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Hilbert/Fock space spanned by non-interacting states

  1. Nov 8, 2007 #1
    Correct me if wrong, but in non-relativistic QM, the Hilbert space of two interacting spins is spanned by the tensor product of non-interacting states (the so called spin-addition). For addition of two 1/2 spins for example:

    |1,1> = |+>|+>
    |1,0> = (|+>|-> + |->|+>) / sqrt(2)
    |1,-1> = |->|->

    |0,0> = (|+>|-> - |->|+>) / sqrt(2)

    Now, why the Fock space of the interacting theory in QFT can't be spanned by the non-interacting states i.e. why the interaction and non-interaction Fock spaces are not the same?
     
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  3. Nov 8, 2007 #2

    nrqed

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    Why do you say they are not the same?
    As far as I know the Fock spaces are built out of all possible non-interacting states. The only difference with the usual Qm approach is that now transitions between different sectors (let's say fixed number states) are allowed.
     
  4. Nov 8, 2007 #3
    Because I have this course in condensed matter, talking about interacting electron gas, mean field theory and later superconductivity and the instructor claimed that

    [tex] <state| a^+ (k) a(k') |state> [/tex]

    could be non-zero even when k is not equal to k'.

    That is impossible, if the sate is spanned in the non-interacting Fock space with the usual occupation number basis and the creation and annihilation operators act on that basis in the usual way. The creation and annihilation operators must create and destroy the same one-particle state (k=k') for the two sides of the < > to match and give non zero result.

    So my natural assumption was that the |state> of the interacting theory CAN'T be spanned by the non-interacting basis, otherwise the instructor's claim would be wrong. Therefore the two Fock spaces, interacting and non interacting, must not be equal.

    Another possible explanation would be that the creation/annihilation operators above are Fourier coefficients of the field operator and in interacting theory do NOT act on the non-interacting basis like the operators in non-interacting theory.

    So which one is true?
     
    Last edited: Nov 8, 2007
  5. Nov 8, 2007 #4

    samalkhaiat

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  6. Nov 8, 2007 #5

    Gokul43201

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    Is it not be sufficient that the set of states spanning this space be complete but not orthogonal?
     
  7. Nov 8, 2007 #6
    Hi nrqed,

    I share exactly the same opinion. The only trouble with this picture is that in realistic QFT theories (e.g., in QED), single-particle states are not eigenstates of the full Hamiltonian, so they cannot be identified with real particles. They are, in fact, so-called "bare" fictitious particles. States of "real" or "dressed" particles can be represented as linear combinations of these "bare" particle states. In the "dressed" particle basis, QFT becomes very similar to ordinary quantum mechanics, the only difference is that "transitions between different sectors" are now allowed.

    Eugene.
     
  8. Nov 8, 2007 #7

    Then how come we believe that the Hilbert space of two interacting spins is exactly spanned by the tensor product of non-interacting states?
     
  9. Nov 8, 2007 #8
    The reason why QFT is such a mess is the following: Most QFT Hamiltonians contain ugly interaction terms with three operators, like [itex] a^{\dag}b^{\dag}c^{\dag}, a^{\dag}c^{\dag}a, a^{\dag}ac, \ldots [/itex]. These terms have a non-trivial action on the vacuum [itex] | 0 \rangle [/itex] and one-particle states [itex] a^{\dag}| 0 \rangle [/itex], [itex] b^{\dag}| 0 \rangle [/itex]. So, in very sharp contrast to ordinary quantum mechanics, QFT particles have self-interactions with themselves. Even vacuum interacts with itself! This is the ultimate reason for ultraviolet divergences in QFT.

    Two points of view are possible. First, one can accept that self-interactions are inevitable ingredients of reality and learn to coexist with this weird picture. This seems to be the current consensus. Second, one can postulate that particles interact only when there are two or more of them. Then tri-linear interaction terms should be abandoned. Of course, we cannot simply drop these terms without altering physical predictions of QFT. However, there exists a "soft" (unitary transformation) approach which allows one to reformulate QFT so that the offending terms get eliminated from the Hamiltonian, and a physically transparent "dressed particle" picture emerges. In this picture, multiparticle states are obtained by tensor products of one-particle states, just as in ordinary quantum mechanics.

    Eugene.
     
  10. Nov 8, 2007 #9
    I still don't understand what is the meaning of the operators in

    [tex] <state| a^+ (k) a(k') |state> [/tex]

    If those are creation/annihilation operators of orthogonal states (doesn't matter bare or dressed) that happen to span the Fock space, then the above expression is zero unless k=k', which contradicts what my instructor said and whole chapters in condensed matter theory.

    So the a's are either not creation/annihilation operators or the states they create are not orthogonal or they don't span the whole space ?????
     
    Last edited: Nov 8, 2007
  11. Nov 9, 2007 #10
    I figured it out. It has nothing to do with interacting or non-interacting theory. Even in non-interacting theory the above scalar product can be nonzero for k not equal to k' if the |state> is not an eigenstate of the number operators i.e. is not an occupation number basis vector but a linear combination of those. For example, a state that is linear combination of one particle states with momenta k and k' :

    [tex] (<k| + <k'| ) a^+ (k) a(k') (|k> + |k'>) =<k|a^+ (k) a(k') |k'>=<0|0>=1 [/tex]
     
    Last edited: Nov 9, 2007
  12. Nov 9, 2007 #11

    Gokul43201

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    Another commonly seen example of a state that will satisfy the above condition is the coherent state of a BEC (which too is a superposition of number eigenstates).

    There's also another example I can think of, but it's something of a cheat: the number eigenstates of a system of n uncoupled harmonic oscillators (what you get when you do a co-ordinate transformation on a system of harmonically coupled lattice sites) are only orthogonal for all k lying within a given Brillouin zone, and if k - k' = G = (2*\pi)/a, then the inner product is 1.
     
  13. Nov 10, 2007 #12
    The example came from BCS superconductivity theory. My understanding is that the number operators are postulated hermitian (or equivalently, the field operator is postulated hermitian). Therefore their eigenstates are orthogonal and span the whole Hilbert space of the theory. The number operators in Heisenberg picture depend on time since they don't commute with the Hamiltonian when it contains terms higher than quadratic. Their eigenstates would depend on time too. The states of the system in Heisenberg picture are fixed with time but their decomposition into the time dependent number eigenstates would depend on time. Therefore the physical interpretation a state with respect to the number eigenstates will change with time - particles are created, decay or scatter.

    So what is the physical interpretation of the eigenstates of the number operators at given time? Are those all possible combinations of interacting non-localized experimentally measurable phonons at this time?

    The ground state of the system, defined as the lowest energy eigenstate of the Hamiltonian, is fixed in Heisenberg picture but changes with time with respect to time-changing number eigenstate basis. It is not the 'vacuum' state of the creation/annihilation operators in general or if it is at certain time, it is not later. Does that mean we should expect physical phonons to appear from the physical vaccum since the occupation numbers of the vacuum state change with time? I mean real physically detectable phonons not 'virtual ones' since I am assuming the number eigenstates at any time are interpreted as the physically detectable states.
     
    Last edited: Nov 10, 2007
  14. Nov 10, 2007 #13

    samalkhaiat

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  15. Nov 11, 2007 #14
    There is nothing strange about this. Consider a single harmonic oscillator. Suppose it is in the state:

    [tex]
    \begin{equation*}
    \newcommand{\lhk}[1]{\!\left | #1\!\right |}
    \newcommand{\lhkr}[1]{\left.#1\right |}
    \newcommand{\lhkl}[1]{\left |#1\right.}
    \newcommand{\gem}[1]{\left\langle #1\right\rangle}
    \newcommand{\geml}[1]{\left\langle #1\right.}
    \newcommand{\gemr}[1]{\left. #1\right\rangle}
    \newcommand{\ket}[1]{\lhkl{\gemr{#1}}}
    \newcommand{\bra}[1]{\lhkr{\geml{#1}}}
    \newcommand{\brak}[2]{\left\langle #1\vphantom{#2}\right | \!\left.\vphantom{#1}#2
    \right\rangle}
    \newcommand{\braket}[3]{\gem{#1\lhk{\vphantom{#1}#2\vphantom{#3}}#3}}

    \ket{s}= \ket{0} + \ket{1}+\ket{2}

    \end{equation}

    [/tex]

    Then, clearly,

    [tex]
    \begin{equation*}
    \newcommand{\lhk}[1]{\!\left | #1\!\right |}
    \newcommand{\lhkr}[1]{\left.#1\right |}
    \newcommand{\gem}[1]{\left\langle #1\right\rangle}
    \newcommand{\geml}[1]{\left\langle #1\right.}
    \newcommand{\gemr}[1]{\left. #1\right\rangle}
    \newcommand{\bra}[1]{\lhkr{\geml{#1}}}
    \newcommand{\brak}[2]{\left\langle #1\vphantom{#2}\right | \!\left.\vphantom{#1}#2
    \right\rangle}
    \newcommand{\braket}[3]{\gem{#1\lhk{\vphantom{#1}#2\vphantom{#3}}#3}}

    \braket{s}{a}{s}\neq 0
    \end{equation}

    [/tex]

    So, you are just dealing with states that are not tensor products of eigenstates of the number operator for each momentum separately.
     
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