# Normalising Quantum States

• latentcorpse

#### latentcorpse

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

but what about $\psi=|1,-1>+2|0,0>+|-1,1>$
i get $\psi^*=<1,-1| +2<0,0| + <-1,1|$

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

$<1,-1|1,-1>+<-1,1|-1,1>+<1,-1|-1,1>+<-1,1|1,-1>=0$ which is impossible

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

so which is right (if either) and why?

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).

yeah i just looked at the answer and it says that $|1,-1>$ is already normalised. two question on this:
(i) why is it already normalised and how did you know this?
(ii) i got $<1,-1|1,-1>=2$ 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 $\frac{1}{\sqrt{6}}$ so that implies $\psi^* \psi=6$ - what's gone wrong there?

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 sort of 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.

I think you're confusing notation. $$|1,-1 \rangle$$ 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 $$|1\rangle$$ and $$|2\rangle$$, and a vector written as a linear combination of these $$|a\rangle = 2|1\rangle + 4|2\rangle$$ , then you say that the components of $$|a\rangle$$ 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.

clearly i am confusing notation but I am 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?

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.

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.

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

You forgot to complex conjugate.

You forgot to complex conjugate.

Yes, true. Hopefully the point was made however.

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 $<+\frac{1}{2},-\frac{1}{2}|-\frac{1}{2},+\frac{1}{2}>$ - could convert to matrices and do it that way but is there a way of doing it directly with the bras and kets?

fo $j=1$ we can have 3 possible $m_j$'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 $|\psi>$ and $|\phi>$?

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).

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 $<+\frac{1}{2},-\frac{1}{2}|-\frac{1}{2},+\frac{1}{2}>$ - 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

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

fo $j=1$ we can have 3 possible $m_j$'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 $|\psi>$ and $|\phi>$?
No, he is representing the bras and kets as vectors first, and taking the scalar product of the vectors.

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 $<+\frac{1}{2},-\frac{1}{2}|-\frac{1}{2},+\frac{1}{2}>$ - could convert to matrices and do it that way but is there a way of doing it directly with the bras and kets?

fo $j=1$ we can have 3 possible $m_j$'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 $|\psi>$ and $|\phi>$?
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
$$\Psi$$> = ($$\Psi$$_1, $$\Psi$$_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.\
$$\Psi$$ = 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 $$\Psi$$ = ($$\Psi$$_1, $$\Psi$$_2),
$$\Psi$$ does not equal to l $$\Psi$$_1, $$\Psi$$_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?
$$\Psi$$_1, $$\Psi$$_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,
$$\Psi$$_1, $$\Psi$$_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

ok. yeah, i remember this now:

it's like $|\psi>=\psi_1|j=\frac{1}{2},m_j=\frac{1}{2}>+\psi_2|j=\frac{1}{2},m_j=-\frac{1}{2}>$ and so we can write $\psi=(\psi_1,\psi_2)$ - 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 $|1,-1>+2|0,0>+|-1,1>$. j is different in each of these kets (lets call them $k_1,k_2,k_3$) and so surely the basis kets are going to be different for each $k_i$.

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

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

similarly for k3

my point is - do i have to normalise each of the k1,k2,k3 seperately?

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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

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

$\left( (1,0,0)+2(1)+(0,0,1) \right) \left( (1,0,0)+2(1)+(0,0,1) \right)^T$
which gives 6 which is fair enough

why can we neglect the $(1,0,0)(1)$ term though - is it just because it's not mathematically acceptable?

I think that "(1)" should actually be "(0,1,0)". millitz just made a typo.

so, for this l=1 system, we write out all the basis kets:

$|1,1>,|1,0>,|1,-1>,|0,1>,|0,0>,|0,-1>,|-1,1>,|-1,0>,|-1,-1>$

(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 $|1,0>+3|-1,0>$

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

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

can someone confirm both those points please?

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.