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States are or aren't unit vectors?

  1. Apr 1, 2014 #1
    I am a little confused by an elementary point. Something must be wrong with the following:

    On one hand, a Hermitian operator (which is not necessarily unitary) takes one state to another state. Hence a state need not be represented as a unit vector; its norm can be greater (or less than) one.

    On the other hand, the probability amplitude for one state to collapse upon measurement into another (or the same) state is the inner product of the two states; therefore the norm of the state's representative vector must be one.

    These contradict each other. What is wrong? Does it have anything to do with a distinction between pure and mixed states?
  2. jcsd
  3. Apr 1, 2014 #2
    Besides the point, but hermitian operators are rarely unitary and unitary operators are rarely hermitian. So the two don't really have much to do with eachother.

    The idea is one of equivalence relations. So, we usually work in some Hilbert space ##H##. If you don't know what a Hilbert space is, don't worry, let's just say we work in ##\mathbb{C}^n## for now. I will (suggestively) denote ##\mathbb{C}^n## with ##H##.

    The idea is that different elements of ##H## represent the same state. So we call nonzero elements ##\psi## and ##\psi^\prime## equivalent if they are "on the same ray". That is, if there is a complex number ##\alpha\in \mathbb{C}## such that ##\psi = \alpha \psi^\prime##. We write ##\psi\sim \psi^\prime##. Then ##\sim## is an equivalence relation on ##H##. This just means that it satisfies three very easy properties: http://en.wikipedia.org/wiki/Equivalence_relation#Definition

    Anyway, the idea is that equivalent elements of ##H## define the same state.

    Now, for many applications we don't just need any state, we need states of norm one. Luckily enough, if ##\psi## is any nonzero element of ##H##, then there is always an element ##\varphi\in H## that is equivalent to ##\psi## and that has norm one. Furthermore, ##\psi## and ##\varphi## define the same state, since they are equivalent. So instead of working with the arbitrary ##\psi##, we might as well work with ##\varphi## which has better properties (norm one) and which defines the same state anyway!
  4. Apr 2, 2014 #3
    Many, many thanks, micromass. That clears it up.
    [As Physics is not my field, I depend on books, and the ones I have do not make this point clear at the outset. I have been spoiled by self-contained mathematics text which are more explicit when dealing with representatives of equivalence classes.]
  5. Apr 2, 2014 #4

    Simon Bridge

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    Note: if ##\psi## is a state vector then ##a\psi## is also a state vector for the same state.
    All state vectors can be represented as unit vectors... but they don't have to be.

    The requirement for the state vector to also be a unit vector to compute probabilities is not part of the definition of a state vector. It is just that unit vectors are so useful for all kinds of reasons that we routinely normalize them.
  6. Apr 2, 2014 #5


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    It's a matter of taste. If the state is a ray, then the Born rule should contain a normalization.

    If the state is a unit vector, then the Born rule need not contain a normalization. This convention is more usual.

    But basically, you can put the normalization wherever you like - in the state, or in the Born rule.

    The time evolution of the state by the Schroedinger equation is unitary, so it doesn't matter which convention you use. For a time-independent Hamiltonian H, which is a Hermitian operator, the unitary time evolution of the state is given by |ψ(t)> = exp(-iHt)|ψ(t=0)>
  7. Apr 2, 2014 #6
    Thanks, Simon Bridge and atty.

    A question on Simon Bridge's response. Does the scalar a in

    need to be real, or can it also be complex? Thanks in advance for a response to this.

    To atty: for the time evolution operator this makes sense. It is the non-unitary ones that cause my confusion. You said that it is a matter of taste where one puts the normalization. I seem to have come across hybrids, where neither normalization is explicitly mentioned.
  8. Apr 2, 2014 #7


    Staff: Mentor

    It can be complex.

    The issue here is that states are in general not vectors. They are really operators, but certain types of operators called pure (mathematically they are of the form |x><x| where x is the element of the underlying vector space the operators act on) are usually looked on as elements of the vector space.

    The thing is these types of operators have an interesting property |ax><ax| = |x><x|. When considered as elements of a vector space one of the fundamental properties of a vector space is if |x> and |y> are any elements a|x> + b|y> where a and b are any complex numbers is also an element - this is the so called principle of superposition. If you superimpose the same state you get the same state again.

    Last edited: Apr 2, 2014
  9. Apr 2, 2014 #8

    Simon Bridge

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    It can be complex as bhobba said.
    BTW: have you got as far as Dirac bra-ket notation yet are are you still working in terms of ##\psi## and differential operators?

    You are getting replies in terms of a more complete (and advanced) picture of QM which may be confusing if you are just starting out.

    This freedom in QM can take a while to get used to.

    There is a lot of sloppy writing in QM so the normalization is not always explicit.
    If you can find an example, please provide a reference.
  10. Apr 2, 2014 #9
    Thanks, very much, Bill, aka bhobba. The concept of closure was helpful, thanks. Nonetheless, I am having a bit of a problem visualizing this. In the example you gave, I can see |ax><ax| = |x><x| because |ax><ax| =||a|| |x><x|, where ||a|| is a real number. However, if I represent a state by a complex number, say for simplicity cosA+i(sinA) for non-zero A, then I can think of this as a state located at angle A from my base; but if I know multiply that times its reciprocal, I get unity, so at my base vector. Thinking of them as operators, as you suggest, one is a rotation of A, the other is the unity operator. ?? I am obviously doing something elementary wrong, but that is one reason that this forum is so useful to non-physicists like me, and why I am so grateful for the help.
  11. Apr 2, 2014 #10
    Thanks again, Simon Bridge. First, to your questions: yes, I have gotten to bra-ket notation. In fact, I like it better than differential operators, despite my lack of mastery of it. I am not at the absolute beginning, but since I am mainly auto-didact in quantum physics (although with a background in mathematics), I still consider myself a beginner.
    It is not really a matter of sloppy writing, but rather I rely a lot on examples, and the examples tend to avoid the issue altogether. On one side, the concept of having states which are not (represented by) unit vectors are used in theorems and definitions yet in the examples only unit vectors are used; my main sources are "An Introduction to Hilbert Space and Quantum Logic" by David W. Cohen, and "Quantum Computation and Quantum Information" by Nielsen and Chuang. However, I may be guilty of unjustly pointing at these books, as I may have missed something: I don't claim that I understand everything in them. This is why I ask questions here.
  12. Apr 2, 2014 #11


    Staff: Mentor

    The thing I forgot to mention is x is a vector of unit norm so |a| is actually 1.

    Simon was spot on though - I am speaking in terms of the more advanced treatment of QM found in the following link:

    IMHO however its the correct treatment because there is nothing that needs to be unlearned from the more elementary texts.

    Those operators that states formally are mathematically, are positive operators of unit trace, and that unit trace bit implies the x in |x><x| is of unit norm.

    See if you can get a copy of Ballentine and read the first 3 chapters.

    You probably wont understand all the math - but don't worry - you likely will get the gist and that's all you really need to have the right concepts from the start. Later as you learn more you can fill in the gaps and eventually understand exactly what's going on.

    That book had a big effect on me. QM really involves two axioms.

    To spell out the details Ballentine bases QM on:

    1. Observables are Hermitian Operators, O, defined on a complex vector space whose eigenvalues give the possible outcomes of observations.
    2. The expected outcome of an observation E(O) = Trace (PO) where P is a positive operator of unit trace and is by definition called the state of the system.

    Axiom 2 is the Born Rule and to some extent is derivable from the first axiom via Gleason's Theorem.

    A quantum state of the form |x><x| is called pure. Convex sums a pure states are called mixed. It can be shown all states are mixed or pure. Pure states can be mapped to a vector space and because of that obey the superposition property because elements of a vector space are linear combinations of other vectors.

    Yep - QM is just two axioms and one of them is not entirely independent of the other. All QM is really the working out of that - Schrodinger's equation etc follows from symmetry - intrigued - read the book.

    Last edited by a moderator: May 6, 2017
  13. Apr 2, 2014 #12
    Thanks, Bill, for the book recommendation. I will definitely get it and read as much of it as I can.
    P.S. Any comment on my example that multiplication by a complex number would seem to add a rotation, thereby changing the vector? If not, I am sure I will get to it as I work through the book.
    Last edited: Apr 2, 2014
  14. Apr 2, 2014 #13


    Staff: Mentor

    Good pick up.

    Indeed it is - its just states are invariant to it.

    This is an interesting symmetry and is an example of a gauge symmetry - specifically its U(1) symmetry. In fact, and I wont explain it yet - you really need to internalise a few things before I let the cat out of the bag - its actually one of the rock bottom foundations of QM - but just to whet you appetite a bit check out:

    I could detail it now, and I have posted about it in the past - but it's really utterly trivial unless you have your thinking cap on a bit to see what's going on. People would basically comment - first they don't get it - then so? That's the problem with this view of physics - you hear it all the time but unless you are alert to it its importance it escapes attention:
    'Although simply written, this is not a book for beginers. On the other hand it doesn't hurt to read it early and think about it for a long time, rereading it from time to time, in order finally to get the main point. Wigner points out that the basis for answering the question posed by him, 'Why is it possible to discover laws of nature?' is explained in every elementary physics text but the point is too subtle, is therefore lost on nearly every reader. The answer, he explains convincingly, lies in invariance principles. As an example, were local Galilean invariance not true it would have been impossible for Galileo to have discovered any law of motion at all. The same holds for local translational, rotational and time-translational invariance. Inherent in Wigner's argument is the explanation why the so-called principle of general covariance is not the foundation of general relativity, which also is grounded in the local invariance principles of special relativity.'

    A deep analysis of fields such as the EM field shows that gauge symmetry is its deep explanation - in fact U(1) symmetry is what's responsible for EM:

    Strange but true.

    Now you are getting to the deep, the really deep, revelations modern physics has told us about the world:

    Once you fully internalise it you will sit there stunned.

    Philosophers always want to sit around arguing what exactly is reality (yawn - they get nowhere - its a wordfest) - what at rock bottom is it really. They get nowhere of course because you always must double check your speculations by asking nature if its true - that's called doing experiments - which is basically what science is about. Quietly, without fan fare physicists and mathematicians have figured it out.

    It should be more widely known. Me telling you about it however will not actually replace you experiencing it yourself - which I invite you to do.

    Start out with Noether's Theorem:

    Last edited by a moderator: May 6, 2017
  15. Apr 2, 2014 #14


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    There are only two time evolutions in quantum mechanics. There's the unitary evolution given by the Schroedinger equation as mentioned above. The other is the collapse of the wave function after a measurement, which is non-unitary. After the collapse, the wave function may be unnormalized if we use the convention that a state is a ray in a vector space, or it may be normalized if we use the convention that a state is a unit vector in a vector space.

    Yes, it is confusing that neither convention is explicitly mentioned in many texts, or that people switch back and forth between conventions without mentioning.

    Here are some notes which do state the normalization explicitly. Both choose to make the state a unit vector.
    http://arxiv.org/abs/1110.6815 (Postulate 1, p3)
    http://www.johnboccio.com/research/quantum/notes/QC10th.pdf [Broken] (Postulate 1, p108).

    Edit: Looking at the discussion of the gauge symmetry, it seems it is formally more complete to make the state a ray, and put the normalization in the Born rule.
    Last edited by a moderator: May 6, 2017
  16. Apr 2, 2014 #15
    To Bill (bhobba) and etty: enormous thanks! I will have to invent a new large cardinal hypothesis to express my thanks. (The name will be a problem: "ineffable", "indescribable", "huge" have all been taken.)

    I have downloaded or bookmarked all these suggestions (including the book "Quantum Mechanics: A modern development") and glanced at them with anticipatory pleasure.

    etty: yes, the importance (and beauty) of Noether's Theorem has often been brought to my attention, and I was just going through the proof when I started this post. I agree with etty about traditional philosophers; they might take a cue from Max Tegmark's theis that mathematics not only represents reality, but IS reality. I know that my first university required philosophy doctoral students to take a course "Introduction to Quantum Mechanics" (as well as a year of Mathematical Logic).

    I stand corrected on my earlier statement that "Quantum Computing and Quantum Information" by Nielsen and Chuang (which I actually have in hard cover) skips over the problem of normalization; it does mention it briefly twice (Postulate 1, as etty pointed out, and between exercises 2.6 and 2.7).

    Now I have lots to work through, and may be bothering the great people on this Forum again on individual points. Thanks again!
  17. Apr 2, 2014 #16


    Staff: Mentor

    That's a very seductive view that I once held along with greats like Roger Penrose.

    Now I hold the weaker view it simply describes reality.

    The reason I changed my mind is some thought provoking stuff by Murray Gell-Mann:

    You have embarked on a fascinating journey - enjoy.

    When I started on it all those moons ago I never expected it to lead to symmetry as the real rock bottom foundation of things. I often tell people about it, but really its only a shadow of experiencing it yourself.

    Last edited by a moderator: Sep 25, 2014
  18. Apr 3, 2014 #17
    to Bill (bhobba), thanks for the video, which I watched. I will say no more about that here, for two reasons:
    (a) as you say, I need to go through the mathematical details myself, and
    (b) it is veering off into philosophy (my fault, I started the detour), which would be off-topic.
    So, off onto the yellow brick road.....
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