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Normalising Quantum States

  • #1
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normalising [itex]\psi=|1,-1>[/itex] is easy as [itex]\psi^*=<1,-1|[/itex]
and then [itex]\psi^* \psi = <1,-1|1,-1>=2[/itex]
which gives [itex]\psi= \frac{1}{\sqrt{2}} |1,-1>[/itex] for the normalised ket.

but what about [itex]\psi=|1,-1>+2|0,0>+|-1,1>[/itex]
i get [itex]\psi^*=<1,-1| +2<0,0| + <-1,1|[/itex]

now im guessing that seeing as i want to normalise the whole wavefunction [itex]\psi[/itex] i can't just normalise the kets individually so multiplying every term by every other term i get non-zero contributions giving

[itex]<1,-1|1,-1>+<-1,1|-1,1>+<1,-1|-1,1>+<-1,1|1,-1>=0[/itex] which is impossible


however if i can normalise them seperately then i would get for my normalised wavefunction
[itex]\psi=\frac{1}{\sqrt{2}} |1,-1> +2|0,0> + \frac{1}{\sqrt{2}} |-1,1>[/itex]

so which is right (if either) and why?
 

Answers and Replies

  • #2
dx
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Are these kets basis kets? Basis kets are assumed to be normalized already, i.e. if |i> is a basis ket, then <i|i> = 1. At the beginning of your post, you wrote "<1,-1|1,-1> = 2". Where did you get that? I hope you didn't get 2 by doing (1)(1) + (-1)(-1) = 2, since the 1 and -1 are labels for the states (unless you're using some weird non-standard notation).
 
  • #3
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yeah i just looked at the answer and it says that [itex]|1,-1>[/itex] is already normalised. two question on this:
(i) why is it already normalised and how did you know this?
(ii) i got [itex]<1,-1|1,-1>=2[/itex] using the method described on page 4 of
http://www.ph.ed.ac.uk/teaching/course-notes/documents/64/850-lecture13.pdf
why doesn't this work here? surely if it's already normalised we should get 1 when we do this?

secondly, for the other state i was asking about (the one with 3 kets), it gets a normalisation factor of [itex]\frac{1}{\sqrt{6}}[/itex] so that implies [itex]\psi^* \psi=6[/itex] - what's gone wrong there?

thanks for your help!
 
  • #4
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Notice that these 1, -1 indicates the j, m_j quantum number (and it is in your lecture note too).
but the ket itself is a normalized function which depends on these two parameters j, and m_j.
I think the lecture note is also a bit confusing.
in the example on the 4th page, s/he use the example of j=1/2 m_j=1/2> and j=1/2 m_j=-1/2>
Notice that there are only 2 possible states for j =1/2, ie, m_j = 1/2 or -1/2
Therefore that is where the (1,0), (0,1) comes from:it just indicates 2 different states.
Now, look back to your questions. If j = 1, how many possible states would you have?
Now, assign one unit vector to each state, what would be the product of the state?
(and yes, I got 1. You can further use it to sorta check that it will equal to 0 if the two vectors have different energy states and therefore are orthogonal).
And about your second question, if you think through what I just said, you can probably get the one over square root of 6: just simply count how many 1 do you get at the end.
 
  • #5
dx
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I think you're confusing notation. [tex] |1,-1 \rangle [/tex] is not a vector with components 1 and -1. These numbers are labels which tell you what its eigenvalues are. If you have two basis kets [tex]|1\rangle[/tex] and [tex]|2\rangle[/tex], and a vector written as a linear combination of these [tex]|a\rangle = 2|1\rangle + 4|2\rangle[/tex] , then you say that the components of [tex]|a\rangle[/tex] in this basis are (2,4). If you have two vectors with components (a,b) and (c,d), then their scalar product is a*b + c*d, where * means complex conjugate.
 
  • #6
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clearly i am confusing notation but im still unsure of what's going on.

in my notes why does he just multiply the 2 matrtices together on page 4 if your telling me you cannot do that? what are the kets if they aren't vectors? and how do we scalar product two kets-isn't it just as on page 4?
 
  • #7
Redbelly98
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Let's look at a different example. Suppose we have a system of two spin-1/2 particles. The basis states can be represented by the kets:

|+½, +½>, |+½, -½>, |-½, +½>, and |-½, -½>​

Note that there are 4 basis states. So we could, instead, use a 4-component vector to represent the basis states:

|+½, +½> → (1,0,0,0)
|+½, -½> → (0,1,0,0)
|-½, +½> → (0,0,1,0)
|-½, -½> → (0,0,0,1)

And to represent an arbitrary state:


a|+½, +½> + b|+½, -½> + c|-½, +½> + d|-½, -½>

→ (a,b,c,d)


You can use the usual matrix operations with (a,b,c,d), but not with the |±½, ±½> as you were essentially doing.

So:

(a1,b1,c1,d1)·(a2,b2,c2,d2)t = a1a2 + b1b2 + c1c2 + d1d2

but

<+½, +½|-½, -½> (½)(½) + (-½)(-½)

Hope that helps.
 
  • #8
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Maybe my explanation is confusing too.
The reason s/he used (0,1), (1,0) in his lecture note is that
the kets has j = 1/2
What could m_j be for j = 1/2?
Well, the rules tell us that m_j could only be +1/2, and -1/2, and nothing else.
so lj m_j> = l1/2 1/2> and l1/2 -1/2> are the only two possible wave function/eigenfunction/vector that are allowed with j = 1/2, right?
And that is the reason why s/he choose 2 orthogonal unit vector to represent the two states.

Now, think back to your question, if j = 1, how many degenerated states do you have? ie, what are the possible m_j values are allowed if j = 1.
Clearly you'll get some #. Now you assign each possible j/m_j pairs to a unit vector, how many vectors do you need? What is the dimension of the vector?
I'll stop here first.
 
  • #9
dx
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(a1,b1,c1,d1)·(a2,b2,c2,d2)t = a1a2 + b1b2 + c1c2 + d1d2
You forgot to complex conjugate.
 
  • #10
Redbelly98
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You forgot to complex conjugate.
Yes, true. Hopefully the point was made however.
 
  • #11
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ok. so if i can use matrix operations on (a,b,c,d) but not on the kets, how would i calculate a bra-ket, say [itex]<+\frac{1}{2},-\frac{1}{2}|-\frac{1}{2},+\frac{1}{2}>[/itex] - could convert to matrices and do it that way but is there a way of doing it directly with the bras and kets?

fo [itex]j=1[/itex] we can have 3 possible [itex]m_j[/itex]'s: 1,0,-1. so we can represent the basis kets with vectors (1,0,0),(0,1,0),(0,0,1). how did you know j=1 here? and how do i proceed from here?


HOWEVER to be honest i still don't see why i can't say <1,-1|1,-1>=2 using matrices - isn't that what he does on page 4 when he takes the scalar product of [itex]|\psi>[/itex] and [itex]|\phi>[/itex]?
 
  • #12
dx
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To convert a vector into a matrix (a row or column matrix), you need a basis. Let's use Redbelly98's example:

|+½, +½>, |+½, -½>, |-½, +½>, and |-½, -½>

This is a basis. Now, what is the matrix corresponding to |+½, +½>? It is (1,0,0,0), because

|+½, +½>, = 1|+½, +½> +0|+½, -½> + 0|-½, +½> +0|-½, -½>

The matrix corresponding to |+½, +½> is NOT (+½, +½).
 
  • #13
Redbelly98
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ok. so if i can use matrix operations on (a,b,c,d) but not on the kets, how would i calculate a bra-ket, say [itex]<+\frac{1}{2},-\frac{1}{2}|-\frac{1}{2},+\frac{1}{2}>[/itex] - could convert to matrices and do it that way but is there a way of doing it directly with the bras and kets?
Yes, we can do it with bras and kets.

The basis states are orthogonal and normalized, so

<basis state A|a different basis state> = 0 always

<basis state A|basis state A> = 1 always

[itex]<+\frac{1}{2},-\frac{1}{2}|-\frac{1}{2},+\frac{1}{2}> \ [/itex] involves different basis states (which are orthogonal), so the result is 0.

fo [itex]j=1[/itex] we can have 3 possible [itex]m_j[/itex]'s: 1,0,-1. so we can represent the basis kets with vectors (1,0,0),(0,1,0),(0,0,1). how did you know j=1 here? and how do i proceed from here?

HOWEVER to be honest i still don't see why i can't say <1,-1|1,-1>=2 using matrices - isn't that what he does on page 4 when he takes the scalar product of [itex]|\psi>[/itex] and [itex]|\phi>[/itex]?
No, he is representing the bras and kets as vectors first, and taking the scalar product of the vectors.
 
  • #14
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ok. so if i can use matrix operations on (a,b,c,d) but not on the kets, how would i calculate a bra-ket, say [itex]<+\frac{1}{2},-\frac{1}{2}|-\frac{1}{2},+\frac{1}{2}>[/itex] - could convert to matrices and do it that way but is there a way of doing it directly with the bras and kets?

fo [itex]j=1[/itex] we can have 3 possible [itex]m_j[/itex]'s: 1,0,-1. so we can represent the basis kets with vectors (1,0,0),(0,1,0),(0,0,1). how did you know j=1 here? and how do i proceed from here?


HOWEVER to be honest i still don't see why i can't say <1,-1|1,-1>=2 using matrices - isn't that what he does on page 4 when he takes the scalar product of [itex]|\psi>[/itex] and [itex]|\phi>[/itex]?
I think that you just answer it yourself =)
You said that with j = 1, m_j =-1, isn't it just the ket l1, -1>?
And didn't you just say that it could be represented by a vector (0,0,1)?
What is (0,0,1)*(0,0,1)T?
Not 2, right?
The reason I choose j = 1 is that you are wondering why <1,-1|1,-1> =1 but not 2. Well, l1,-1> is lj = 1, m_j=-1>, right? The reason I choose j =1 is because you chose it to be 1.
And as you say, l1,-1> is a vector (0,0,1) but not (1,-1).
If it is (1,-1), you'll obviously get something really weird.
And notice that in his lecture note
[tex]\Psi[/tex]> = ([tex]\Psi[/tex]_1, [tex]\Psi[/tex]_2) is nothing more than a linear combination of 2 basis states
(1,0), and (0,1).
And the reason that you only have 2 basis states is that there are only 2 possible states for j=1/2
Whereas j=1, there are 3 possible states, and for an arbitrary function.\
[tex]\Psi[/tex] = a (1,0,0)+b(0,1,0)+c(0,0,1)
And these 3 basis functions represent the m_j = 1,0,-1.
And while [tex]\Psi[/tex] = ([tex]\Psi[/tex]_1, [tex]\Psi[/tex]_2),
[tex]\Psi[/tex] does not equal to l [tex]\Psi[/tex]_1, [tex]\Psi[/tex]_2>.
Read his note a bit more careful for the previous part.
How did he or she defines the ket? s/he defines it by j, mj value, right?
[tex]\Psi[/tex]_1, [tex]\Psi[/tex]_2 is simply the amount of lj=1/2, m_j=1/2> and lj = 1/2, m_j = -1/2>
And since we know that the probability must add up to one,
[tex]\Psi[/tex]_1, [tex]\Psi[/tex]_2 needed to be normalized (ie, that is the number in front your brakets instead of the number inside the braket).
Hope if it clears some of your question
 
  • #15
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ok. yeah, i remember this now:

it's like [itex]|\psi>=\psi_1|j=\frac{1}{2},m_j=\frac{1}{2}>+\psi_2|j=\frac{1}{2},m_j=-\frac{1}{2}>[/itex] and so we can write [itex]\psi=(\psi_1,\psi_2)[/itex] - that's one preoblem solved.

but how did you know that what i'd written down was a basis ket in the first place when even i didn't know that let alone write it in part of the question?

and finally,

for normalising [itex]|1,-1>+2|0,0>+|-1,1>[/itex]. j is different in each of these kets (lets call them [itex]k_1,k_2,k_3[/itex]) and so surely the basis kets are going to be different for each [itex]k_i[/itex].

e.g. in k1, j=1 and so the basis kets are
[itex]|1,1>[/itex]
[itex]|1,0>[/itex]
[itex]|1,-1>[/itex]
and so k1=(1,0,0)

but then for k2, j=0 and so there is only one possible basis ket - it is [itex]|0,0>[/itex] and so k2=(1)

similarly for k3

my point is - do i have to normalise each of the k1,k2,k3 seperately?
 
Last edited:
  • #16
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For your final question, I was concerning about that, too.
j = 1 state mix with j = 0 state mix with j = -1 state (j = -1 doesn't even make sense to me)!?

However, the math is still the same. Make sure that all three different states are represented by 3 different unit vector.
say, (1,0,0), (1), (0,0,1) and what not
Do the product. And you'll see that the cross terms either doesn't make sense or equal to 0
and at the and, you'll get 6 one (be careful while doing the l0,0>).
And that is where root 6 comes from
 
  • #17
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it's question 2 on the following
http://www.ph.ed.ac.uk/teaching/course-notes/documents/64/889-tut8qm05.pdf
so clearly they have its actually m1=-1 in the third ket which is ok.

so is it like this

[itex]\left( (1,0,0)+2(1)+(0,0,1) \right) \left( (1,0,0)+2(1)+(0,0,1) \right)^T[/itex]
which gives 6 which is fair enough

why can we neglect the [itex](1,0,0)(1)[/itex] term though - is it just because it's not mathematically acceptable?
 
  • #18
Redbelly98
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I think that "(1)" should actually be "(0,1,0)". millitz just made a typo.
 
  • #19
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so, for this l=1 system, we write out all the basis kets:

[itex]|1,1>,|1,0>,|1,-1>,|0,1>,|0,0>,|0,-1>,|-1,1>,|-1,0>,|-1,-1>[/itex]

(i) do we just ASSUME that all the basis kets are normalised to 1?
(ii) we can just work from here yes e.g.

normalise [itex]|1,0>+3|-1,0>[/itex]

we get [itex] \left( <1,0|+3<-1,0| \right) \left(|1,0>+3|-1,0> \right) = <1,0|1,0> +3<1,0|-1,0> + 3<-1,0|1,0> + 9<-1,0|-1,0> = 1 + 9=10[/itex] using the orthogonality of eigenstates

and so the normalised wavefunction is [itex]\psi=\frac{1}{\sqrt{10}}(|1,0>+3|-1,0>)[/itex]

can someone confirm both those points please?
 
  • #20
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After reading your problem, it makes more sense to me right now.
Now I finally understand that the first and second # represent the 2 m value for 2 particles.
So actually this'll give us 9 basic vectors (which you just listed).
So theoretically, you should use some vector like (1,0,0,0,0,0,0,0,0) instead of (1,0,0). But I think you get the idea ;)
Yes, I made a wrong assumption. I thought the 2 numbers represent j, and m_j value; which is not true.
And for 1/(10)^(1/2)..., I don't see anything wrong with it.
 
  • #21
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sweet. thanks for your help.
 

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