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Normalising Quantum States

  1. Apr 10, 2009 #1
    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?
  2. jcsd
  3. Apr 10, 2009 #2


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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).
  4. Apr 10, 2009 #3
    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
    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!
  5. Apr 10, 2009 #4
    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.
  6. Apr 11, 2009 #5


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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.
  7. Apr 11, 2009 #6
    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?
  8. Apr 11, 2009 #7


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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.


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


    <+½, +½|-½, -½> (½)(½) + (-½)(-½)

    Hope that helps.
  9. Apr 11, 2009 #8
    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.
  10. Apr 11, 2009 #9


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    You forgot to complex conjugate.
  11. Apr 11, 2009 #10


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    Yes, true. Hopefully the point was made however.
  12. Apr 11, 2009 #11
    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]?
  13. Apr 11, 2009 #12


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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).
  14. Apr 11, 2009 #13


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    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.

    No, he is representing the bras and kets as vectors first, and taking the scalar product of the vectors.
  15. Apr 11, 2009 #14
    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
  16. Apr 11, 2009 #15
    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
    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: Apr 11, 2009
  17. Apr 12, 2009 #16
    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
  18. Apr 12, 2009 #17
    it's question 2 on the following
    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?
  19. Apr 12, 2009 #18


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    I think that "(1)" should actually be "(0,1,0)". millitz just made a typo.
  20. Apr 12, 2009 #19
    so, for this l=1 system, we write out all the basis kets:


    (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?
  21. Apr 12, 2009 #20
    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.
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