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The basic math of quantum mechanics

  1. Dec 15, 2014 #1
    From another thread, this link was posted.

    I would love to read this paper, however I'm stumped in the very first equation.

    [itex]|\psi \rangle =\sum|a_k\rangle\langle a_k|\psi\rangle[/itex]

    I actually have some basis in the ideas behind QM, and a general love of math, but I just have no idea how to approach it mathematically. I have tried to look up the Dirac notation before and got some answer, but it was as equally hard to understand as my original question. This seems to be my main problem with even beginning to try and teach myself some semblance of QM, all I find as explanations to my questions are more confusing than my original question!

    I wouldn't be against buying a book, but I suspect that any book I buy will have equations similar to this and I will be in the exact same predicament, but a 100$ poorer. I don't think this math is hard, I just don't know what it is. Like if you tried to teach yourself Einstein Notation but had no idea what a summation over i was; it just wouldn't work. If I could get some basic understanding of the notation used throughout the linked paper, I think I would be well on my way to being able to more fully understand the real quantities of QM and not just some 'kind of' answers that I am forced to search for because of my lack of mathematical understanding.

    A little about me background: I just graduated with my masters in civil - structural engineering, so I have some basic understanding of tensors and higher level analysis etc, I just have never needed to be exposed to the higher mathematics which are so crucial for things like QM and GR (And finally, string theory or LQG). I hope to be able to pursue a degree in theoretical physics next, as my currently chosen career progresses, but in the mean time I would love to be able to pursue these topics without being completely overwhelmed in simple (or maybe not so simple?) notation.
     
  2. jcsd
  3. Dec 15, 2014 #2

    jedishrfu

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    Last edited by a moderator: May 7, 2017
  4. Dec 16, 2014 #3

    Stephen Tashi

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    Setting the budget at less than $100, where is the first placed you are stumped in the Wikipedia article on the notation? http://en.wikipedia.org/wiki/Bra–ket_notation
     
  5. Dec 16, 2014 #4
    Find V.Subrmanium IIT Madras lectures on YOUTUBE.I have similar interest and found these about 30 lectures very good.So far I studied only first five which contain mathematics background.
     
  6. Dec 16, 2014 #5
    If you know linear algebra, bra-ket notation is just a slightly esoteric way of writing out vectors and linear operators. ##|v\rangle## is just a vector, e.g., ##\vec{v}##. ##\langle u|## is the linear functional dual the vector ##\vec{u}## via an inner product defined on the vector space. In other words, ##\langle u | v \rangle##, which we interpret as the functional ##\langle u |## applied to the vector ##|v\rangle##, is just another way of writing ##\vec{u} \cdot \vec{v}##. In fact, that's where the notation comes from in the first place. One sometimes write a general inner product as ##(\vec{u},\vec{v})## or ##\langle\vec{u},\vec{v}\rangle## so just Dirac looked at that and said, "Well we've got a bracket of two vectors. Let's just use the right side of that as the notation for vectors themselves and call them "kets" and the left side as the notation for dual vectors and call them (hehe) "bras". Then I can make an inner product bracket just multiplying a bra and a ket. Get it guys? Bra-ket? Bracket? Guys?" Pretty much everything else follows from this. If you've got a linear operator ##\hat{O}## on your vector space then ##\langle u | \hat{O} | v \rangle## is ##\vec{u} \cdot (\hat{O} \vec{v})##. If I've got an orthonormal basis of vectors (which we use almost exclusively in quantum mechanics) where I label the i-th vector ##|a_i\rangle## then ##\langle a_i | \hat{O} | a_j \rangle## is the ij-th element of the matrix representative of ##\hat{O}## with respect to this basis, ##O_{ij}##. Like I said, exactly the same as standard linear algebra. The expression you're having trouble with is the analogous result for vectors themselves: the i-th component of the coordinate vector of some vector ##\vec{u}## in this basis is just ##\vec{a}_i \cdot \vec{u} = \langle a_i | u \rangle##.

    It seems a bit silly at first, but once you get used to it bra-ket notation has very nice properties. You can only put vectors together in so many ways in linear algebra (inner products, tensor products, and linear combinations basically exhaust the list) and when you use bra-ket notation you can usually see at a glance if you've done something that doesn't make sense.
     
  7. Dec 16, 2014 #6
    I'll give a concrete example. It might make things clearer.

    If you have a two dimension vector you can pick a coordinate system (which doesn't have to be at right angles) then you can define two basic vectors a1=(1,0) and a2=(0,1) from which all other vectors can be made. So choose a vector like ψ=(c,d)

    (c,d) = (1,0) * c + (0,1) * d

    but c = (1,0).(c,d)
    and d= (0,1).(c,d)

    so

    (c,d) = (1,0) * (1,0).(c,d) + (0,1) * (0,1).(c,d)

    or

    Ψ =∑ a * (a.Ψ)

    which is basically the equation you were after. All it says is that any vector can be built out of basic vectors.
     
  8. Dec 16, 2014 #7

    Fredrik

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    The Riesz representation theorem says that if ##\mathcal H## is a Hilbert space over ##\mathbb C##, then for each bounded linear functional ##\phi:\mathcal H\to\mathbb C##, there's a unique ##x\in\mathcal H## such that ##\phi=\langle x,\cdot\rangle##, i.e. a unique ##x\in\mathcal H## such that ##\phi(y)=\langle x,y\rangle## for all ##y\in\mathcal H##.

    Bra-ket notation is essentially just the convention to denote a typical vector by ##|x\rangle## instead of ##x##, and the corresponding linear functional by ##\langle x|## instead of by ##\langle x,\cdot\rangle##. (As already indicated above, this denotes the map that takes an arbitrary ##y\in\mathcal H## to ##\langle x,y\rangle##).

    So bras are linear functionals that take vectors (that we now call kets) to complex numbers. The translation from bra-ket notation to standard notation is pretty simple: ##\langle x|y\rangle## is defined as ##\langle x||y\rangle## (this is the value of ##\langle x|## at ##|y\rangle##, i.e. the output produced by ##\langle x|## when it takes ##|y\rangle## as input), so we have
    $$\langle x|y\rangle = \langle x||y\rangle =\langle x,\cdot\rangle(y)=\langle x,y\rangle.$$ The formula you asked about is just a vector expressed using an orthonormal basis. Suppose that ##x\in\mathcal H## and that ##\{e_i\}_{i=1}^\infty## is an orthonormal basis for ##\mathcal H##. Then there's a unique sequence ##(c_i)_{i=1}^\infty## in ##\mathbb C## such that ##x=\sum_{i=1}^\infty c_i e_i##. It's very easy to find a formula for those numbers. For all positive integers ##i##, we have
    $$\langle e_i,x\rangle =\left\langle e_i,\sum_{j=1}^\infty c_j e_j\right\rangle =\sum_{j=1}^\infty c_j \langle e_i,e_j\rangle = c_i.$$ So we can always write
    $$x=\sum_{i=1}^\infty \langle e_i,x\rangle e_i.$$ If we use bra-ket notation as discussed above, and also the convention that it's OK to write the product of a number c and a vector x as xc rather than cx, then we can rewrite this as
    $$|x\rangle =\sum_{i=1}^\infty |e_i\rangle\langle e_i|x\rangle.$$
     
    Last edited: Dec 16, 2014
  9. Dec 16, 2014 #8

    Strilanc

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    Symbols bracketed like [itex]|X\rangle[/itex] are called kets and represent an entry in a column vector (or a fixed linear combination of entries). So, for example, [itex]|X\rangle[/itex] might be shorthand for [itex]\left[ \begin{array}{c} 1 \\ 0 \\0 \end{array} \right][/itex] or for [itex]\left[ \begin{array}{c} 0 \\ 1 \\0 \end{array} \right][/itex] or for [itex]\left[ \begin{array}{c} 0 \\ 0 \\ -1 \end{array} \right][/itex] or for [itex]\left[ \begin{array}{c} \sqrt{1/2} \\ i \sqrt{1/2} \\0 \end{array} \right][/itex].

    Symbols bracketed like [itex]\langle X |[/itex] are called bras and represent an entry in a row vector. For example, [itex]\langle X |[/itex] could be [itex]\left[ \begin{array}{cccc} 0 & 0 & 0 & 1 \end{array} \right][/itex] or [itex]\left[ \begin{array}{cccc} \sqrt{1/2} & i \sqrt{1/2} & 0 & 0 \end{array} \right][/itex].

    Placing a bra next to a ket (resulting in a "bracket", har har) means "multiply these two things". So [itex]|X\rangle\langle Y|[/itex] means multiply some column vector pointing along X by a row vector pointing along Y. Note that the order matters quite a bit here: row-times-column (bra-ket) gives you a single complex value as the result while column-times-row (ket-bra) gives you an NxN matrix as the result. When you multiply a ket [itex]| X \rangle[/itex] by its corresponding bra [itex]\langle X |[/itex], the 1x1 ordering gives you the inner product and the NxN ordering gives you the outer product.

    So what [itex]|\psi \rangle =\sum|a_k\rangle\langle a_k|\psi\rangle[/itex] is saying is something like "you can break any wavefunction into parts corresponding to the outer-products of each [itex]a_k[/itex]". Alternatively, you can think of it as "if you keep the stuff along [itex]a_k[/itex] for each k, you've kept everything (without repetition)". In other words, [itex]a[/itex] is a basis for the state space.
     
    Last edited: Dec 16, 2014
  10. Dec 16, 2014 #9

    bhobba

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    Although it would not be my first port of call - Susskinds books and lectures are much better for that - but after you get a smattering I would go through Chapter 2 of Ballentine:
    https://www.amazon.com/Quantum-Mechanics-Modern-Development-Edition/dp/9814578584

    Things should be a lot clearer after that.

    Its a bit more advanced but continue on to Chapter 3 to understand the true basis of Schroedingers equation etc.

    Thanks
    Bill
     
    Last edited by a moderator: May 7, 2017
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