Anti-Humor

[Razgat]

[tuberjustin]

Now the definition of the inner product <A|B> has to be modified to apply to complex spaces. Let c* be the complex conjugate of c. (When c is a complex number of the form a ± bi, then the complex conjugate c* of c is defined as follows:

* = a − bi
* = a + bi

So, for all complex numbers c, [c*]* = c, but c* = c just in case c is real.) Now definition of the inner product of |A> and |B> for complex spaces can be given in terms of the conjugates of complex coefficients as follows. Where |A1> and |A2> are the unit vectors described earlier, |A> = a1|A1> + a2|A2> and |B> = b1|A1> + b2|A2>, then

<A|B> = (a1*)(b1) + (a2*)(b2)

The most general and abstract notion of an inner product, of which we've now defined two special cases, is as follows. <A|B> is an inner product on a vector space V just in case

1. <A|A> = |A|2, and <A|A>=0 if and only if A=0
2. <B|A> = <A|B>*
3. <B|A+C> = <B|A> + <B|C>.

It follows from this that

1. the length of |A> is the square root of inner product of |A> with itself, i.e.,

|A| = √<A|A>,

and

2. |A> and |B> are mutually perpendicular, or orthogonal, if, and only if, <A|B> = 0.

A vector space is a set of vectors closed under addition, and multiplication by constants, an inner product space is a vector space on which the operation of vector multiplication has been defined, and the dimension of such a space is the maximum number of nonzero, mutually orthogonal vectors it contains.

Any collection of N mutually orthogonal vectors of length 1 in an N-dimensional vector space constitutes an orthonormal basis for that space. Let |A1>, … , |AN> be such a collection of unit vectors. Then every vector in the space can be expressed as a sum of the form:

|B> = b1|A1> + b2|A2> + … + bN|AN>,

where bi = <B|Ai>. The bi's here are known as B's expansion coefficients in the A-basis.[5]

Notice that:

1. for all vectors A, B, and C in a given space,

<A|B+C> = <A|B> + <A|C>

2. for any vectors M and Q, expressed in terms of the A-basis,

|M> + |Q> = (m1 + q1)|A1> + (m2 + q2)|A2> + ... + (mN + qN)|AN>,

and

<M|Q> = m1q1 + m2q2 + ... + mnqn

There is another way of writing vectors, namely by writing their expansion coefficients (relative to a given basis) in a column, like so:

|Q> = [ q1
q2 ]

where qi = <Q|Ai> and the Ai are the chosen basis vectors.

When we are dealing with vector spaces of infinite dimension, since we can't write the whole column of expansion coefficients needed to pick out a vector since it would have to be infinitely long, so instead we write down the function (called the ‘wave function’ for Q, usually represented ψ(i)) which has those coefficients as values. We write down, that is, the function:

ψ(i) = qi = <Q|Ai>

Given any vector in, and any basis for, a vector space, we can obtain the wave-function of the vector in that basis; and given a wave-function for a vector, in a particular basis, we can construct the vector whose wave-function it is. Since it turns out that most of the important operations on vectors correspond to simple algebraic operations on their wave-functions, this is the usual way to represent state-vectors.

When a pair of physical systems interact, they form a composite system, and, in quantum mechanics as in classical mechanics, there is a rule for constructing the state-space of a composite system from those of its components, a rule that tells us how to obtain, from the state-spaces, HA and HB for A and B, respectively, the state-space — called the ‘tensor product’ of HA and HB, and written HA⊗HB — of the pair. There are two important things about the rule; first, so long as HA and HB are Hilbert spaces, HA⊗HB will be as well, and second, there are some facts about the way HA⊗HB relates to HA and HB, that have surprising consequences for the relations between the complex system and its parts. In particular, it turns out that the state of a composite system is not uniquely defined by those of its components. What this means, or at least what it appears to mean, is that there are, according to quantum mechanics, facts about composite systems (and not just facts about their spatial configuration) that don't supervene on facts about their components; it means that there are facts about systems as wholes that don't supervene on facts about their parts and the way those parts are arranged in space. The significance of this feature of the theory cannot be overplayed; it is, in one way or another, implicated in most of its most difficult problems.

In a little more detail: if {viA} is an orthonormal basis for HA and {ujB} is an orthonormal basis for HB, then the set of pairs (viA, ujB) is taken to form an orthonormal basis for the tensor product space HA⊗HB. The notation viA⊗ujB is used for the pair (viA,uj B), and inner product on HA⊗HB is defined as:[6]

<viA⊗umB | vjA⊗unB> = <viA | vjA> <umB | unB>

It is a result of this construction that although every vector in HA⊗HB is a linear sum of vectors expressible in the form vA⊗uB, not every vector in the space is itself expressible in that form, and it turns out that

1. any composite state defines uniquely the states of its components.
2. if the states of A and B are pure (i.e., representable by vectors vA and uB, respectively), then the state of (A+B) is pure and represented by vA⊗uB, and
3. if the state of (A+B) is pure and expressible in the form vA⊗uB, then the states of A and B are pure, but
4. if the states of A and B are not pure, i.e., if they are mixed states (these are defined below), they do not uniquely define the state of (A+B); in particular, it may be a pure state not expressible in the form vA⊗u





Example 1 Solve .



Solution

There’s really not a whole lot to do in solving this kind of trig equation. All we need to do is divide both sides by 2 and the go to the unit circle.



So, we are looking for all the values of t for which cosine will have the value of . So, let’s take a look at the following unit circle.



Unit_Quarter_With_Angles

From quick inspection we can see that is a solution. However, as I have shown on the unit circle there is another angle which will also be a solution. We need to determine what this angle is. When we look for these angles we typically want positive angles that lie between 0 and . This angle will not be the only possibility of course, but by convention we typically look for angles that meet these conditions.



To find this angle for this problem all we need to do is use a little geometry. The angle in the first quadrant makes an angle of with the positive x-axis, then so must the angle in the fourth quadrant. So we could use , but again, it’s more common to use positive angles so, we’ll use .



We aren’t done with this problem. As the discussion about finding the second angle has shown there are many ways to write any given angle on the unit circle. Sometimes it will be that we want for the solution and sometimes we will want both (or neither) of the listed angles. Therefore, since there isn’t anything in this problem (contrast this with the next problem) to tell us which is the correct solution we will need to list ALL possible solutions.



This is very easy to do. Recall from the previous section and you’ll see there that I used



to represent all the possible angles that can end at the same location on the unit circle, i.e. angles that end at . Remember that all this says is that we start at then rotate around in the counter-clockwise direction (n is positive) or clockwise direction (n is negative) for n complete rotations. The same thing can be done for the second solution.



So, all together the complete solution to this problem is





As a final thought, notice that we can get by using in the second solution.

[Xelwes]

A man steals an old ladies purse and runs. Promptly the cops are called. A black man in a suit with a briefcase, carrying a stuffed bear for his three year old daughter is arrested. The trial goes through and the man is sent to jail for life with no evidence against him. The real culprit is found a day later, and gets a slap on the wrist. The black man dies in prison from continuous STD infections.

[tuberjustin]

Jul 27, 2009 10:07:46 GMT -5 Xelwes said:

A man steals an old ladies purse and runs. Promptly the cops are called. A black man in a suit with a briefcase, carrying a stuffed bear for his three year old daughter is arrested. The trial goes through and the man is sent to jail for life with no evidence against him. The real culprit is found a day later, and gets a slap on the wrist. The black man dies in prison from continuous STD infections.

it was good until "black man dies in prison from continuous STD infections"

then it turns funny

[Xelwes]

How is getting raped by big burly jail bait funny?

[thecheat]

How is it NOT funny?

[Mizagium]

Jul 27, 2009 12:53:46 GMT -5 thecheat said:

How is it NOT funny?

this.

[TEAM_DERRICK]

Q: How many penguins does it take to cover a doghouse?

A: Purple, because ice cream has no bones.

[Mizagium]

Jul 27, 2009 10:07:46 GMT -5 Xelwes said:

A man steals an old ladies purse and runs. Promptly the cops are called. A black man in a suit with a briefcase, carrying a stuffed bear for his three year old daughter is arrested. The trial goes through and the man is sent to jail for life with no evidence against him. The real culprit is found a day later, and gets a slap on the wrist. The black man dies in prison from continuous STD infections.

Actually, this belongs in "It's funny because it's true"

[TEAM_DERRICK]

* A duck walks into a bar...

Animal control is promptly called, the duck is then taken to a nearby park and released.

A chicken begins crossing the street and is hit by several fast moving metal boxes with wheels.

* What do you get when you cross a chicken with a centipede?

A media circus about the debate over the morals and ethics of genetic engineering.

* What's worse than finding a worm in your apple?

The Holocaust.

* What do a duck and a bicycle have in common?

They both have handlebars. Except the duck doesn't.


* What did Tarzan say when he saw the elephants coming over the hill?

Here come the elephants over the hill.

* What is the difference between elephants and grapes?

Elephants are grey, grapes are purple.

* What did Jane say when she saw the elephants coming over the hill?

Here come the grapes over the hill (Jane was colorblind).

* How do you kill a blue elephant?

You shoot it with a blue elephant gun.

* How do you kill a red elephant?

You hold its trunk until it turns blue and shoot it with a blue elephant gun.

* How do you kill a green elephant?

You embarass it until it turns red, hold its trunk until it turns blue and shoot it with a blue elephant gun.

* How do you kill a white elephant?

You paint it green, embarass it until it turns red, hold its trunk until it turns blue and shoot it with a blue elephant gun.

* How do you kill a pink elephant?

You scare it until it turn white, paint it green, embarass it until it turns red, hold its trunk untils it turns blue and shoot it with a blue elephant gun.

* How do you kill a yellow elephant?

There's no such thing, dumb-shit!

[tuberjustin]

Jul 27, 2009 2:45:51 GMT -5 @tuberjustin said:

Now the definition of the inner product <A|B> has to be modified to apply to complex spaces. Let c* be the complex conjugate of c. (When c is a complex number of the form a ± bi, then the complex conjugate c* of c is defined as follows:

* = a − bi
* = a + bi

So, for all complex numbers c, [c*]* = c, but c* = c just in case c is real.) Now definition of the inner product of |A> and |B> for complex spaces can be given in terms of the conjugates of complex coefficients as follows. Where |A1> and |A2> are the unit vectors described earlier, |A> = a1|A1> + a2|A2> and |B> = b1|A1> + b2|A2>, then

<A|B> = (a1*)(b1) + (a2*)(b2)

The most general and abstract notion of an inner product, of which we've now defined two special cases, is as follows. <A|B> is an inner product on a vector space V just in case

1. <A|A> = |A|2, and <A|A>=0 if and only if A=0
2. <B|A> = <A|B>*
3. <B|A+C> = <B|A> + <B|C>.

It follows from this that

1. the length of |A> is the square root of inner product of |A> with itself, i.e.,

|A| = √<A|A>,

and

2. |A> and |B> are mutually perpendicular, or orthogonal, if, and only if, <A|B> = 0.

A vector space is a set of vectors closed under addition, and multiplication by constants, an inner product space is a vector space on which the operation of vector multiplication has been defined, and the dimension of such a space is the maximum number of nonzero, mutually orthogonal vectors it contains.

Any collection of N mutually orthogonal vectors of length 1 in an N-dimensional vector space constitutes an orthonormal basis for that space. Let |A1>, … , |AN> be such a collection of unit vectors. Then every vector in the space can be expressed as a sum of the form:

|B> = b1|A1> + b2|A2> + … + bN|AN>,

where bi = <B|Ai>. The bi's here are known as B's expansion coefficients in the A-basis.[5]

Notice that:

1. for all vectors A, B, and C in a given space,

<A|B+C> = <A|B> + <A|C>

2. for any vectors M and Q, expressed in terms of the A-basis,

|M> + |Q> = (m1 + q1)|A1> + (m2 + q2)|A2> + ... + (mN + qN)|AN>,

and

<M|Q> = m1q1 + m2q2 + ... + mnqn

There is another way of writing vectors, namely by writing their expansion coefficients (relative to a given basis) in a column, like so:

|Q> = [ q1
q2 ]

where qi = <Q|Ai> and the Ai are the chosen basis vectors.

When we are dealing with vector spaces of infinite dimension, since we can't write the whole column of expansion coefficients needed to pick out a vector since it would have to be infinitely long, so instead we write down the function (called the ‘wave function’ for Q, usually represented ψ(i)) which has those coefficients as values. We write down, that is, the function:

ψ(i) = qi = <Q|Ai>

Given any vector in, and any basis for, a vector space, we can obtain the wave-function of the vector in that basis; and given a wave-function for a vector, in a particular basis, we can construct the vector whose wave-function it is. Since it turns out that most of the important operations on vectors correspond to simple algebraic operations on their wave-functions, this is the usual way to represent state-vectors.

When a pair of physical systems interact, they form a composite system, and, in quantum mechanics as in classical mechanics, there is a rule for constructing the state-space of a composite system from those of its components, a rule that tells us how to obtain, from the state-spaces, HA and HB for A and B, respectively, the state-space — called the ‘tensor product’ of HA and HB, and written HA⊗HB — of the pair. There are two important things about the rule; first, so long as HA and HB are Hilbert spaces, HA⊗HB will be as well, and second, there are some facts about the way HA⊗HB relates to HA and HB, that have surprising consequences for the relations between the complex system and its parts. In particular, it turns out that the state of a composite system is not uniquely defined by those of its components. What this means, or at least what it appears to mean, is that there are, according to quantum mechanics, facts about composite systems (and not just facts about their spatial configuration) that don't supervene on facts about their components; it means that there are facts about systems as wholes that don't supervene on facts about their parts and the way those parts are arranged in space. The significance of this feature of the theory cannot be overplayed; it is, in one way or another, implicated in most of its most difficult problems.

In a little more detail: if {viA} is an orthonormal basis for HA and {ujB} is an orthonormal basis for HB, then the set of pairs (viA, ujB) is taken to form an orthonormal basis for the tensor product space HA⊗HB. The notation viA⊗ujB is used for the pair (viA,uj B), and inner product on HA⊗HB is defined as:[6]

<viA⊗umB | vjA⊗unB> = <viA | vjA> <umB | unB>

It is a result of this construction that although every vector in HA⊗HB is a linear sum of vectors expressible in the form vA⊗uB, not every vector in the space is itself expressible in that form, and it turns out that

1. any composite state defines uniquely the states of its components.
2. if the states of A and B are pure (i.e., representable by vectors vA and uB, respectively), then the state of (A+B) is pure and represented by vA⊗uB, and
3. if the state of (A+B) is pure and expressible in the form vA⊗uB, then the states of A and B are pure, but
4. if the states of A and B are not pure, i.e., if they are mixed states (these are defined below), they do not uniquely define the state of (A+B); in particular, it may be a pure state not expressible in the form vA⊗u





Example 1 Solve .



Solution

There’s really not a whole lot to do in solving this kind of trig equation. All we need to do is divide both sides by 2 and the go to the unit circle.



So, we are looking for all the values of t for which cosine will have the value of . So, let’s take a look at the following unit circle.



Unit_Quarter_With_Angles

From quick inspection we can see that is a solution. However, as I have shown on the unit circle there is another angle which will also be a solution. We need to determine what this angle is. When we look for these angles we typically want positive angles that lie between 0 and . This angle will not be the only possibility of course, but by convention we typically look for angles that meet these conditions.



To find this angle for this problem all we need to do is use a little geometry. The angle in the first quadrant makes an angle of with the positive x-axis, then so must the angle in the fourth quadrant. So we could use , but again, it’s more common to use positive angles so, we’ll use .



We aren’t done with this problem. As the discussion about finding the second angle has shown there are many ways to write any given angle on the unit circle. Sometimes it will be that we want for the solution and sometimes we will want both (or neither) of the listed angles. Therefore, since there isn’t anything in this problem (contrast this with the next problem) to tell us which is the correct solution we will need to list ALL possible solutions.



This is very easy to do. Recall from the previous section and you’ll see there that I used



to represent all the possible angles that can end at the same location on the unit circle, i.e. angles that end at . Remember that all this says is that we start at then rotate around in the counter-clockwise direction (n is positive) or clockwise direction (n is negative) for n complete rotations. The same thing can be done for the second solution.



So, all together the complete solution to this problem is





As a final thought, notice that we can get by using in the second solution.

read all of this, i guarantee that none of it will make you laugh

[Mizagium]

Jul 27, 2009 14:26:29 GMT -5 @tuberjustin said:

Jul 27, 2009 2:45:51 GMT -5 @tuberjustin said:

Now the definition of the inner product <A|B> has to be modified to apply to complex spaces. Let c* be the complex conjugate of c. (When c is a complex number of the form a ± bi, then the complex conjugate c* of c is defined as follows:

* = a − bi
* = a + bi

So, for all complex numbers c, [c*]* = c, but c* = c just in case c is real.) Now definition of the inner product of |A> and |B> for complex spaces can be given in terms of the conjugates of complex coefficients as follows. Where |A1> and |A2> are the unit vectors described earlier, |A> = a1|A1> + a2|A2> and |B> = b1|A1> + b2|A2>, then

<A|B> = (a1*)(b1) + (a2*)(b2)

The most general and abstract notion of an inner product, of which we've now defined two special cases, is as follows. <A|B> is an inner product on a vector space V just in case

1. <A|A> = |A|2, and <A|A>=0 if and only if A=0
2. <B|A> = <A|B>*
3. <B|A+C> = <B|A> + <B|C>.

It follows from this that

1. the length of |A> is the square root of inner product of |A> with itself, i.e.,

|A| = √<A|A>,

and

2. |A> and |B> are mutually perpendicular, or orthogonal, if, and only if, <A|B> = 0.

A vector space is a set of vectors closed under addition, and multiplication by constants, an inner product space is a vector space on which the operation of vector multiplication has been defined, and the dimension of such a space is the maximum number of nonzero, mutually orthogonal vectors it contains.

Any collection of N mutually orthogonal vectors of length 1 in an N-dimensional vector space constitutes an orthonormal basis for that space. Let |A1>, … , |AN> be such a collection of unit vectors. Then every vector in the space can be expressed as a sum of the form:

|B> = b1|A1> + b2|A2> + … + bN|AN>,

where bi = <B|Ai>. The bi's here are known as B's expansion coefficients in the A-basis.[5]

Notice that:

1. for all vectors A, B, and C in a given space,

<A|B+C> = <A|B> + <A|C>

2. for any vectors M and Q, expressed in terms of the A-basis,

|M> + |Q> = (m1 + q1)|A1> + (m2 + q2)|A2> + ... + (mN + qN)|AN>,

and

<M|Q> = m1q1 + m2q2 + ... + mnqn

There is another way of writing vectors, namely by writing their expansion coefficients (relative to a given basis) in a column, like so:

|Q> = [ q1
q2 ]

where qi = <Q|Ai> and the Ai are the chosen basis vectors.

When we are dealing with vector spaces of infinite dimension, since we can't write the whole column of expansion coefficients needed to pick out a vector since it would have to be infinitely long, so instead we write down the function (called the ‘wave function’ for Q, usually represented ψ(i)) which has those coefficients as values. We write down, that is, the function:

ψ(i) = qi = <Q|Ai>

Given any vector in, and any basis for, a vector space, we can obtain the wave-function of the vector in that basis; and given a wave-function for a vector, in a particular basis, we can construct the vector whose wave-function it is. Since it turns out that most of the important operations on vectors correspond to simple algebraic operations on their wave-functions, this is the usual way to represent state-vectors.

When a pair of physical systems interact, they form a composite system, and, in quantum mechanics as in classical mechanics, there is a rule for constructing the state-space of a composite system from those of its components, a rule that tells us how to obtain, from the state-spaces, HA and HB for A and B, respectively, the state-space — called the ‘tensor product’ of HA and HB, and written HA⊗HB — of the pair. There are two important things about the rule; first, so long as HA and HB are Hilbert spaces, HA⊗HB will be as well, and second, there are some facts about the way HA⊗HB relates to HA and HB, that have surprising consequences for the relations between the complex system and its parts. In particular, it turns out that the state of a composite system is not uniquely defined by those of its components. What this means, or at least what it appears to mean, is that there are, according to quantum mechanics, facts about composite systems (and not just facts about their spatial configuration) that don't supervene on facts about their components; it means that there are facts about systems as wholes that don't supervene on facts about their parts and the way those parts are arranged in space. The significance of this feature of the theory cannot be overplayed; it is, in one way or another, implicated in most of its most difficult problems.

In a little more detail: if {viA} is an orthonormal basis for HA and {ujB} is an orthonormal basis for HB, then the set of pairs (viA, ujB) is taken to form an orthonormal basis for the tensor product space HA⊗HB. The notation viA⊗ujB is used for the pair (viA,uj B), and inner product on HA⊗HB is defined as:[6]

<viA⊗umB | vjA⊗unB> = <viA | vjA> <umB | unB>

It is a result of this construction that although every vector in HA⊗HB is a linear sum of vectors expressible in the form vA⊗uB, not every vector in the space is itself expressible in that form, and it turns out that

1. any composite state defines uniquely the states of its components.
2. if the states of A and B are pure (i.e., representable by vectors vA and uB, respectively), then the state of (A+B) is pure and represented by vA⊗uB, and
3. if the state of (A+B) is pure and expressible in the form vA⊗uB, then the states of A and B are pure, but
4. if the states of A and B are not pure, i.e., if they are mixed states (these are defined below), they do not uniquely define the state of (A+B); in particular, it may be a pure state not expressible in the form vA⊗u





Example 1 Solve .



Solution

There’s really not a whole lot to do in solving this kind of trig equation. All we need to do is divide both sides by 2 and the go to the unit circle.



So, we are looking for all the values of t for which cosine will have the value of . So, let’s take a look at the following unit circle.



Unit_Quarter_With_Angles

From quick inspection we can see that is a solution. However, as I have shown on the unit circle there is another angle which will also be a solution. We need to determine what this angle is. When we look for these angles we typically want positive angles that lie between 0 and . This angle will not be the only possibility of course, but by convention we typically look for angles that meet these conditions.



To find this angle for this problem all we need to do is use a little geometry. The angle in the first quadrant makes an angle of with the positive x-axis, then so must the angle in the fourth quadrant. So we could use , but again, it’s more common to use positive angles so, we’ll use .



We aren’t done with this problem. As the discussion about finding the second angle has shown there are many ways to write any given angle on the unit circle. Sometimes it will be that we want for the solution and sometimes we will want both (or neither) of the listed angles. Therefore, since there isn’t anything in this problem (contrast this with the next problem) to tell us which is the correct solution we will need to list ALL possible solutions.



This is very easy to do. Recall from the previous section and you’ll see there that I used



to represent all the possible angles that can end at the same location on the unit circle, i.e. angles that end at . Remember that all this says is that we start at then rotate around in the counter-clockwise direction (n is positive) or clockwise direction (n is negative) for n complete rotations. The same thing can be done for the second solution.



So, all together the complete solution to this problem is





As a final thought, notice that we can get by using in the second solution.

read all of this, i guarantee that none of it will make you laugh

I did. It didn't.

[tuberjustin]

its quantum mechanics

[Mizagium]

decision: Quantum Mechanics can never be funny.

[tuberjustin]

in SOVIET RUSSIA, quantum mechanics make everybody chuckle!

[thecheat]

Quantum physics can be funny.

[tuberjustin]

Jul 27, 2009 23:05:54 GMT -5 thecheat said:

Quantum physics can be funny.

explain

[Mizagium]

Jul 27, 2009 23:07:55 GMT -5 @tuberjustin said:

Jul 27, 2009 23:05:54 GMT -5 thecheat said:

Quantum physics can be funny.

explain

Only is explained by a retarded clown.

with balloon animals.

and a pie.

[Myrdraxxis]

it hard to typ right now...

[Mizagium]

why?