Mixing
![Full explanation on the cancellation property: $;sinbig(arcsin(x)big)=x,; forall; xin [-1,1]$](https://blogger.googleusercontent.com/img/b/R29vZ2xl/AVvXsEhZCTB4-bb-s-32SAm6ytjzFOU06cH9o8q-a_wq_UnH6lcd8hvws23CmBWHM6g256LzTX9yK1EiCCzTC1NSgtxlNk_J9hdr9bA3tD0Nqntbl521j4bLAm_uXPANuWLBhcCKTKzns2gaUodx/s72-c/1.jpg)
Can complex vectors be orthogonal even if none of the components are? [closed]
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I tried researching this on my own but I couldn't find a concise answer that didn't bog me down with terms and math I don't understand. I'm attempting to read Leonard Susskind's Theoretical Minimum book on Quantum Mechanics and I'm just getting into ket-vectors, inner products, etc. but I had a question I couldnt wrap my head around. (Just assume I know very little and go from there, also I have no idea how to properly format the mathematical expressions so I'll use plain English).
If the definition of orthogonal is that the inner product of two vectors is 0, there are hypothetically multiple ways this could be achieved, in my mind at least. What I want to know is, which of these is actually true.
The first case is if every component (B* times A) is 0, then the sum of all those 0's is of course 0.
What if two components are positive and negative pairs? Say +1, -1, and 0? The sum is 0 but some or all components are non-zero.
The third case, based on the previous one, is what if there are no +/- pairs but the sum is 0? Say -0.5, -0.5, and +1. Can either of these last two cases actually happen?
Is there something I'm missing that makes my interpretation too simplistic to be accurate? Again, I don't have a strong math or physics background so I wouldn't understand complicated mathematical proofs or explanations, just a simple answer (if that's even possible).
complex-numbers vectors orthogonality quantum-mechanics
asked Jul 31 at 16:47
Mercurae
82
closed as unclear what you're asking by amWhy, Xander Henderson, Lord Shark the Unknown, John Ma, Mostafa Ayaz Aug 1 at 18:32
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
 |Â
show 1 more comment
up vote
0
down vote
favorite
I tried researching this on my own but I couldn't find a concise answer that didn't bog me down with terms and math I don't understand. I'm attempting to read Leonard Susskind's Theoretical Minimum book on Quantum Mechanics and I'm just getting into ket-vectors, inner products, etc. but I had a question I couldnt wrap my head around. (Just assume I know very little and go from there, also I have no idea how to properly format the mathematical expressions so I'll use plain English).
If the definition of orthogonal is that the inner product of two vectors is 0, there are hypothetically multiple ways this could be achieved, in my mind at least. What I want to know is, which of these is actually true.
The first case is if every component (B* times A) is 0, then the sum of all those 0's is of course 0.
What if two components are positive and negative pairs? Say +1, -1, and 0? The sum is 0 but some or all components are non-zero.
The third case, based on the previous one, is what if there are no +/- pairs but the sum is 0? Say -0.5, -0.5, and +1. Can either of these last two cases actually happen?
Is there something I'm missing that makes my interpretation too simplistic to be accurate? Again, I don't have a strong math or physics background so I wouldn't understand complicated mathematical proofs or explanations, just a simple answer (if that's even possible).
complex-numbers vectors orthogonality quantum-mechanics
asked Jul 31 at 16:47
Mercurae
82
closed as unclear what you're asking by amWhy, Xander Henderson, Lord Shark the Unknown, John Ma, Mostafa Ayaz Aug 1 at 18:32
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
Please clarify what you mean by the title. It is really misleading.
– Ahmad Bazzi
Jul 31 at 16:53
(1) You have written a wall of text, which is very hard to read. A few paragraph breaks would be very helpful. (2) You are asking a question about something that is fairly deep, yet you want a "simple" answer. There is something amiss here...
– Xander Henderson
Jul 31 at 16:54
1
You seem to be confused regarding what a definition is. Yes, the definition says two vectors are orthogonal if their inner product is zero. So if the inner product is zero they're orthogonal, period. For example $(1,1)$ and $(1,-1)$ are orthogonal.
– David C. Ullrich
Jul 31 at 16:59
If the sum of the products of the corresponding elements is $0,$ then the vectors are orthogonal. That's the definition. It isn't necessary that anything but the sum be $0$.
– saulspatz
Jul 31 at 17:00
orthogonality is the greek etymology version of perpendicularity. Take your favourite cardboard box ;), move it around in space, any three edges that meet at a corner are mutually perpendicular, no matter if their direction co-ordinates are moving around all over the place against some $(x,y,z)$ axes you've fixed. Am not a physicist, whom you'll have to consult for interpretations on what orthogonality means in quantum world. Sorry if that's too simplistic!
– Mehness
Jul 31 at 17:06
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I tried researching this on my own but I couldn't find a concise answer that didn't bog me down with terms and math I don't understand. I'm attempting to read Leonard Susskind's Theoretical Minimum book on Quantum Mechanics and I'm just getting into ket-vectors, inner products, etc. but I had a question I couldnt wrap my head around. (Just assume I know very little and go from there, also I have no idea how to properly format the mathematical expressions so I'll use plain English).
If the definition of orthogonal is that the inner product of two vectors is 0, there are hypothetically multiple ways this could be achieved, in my mind at least. What I want to know is, which of these is actually true.
The first case is if every component (B* times A) is 0, then the sum of all those 0's is of course 0.
What if two components are positive and negative pairs? Say +1, -1, and 0? The sum is 0 but some or all components are non-zero.
The third case, based on the previous one, is what if there are no +/- pairs but the sum is 0? Say -0.5, -0.5, and +1. Can either of these last two cases actually happen?
Is there something I'm missing that makes my interpretation too simplistic to be accurate? Again, I don't have a strong math or physics background so I wouldn't understand complicated mathematical proofs or explanations, just a simple answer (if that's even possible).
complex-numbers vectors orthogonality quantum-mechanics
asked Jul 31 at 16:47
Mercurae
82
I tried researching this on my own but I couldn't find a concise answer that didn't bog me down with terms and math I don't understand. I'm attempting to read Leonard Susskind's Theoretical Minimum book on Quantum Mechanics and I'm just getting into ket-vectors, inner products, etc. but I had a question I couldnt wrap my head around. (Just assume I know very little and go from there, also I have no idea how to properly format the mathematical expressions so I'll use plain English).
If the definition of orthogonal is that the inner product of two vectors is 0, there are hypothetically multiple ways this could be achieved, in my mind at least. What I want to know is, which of these is actually true.
The first case is if every component (B* times A) is 0, then the sum of all those 0's is of course 0.
What if two components are positive and negative pairs? Say +1, -1, and 0? The sum is 0 but some or all components are non-zero.
The third case, based on the previous one, is what if there are no +/- pairs but the sum is 0? Say -0.5, -0.5, and +1. Can either of these last two cases actually happen?
Is there something I'm missing that makes my interpretation too simplistic to be accurate? Again, I don't have a strong math or physics background so I wouldn't understand complicated mathematical proofs or explanations, just a simple answer (if that's even possible).
complex-numbers vectors orthogonality quantum-mechanics
asked Jul 31 at 16:47
Mercurae
82
edited Aug 1 at 2:12
asked Jul 31 at 16:47
Mercurae
82
asked Jul 31 at 16:47
Mercurae
82
asked Jul 31 at 16:47
Mercurae
82
82
closed as unclear what you're asking by amWhy, Xander Henderson, Lord Shark the Unknown, John Ma, Mostafa Ayaz Aug 1 at 18:32
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by amWhy, Xander Henderson, Lord Shark the Unknown, John Ma, Mostafa Ayaz Aug 1 at 18:32
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
1
Please clarify what you mean by the title. It is really misleading.
– Ahmad Bazzi
Jul 31 at 16:53
(1) You have written a wall of text, which is very hard to read. A few paragraph breaks would be very helpful. (2) You are asking a question about something that is fairly deep, yet you want a "simple" answer. There is something amiss here...
– Xander Henderson
Jul 31 at 16:54
1
You seem to be confused regarding what a definition is. Yes, the definition says two vectors are orthogonal if their inner product is zero. So if the inner product is zero they're orthogonal, period. For example $(1,1)$ and $(1,-1)$ are orthogonal.
– David C. Ullrich
Jul 31 at 16:59
If the sum of the products of the corresponding elements is $0,$ then the vectors are orthogonal. That's the definition. It isn't necessary that anything but the sum be $0$.
– saulspatz
Jul 31 at 17:00
orthogonality is the greek etymology version of perpendicularity. Take your favourite cardboard box ;), move it around in space, any three edges that meet at a corner are mutually perpendicular, no matter if their direction co-ordinates are moving around all over the place against some $(x,y,z)$ axes you've fixed. Am not a physicist, whom you'll have to consult for interpretations on what orthogonality means in quantum world. Sorry if that's too simplistic!
– Mehness
Jul 31 at 17:06
 |Â
show 1 more comment
1
Please clarify what you mean by the title. It is really misleading.
– Ahmad Bazzi
Jul 31 at 16:53
(1) You have written a wall of text, which is very hard to read. A few paragraph breaks would be very helpful. (2) You are asking a question about something that is fairly deep, yet you want a "simple" answer. There is something amiss here...
– Xander Henderson
Jul 31 at 16:54
1
You seem to be confused regarding what a definition is. Yes, the definition says two vectors are orthogonal if their inner product is zero. So if the inner product is zero they're orthogonal, period. For example $(1,1)$ and $(1,-1)$ are orthogonal.
– David C. Ullrich
Jul 31 at 16:59
If the sum of the products of the corresponding elements is $0,$ then the vectors are orthogonal. That's the definition. It isn't necessary that anything but the sum be $0$.
– saulspatz
Jul 31 at 17:00
orthogonality is the greek etymology version of perpendicularity. Take your favourite cardboard box ;), move it around in space, any three edges that meet at a corner are mutually perpendicular, no matter if their direction co-ordinates are moving around all over the place against some $(x,y,z)$ axes you've fixed. Am not a physicist, whom you'll have to consult for interpretations on what orthogonality means in quantum world. Sorry if that's too simplistic!
– Mehness
Jul 31 at 17:06
1
1
Please clarify what you mean by the title. It is really misleading.
– Ahmad Bazzi
Jul 31 at 16:53
Please clarify what you mean by the title. It is really misleading.
– Ahmad Bazzi
Jul 31 at 16:53
(1) You have written a wall of text, which is very hard to read. A few paragraph breaks would be very helpful. (2) You are asking a question about something that is fairly deep, yet you want a "simple" answer. There is something amiss here...
– Xander Henderson
Jul 31 at 16:54
(1) You have written a wall of text, which is very hard to read. A few paragraph breaks would be very helpful. (2) You are asking a question about something that is fairly deep, yet you want a "simple" answer. There is something amiss here...
– Xander Henderson
Jul 31 at 16:54
1
1
You seem to be confused regarding what a definition is. Yes, the definition says two vectors are orthogonal if their inner product is zero. So if the inner product is zero they're orthogonal, period. For example $(1,1)$ and $(1,-1)$ are orthogonal.
– David C. Ullrich
Jul 31 at 16:59
You seem to be confused regarding what a definition is. Yes, the definition says two vectors are orthogonal if their inner product is zero. So if the inner product is zero they're orthogonal, period. For example $(1,1)$ and $(1,-1)$ are orthogonal.
– David C. Ullrich
Jul 31 at 16:59
If the sum of the products of the corresponding elements is $0,$ then the vectors are orthogonal. That's the definition. It isn't necessary that anything but the sum be $0$.
– saulspatz
Jul 31 at 17:00
If the sum of the products of the corresponding elements is $0,$ then the vectors are orthogonal. That's the definition. It isn't necessary that anything but the sum be $0$.
– saulspatz
Jul 31 at 17:00
orthogonality is the greek etymology version of perpendicularity. Take your favourite cardboard box ;), move it around in space, any three edges that meet at a corner are mutually perpendicular, no matter if their direction co-ordinates are moving around all over the place against some $(x,y,z)$ axes you've fixed. Am not a physicist, whom you'll have to consult for interpretations on what orthogonality means in quantum world. Sorry if that's too simplistic!
– Mehness
Jul 31 at 17:06
orthogonality is the greek etymology version of perpendicularity. Take your favourite cardboard box ;), move it around in space, any three edges that meet at a corner are mutually perpendicular, no matter if their direction co-ordinates are moving around all over the place against some $(x,y,z)$ axes you've fixed. Am not a physicist, whom you'll have to consult for interpretations on what orthogonality means in quantum world. Sorry if that's too simplistic!
– Mehness
Jul 31 at 17:06
 |Â
show 1 more comment
3 Answers
3
active
oldest
votes
up vote
1
down vote
accepted
You don't need to use complex vectors to see this behaviour, a real vector space suffices. And in fact I'm willing to bet there's a precise sense in which "most of the time", $acdot b = 0$ without all $a_ib_i = 0$.
Examples in 3D matching your exact example numbers-
$$ beginpmatrix 1 \ 1\ 0endpmatrix cdot beginpmatrix 1 \ -1\ 0endpmatrix = 0, text with one pm textpair, quad beginpmatrix 1 \ 1\ 1endpmatrix cdot beginpmatrix 1 \ -1/2\ -1/2endpmatrix = 0, text with no pm textpair$$
answered Jul 31 at 17:00
Calvin Khor
7,99411132
1
In your displayed equations, you don't seem to be using $=$ to mean "equals", which is a little confusing.
– Mark S.
Jul 31 at 17:01
@MarkS. Thanks for the feedback, hopefully this is clearer
– Calvin Khor
Jul 31 at 18:02
This is what I was asking, thanks! In your (or "my") example the two "x" components multiply to one rather than zero. But it doesn't really matter, we could just change the coordinate system and get one back to zero if they are actually orthogonal. I don't know why I didn't get that. I think I just got too caught up in the abstract mathematics. I think I can do much better with things I can visualize. Do you have any advice for thinking about abstract math?
– Mercurae
Aug 1 at 2:30
@Mercurae one good thing would be to try to see many examples, you would have answered your own question if you started writing out all the dot products of all vectors with entries $0,pm1,pm2$ :) Also, I don't know any QM but perhaps it can be helpful to abtract /further/ by picking up a book on pure linear algebra. Hope that helps
– Calvin Khor
Aug 1 at 7:14
@CalvinKhor Thanks, I'll keep that in mind
– Mercurae
Aug 1 at 13:09
add a comment |Â
up vote
1
down vote
To illustrate the flexibility of orthogonality: Given any numbers $a_1,ldots,a_n$ and $b_1,ldots,b_n-1$, as long as $a_nneq0$, there is a number $b_n$ such that the vectors
$$
A=(a_1,ldots,a_n)\
B=(b_1,ldots,b_n)
$$
are orthogonal. This is true regardless of whether the numbers are real, complex, or from any other field.
answered Jul 31 at 17:01
Arthur
98.3k793174
add a comment |Â
up vote
0
down vote
Orhogonality as nothing to do with components as a singular entity but with components as a whole. You cannot say that a component is orthogonal because components are, to put it simple, just numbers. Let me make a simple example: let's choose the following vectors with the standard base (the $x,y,z$ Cartesian plane) $$mathbfv_1 = (1,0,0);;;mathbfv_2=(0,0,1)$$ so you can see that the $x$-component is $1$ for the first vector and $0$ for the second, the $y$-component is zero for both and the $z$-component is $0$ for the first one and $1$ for the second. This two vector are orthogonal as you can prove by simply evaluating their inner product $$mathbfv_1 cdot mathbfv_2 = (1times0) + (0times0) + (0times 1) = 0$$ Orthogonality is pretty simple to prove just with an inner product, there's no constraint on the single components of a vector, the only constraint is that the inner product is zero
answered Jul 31 at 17:03
Davide Morgante
1,654220
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
You don't need to use complex vectors to see this behaviour, a real vector space suffices. And in fact I'm willing to bet there's a precise sense in which "most of the time", $acdot b = 0$ without all $a_ib_i = 0$.
Examples in 3D matching your exact example numbers-
$$ beginpmatrix 1 \ 1\ 0endpmatrix cdot beginpmatrix 1 \ -1\ 0endpmatrix = 0, text with one pm textpair, quad beginpmatrix 1 \ 1\ 1endpmatrix cdot beginpmatrix 1 \ -1/2\ -1/2endpmatrix = 0, text with no pm textpair$$
answered Jul 31 at 17:00
Calvin Khor
7,99411132
1
In your displayed equations, you don't seem to be using $=$ to mean "equals", which is a little confusing.
– Mark S.
Jul 31 at 17:01
@MarkS. Thanks for the feedback, hopefully this is clearer
– Calvin Khor
Jul 31 at 18:02
This is what I was asking, thanks! In your (or "my") example the two "x" components multiply to one rather than zero. But it doesn't really matter, we could just change the coordinate system and get one back to zero if they are actually orthogonal. I don't know why I didn't get that. I think I just got too caught up in the abstract mathematics. I think I can do much better with things I can visualize. Do you have any advice for thinking about abstract math?
– Mercurae
Aug 1 at 2:30
@Mercurae one good thing would be to try to see many examples, you would have answered your own question if you started writing out all the dot products of all vectors with entries $0,pm1,pm2$ :) Also, I don't know any QM but perhaps it can be helpful to abtract /further/ by picking up a book on pure linear algebra. Hope that helps
– Calvin Khor
Aug 1 at 7:14
@CalvinKhor Thanks, I'll keep that in mind
– Mercurae
Aug 1 at 13:09
add a comment |Â
up vote
1
down vote
accepted
You don't need to use complex vectors to see this behaviour, a real vector space suffices. And in fact I'm willing to bet there's a precise sense in which "most of the time", $acdot b = 0$ without all $a_ib_i = 0$.
Examples in 3D matching your exact example numbers-
$$ beginpmatrix 1 \ 1\ 0endpmatrix cdot beginpmatrix 1 \ -1\ 0endpmatrix = 0, text with one pm textpair, quad beginpmatrix 1 \ 1\ 1endpmatrix cdot beginpmatrix 1 \ -1/2\ -1/2endpmatrix = 0, text with no pm textpair$$
answered Jul 31 at 17:00
Calvin Khor
7,99411132
1
In your displayed equations, you don't seem to be using $=$ to mean "equals", which is a little confusing.
– Mark S.
Jul 31 at 17:01
@MarkS. Thanks for the feedback, hopefully this is clearer
– Calvin Khor
Jul 31 at 18:02
This is what I was asking, thanks! In your (or "my") example the two "x" components multiply to one rather than zero. But it doesn't really matter, we could just change the coordinate system and get one back to zero if they are actually orthogonal. I don't know why I didn't get that. I think I just got too caught up in the abstract mathematics. I think I can do much better with things I can visualize. Do you have any advice for thinking about abstract math?
– Mercurae
Aug 1 at 2:30
@Mercurae one good thing would be to try to see many examples, you would have answered your own question if you started writing out all the dot products of all vectors with entries $0,pm1,pm2$ :) Also, I don't know any QM but perhaps it can be helpful to abtract /further/ by picking up a book on pure linear algebra. Hope that helps
– Calvin Khor
Aug 1 at 7:14
@CalvinKhor Thanks, I'll keep that in mind
– Mercurae
Aug 1 at 13:09
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
You don't need to use complex vectors to see this behaviour, a real vector space suffices. And in fact I'm willing to bet there's a precise sense in which "most of the time", $acdot b = 0$ without all $a_ib_i = 0$.
Examples in 3D matching your exact example numbers-
$$ beginpmatrix 1 \ 1\ 0endpmatrix cdot beginpmatrix 1 \ -1\ 0endpmatrix = 0, text with one pm textpair, quad beginpmatrix 1 \ 1\ 1endpmatrix cdot beginpmatrix 1 \ -1/2\ -1/2endpmatrix = 0, text with no pm textpair$$
answered Jul 31 at 17:00
Calvin Khor
7,99411132
You don't need to use complex vectors to see this behaviour, a real vector space suffices. And in fact I'm willing to bet there's a precise sense in which "most of the time", $acdot b = 0$ without all $a_ib_i = 0$.
Examples in 3D matching your exact example numbers-
$$ beginpmatrix 1 \ 1\ 0endpmatrix cdot beginpmatrix 1 \ -1\ 0endpmatrix = 0, text with one pm textpair, quad beginpmatrix 1 \ 1\ 1endpmatrix cdot beginpmatrix 1 \ -1/2\ -1/2endpmatrix = 0, text with no pm textpair$$
answered Jul 31 at 17:00
Calvin Khor
7,99411132
edited Jul 31 at 18:01
answered Jul 31 at 17:00
Calvin Khor
7,99411132
answered Jul 31 at 17:00
Calvin Khor
7,99411132
answered Jul 31 at 17:00
Calvin Khor
7,99411132
7,99411132
1
In your displayed equations, you don't seem to be using $=$ to mean "equals", which is a little confusing.
– Mark S.
Jul 31 at 17:01
@MarkS. Thanks for the feedback, hopefully this is clearer
– Calvin Khor
Jul 31 at 18:02
This is what I was asking, thanks! In your (or "my") example the two "x" components multiply to one rather than zero. But it doesn't really matter, we could just change the coordinate system and get one back to zero if they are actually orthogonal. I don't know why I didn't get that. I think I just got too caught up in the abstract mathematics. I think I can do much better with things I can visualize. Do you have any advice for thinking about abstract math?
– Mercurae
Aug 1 at 2:30
@Mercurae one good thing would be to try to see many examples, you would have answered your own question if you started writing out all the dot products of all vectors with entries $0,pm1,pm2$ :) Also, I don't know any QM but perhaps it can be helpful to abtract /further/ by picking up a book on pure linear algebra. Hope that helps
– Calvin Khor
Aug 1 at 7:14
@CalvinKhor Thanks, I'll keep that in mind
– Mercurae
Aug 1 at 13:09
add a comment |Â
1
In your displayed equations, you don't seem to be using $=$ to mean "equals", which is a little confusing.
– Mark S.
Jul 31 at 17:01
@MarkS. Thanks for the feedback, hopefully this is clearer
– Calvin Khor
Jul 31 at 18:02
This is what I was asking, thanks! In your (or "my") example the two "x" components multiply to one rather than zero. But it doesn't really matter, we could just change the coordinate system and get one back to zero if they are actually orthogonal. I don't know why I didn't get that. I think I just got too caught up in the abstract mathematics. I think I can do much better with things I can visualize. Do you have any advice for thinking about abstract math?
– Mercurae
Aug 1 at 2:30
@Mercurae one good thing would be to try to see many examples, you would have answered your own question if you started writing out all the dot products of all vectors with entries $0,pm1,pm2$ :) Also, I don't know any QM but perhaps it can be helpful to abtract /further/ by picking up a book on pure linear algebra. Hope that helps
– Calvin Khor
Aug 1 at 7:14
@CalvinKhor Thanks, I'll keep that in mind
– Mercurae
Aug 1 at 13:09
1
1
In your displayed equations, you don't seem to be using $=$ to mean "equals", which is a little confusing.
– Mark S.
Jul 31 at 17:01
In your displayed equations, you don't seem to be using $=$ to mean "equals", which is a little confusing.
– Mark S.
Jul 31 at 17:01
@MarkS. Thanks for the feedback, hopefully this is clearer
– Calvin Khor
Jul 31 at 18:02
@MarkS. Thanks for the feedback, hopefully this is clearer
– Calvin Khor
Jul 31 at 18:02
This is what I was asking, thanks! In your (or "my") example the two "x" components multiply to one rather than zero. But it doesn't really matter, we could just change the coordinate system and get one back to zero if they are actually orthogonal. I don't know why I didn't get that. I think I just got too caught up in the abstract mathematics. I think I can do much better with things I can visualize. Do you have any advice for thinking about abstract math?
– Mercurae
Aug 1 at 2:30
This is what I was asking, thanks! In your (or "my") example the two "x" components multiply to one rather than zero. But it doesn't really matter, we could just change the coordinate system and get one back to zero if they are actually orthogonal. I don't know why I didn't get that. I think I just got too caught up in the abstract mathematics. I think I can do much better with things I can visualize. Do you have any advice for thinking about abstract math?
– Mercurae
Aug 1 at 2:30
@Mercurae one good thing would be to try to see many examples, you would have answered your own question if you started writing out all the dot products of all vectors with entries $0,pm1,pm2$ :) Also, I don't know any QM but perhaps it can be helpful to abtract /further/ by picking up a book on pure linear algebra. Hope that helps
– Calvin Khor
Aug 1 at 7:14
@Mercurae one good thing would be to try to see many examples, you would have answered your own question if you started writing out all the dot products of all vectors with entries $0,pm1,pm2$ :) Also, I don't know any QM but perhaps it can be helpful to abtract /further/ by picking up a book on pure linear algebra. Hope that helps
– Calvin Khor
Aug 1 at 7:14
@CalvinKhor Thanks, I'll keep that in mind
– Mercurae
Aug 1 at 13:09
@CalvinKhor Thanks, I'll keep that in mind
– Mercurae
Aug 1 at 13:09
add a comment |Â
up vote
1
down vote
To illustrate the flexibility of orthogonality: Given any numbers $a_1,ldots,a_n$ and $b_1,ldots,b_n-1$, as long as $a_nneq0$, there is a number $b_n$ such that the vectors
$$
A=(a_1,ldots,a_n)\
B=(b_1,ldots,b_n)
$$
are orthogonal. This is true regardless of whether the numbers are real, complex, or from any other field.
answered Jul 31 at 17:01
Arthur
98.3k793174
add a comment |Â
up vote
1
down vote
To illustrate the flexibility of orthogonality: Given any numbers $a_1,ldots,a_n$ and $b_1,ldots,b_n-1$, as long as $a_nneq0$, there is a number $b_n$ such that the vectors
$$
A=(a_1,ldots,a_n)\
B=(b_1,ldots,b_n)
$$
are orthogonal. This is true regardless of whether the numbers are real, complex, or from any other field.
answered Jul 31 at 17:01
Arthur
98.3k793174
add a comment |Â
up vote
1
down vote
up vote
1
down vote
To illustrate the flexibility of orthogonality: Given any numbers $a_1,ldots,a_n$ and $b_1,ldots,b_n-1$, as long as $a_nneq0$, there is a number $b_n$ such that the vectors
$$
A=(a_1,ldots,a_n)\
B=(b_1,ldots,b_n)
$$
are orthogonal. This is true regardless of whether the numbers are real, complex, or from any other field.
answered Jul 31 at 17:01
Arthur
98.3k793174
To illustrate the flexibility of orthogonality: Given any numbers $a_1,ldots,a_n$ and $b_1,ldots,b_n-1$, as long as $a_nneq0$, there is a number $b_n$ such that the vectors
$$
A=(a_1,ldots,a_n)\
B=(b_1,ldots,b_n)
$$
are orthogonal. This is true regardless of whether the numbers are real, complex, or from any other field.
answered Jul 31 at 17:01
Arthur
98.3k793174
answered Jul 31 at 17:01
Arthur
98.3k793174
answered Jul 31 at 17:01
Arthur
98.3k793174
answered Jul 31 at 17:01
Arthur
98.3k793174
98.3k793174
add a comment |Â
add a comment |Â
up vote
0
down vote
Orhogonality as nothing to do with components as a singular entity but with components as a whole. You cannot say that a component is orthogonal because components are, to put it simple, just numbers. Let me make a simple example: let's choose the following vectors with the standard base (the $x,y,z$ Cartesian plane) $$mathbfv_1 = (1,0,0);;;mathbfv_2=(0,0,1)$$ so you can see that the $x$-component is $1$ for the first vector and $0$ for the second, the $y$-component is zero for both and the $z$-component is $0$ for the first one and $1$ for the second. This two vector are orthogonal as you can prove by simply evaluating their inner product $$mathbfv_1 cdot mathbfv_2 = (1times0) + (0times0) + (0times 1) = 0$$ Orthogonality is pretty simple to prove just with an inner product, there's no constraint on the single components of a vector, the only constraint is that the inner product is zero
answered Jul 31 at 17:03
Davide Morgante
1,654220
add a comment |Â
up vote
0
down vote
Orhogonality as nothing to do with components as a singular entity but with components as a whole. You cannot say that a component is orthogonal because components are, to put it simple, just numbers. Let me make a simple example: let's choose the following vectors with the standard base (the $x,y,z$ Cartesian plane) $$mathbfv_1 = (1,0,0);;;mathbfv_2=(0,0,1)$$ so you can see that the $x$-component is $1$ for the first vector and $0$ for the second, the $y$-component is zero for both and the $z$-component is $0$ for the first one and $1$ for the second. This two vector are orthogonal as you can prove by simply evaluating their inner product $$mathbfv_1 cdot mathbfv_2 = (1times0) + (0times0) + (0times 1) = 0$$ Orthogonality is pretty simple to prove just with an inner product, there's no constraint on the single components of a vector, the only constraint is that the inner product is zero
answered Jul 31 at 17:03
Davide Morgante
1,654220
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Orhogonality as nothing to do with components as a singular entity but with components as a whole. You cannot say that a component is orthogonal because components are, to put it simple, just numbers. Let me make a simple example: let's choose the following vectors with the standard base (the $x,y,z$ Cartesian plane) $$mathbfv_1 = (1,0,0);;;mathbfv_2=(0,0,1)$$ so you can see that the $x$-component is $1$ for the first vector and $0$ for the second, the $y$-component is zero for both and the $z$-component is $0$ for the first one and $1$ for the second. This two vector are orthogonal as you can prove by simply evaluating their inner product $$mathbfv_1 cdot mathbfv_2 = (1times0) + (0times0) + (0times 1) = 0$$ Orthogonality is pretty simple to prove just with an inner product, there's no constraint on the single components of a vector, the only constraint is that the inner product is zero
answered Jul 31 at 17:03
Davide Morgante
1,654220
Orhogonality as nothing to do with components as a singular entity but with components as a whole. You cannot say that a component is orthogonal because components are, to put it simple, just numbers. Let me make a simple example: let's choose the following vectors with the standard base (the $x,y,z$ Cartesian plane) $$mathbfv_1 = (1,0,0);;;mathbfv_2=(0,0,1)$$ so you can see that the $x$-component is $1$ for the first vector and $0$ for the second, the $y$-component is zero for both and the $z$-component is $0$ for the first one and $1$ for the second. This two vector are orthogonal as you can prove by simply evaluating their inner product $$mathbfv_1 cdot mathbfv_2 = (1times0) + (0times0) + (0times 1) = 0$$ Orthogonality is pretty simple to prove just with an inner product, there's no constraint on the single components of a vector, the only constraint is that the inner product is zero
answered Jul 31 at 17:03
Davide Morgante
1,654220
answered Jul 31 at 17:03
Davide Morgante
1,654220
answered Jul 31 at 17:03
Davide Morgante
1,654220
answered Jul 31 at 17:03
Davide Morgante
1,654220
1,654220
add a comment |Â
add a comment |Â
1
Please clarify what you mean by the title. It is really misleading.
– Ahmad Bazzi
Jul 31 at 16:53
(1) You have written a wall of text, which is very hard to read. A few paragraph breaks would be very helpful. (2) You are asking a question about something that is fairly deep, yet you want a "simple" answer. There is something amiss here...
– Xander Henderson
Jul 31 at 16:54
1
You seem to be confused regarding what a definition is. Yes, the definition says two vectors are orthogonal if their inner product is zero. So if the inner product is zero they're orthogonal, period. For example $(1,1)$ and $(1,-1)$ are orthogonal.
– David C. Ullrich
Jul 31 at 16:59
If the sum of the products of the corresponding elements is $0,$ then the vectors are orthogonal. That's the definition. It isn't necessary that anything but the sum be $0$.
– saulspatz
Jul 31 at 17:00
orthogonality is the greek etymology version of perpendicularity. Take your favourite cardboard box ;), move it around in space, any three edges that meet at a corner are mutually perpendicular, no matter if their direction co-ordinates are moving around all over the place against some $(x,y,z)$ axes you've fixed. Am not a physicist, whom you'll have to consult for interpretations on what orthogonality means in quantum world. Sorry if that's too simplistic!
– Mehness
Jul 31 at 17:06