if it were a Cartesian space. It has different rules. The fact that we use a notation that looks like that for a Cartesian 2-vector is beside the point (or maybe it is the point, as this seems to be causing some trouble). $\endgroup$
    – dmckee --- ex-moderator kitten
    Commented Jun 13, 2013 at 19:49

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