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BCS theory by canonical transformation

  1. Jun 24, 2010 #1
    I am reading Tinkham's "introduction to superconductivity" 1975 by McGraw-Hill, Inc.

    Tinkham derives the BCS theory by canonical transformation. At the beginning of the chapter he writes:
    "We start with the observation that the characteristic BCS pair interaction Hamiltonian will lead to a ground state which is some phase-coherent superposition of many-body states with pairs of Bloch states $(k\uparrow, -k\downarrow)$ occupied or unoccupied as units. Because of the coherence, operators such as $c_{-k\downarrow}c_{k\uparrow}$ can have nonzero expectation values b_k in such a state, rather than averaging to zero as in a normal metal, where the phases are random. "

    I have some questions:
    1. Is the characteristic BCS pair interaction Hamiltonian the so-called "reduced Hamiltonian"?
    2. The BCS ground state is a state with pairs of Bloch states $(k\uparrow, -k\downarrow)$ either occupied or unoccupied, but what does the phase-coherence mean?
    3. Why does it follow from coherence, that operators such as $c_{-k\downarrow}c_{k\uparrow}$ can have nonzero expectation values? When he writes phase, he means the phase of what?

    It is the first time posting here, so bear with me.^^
  2. jcsd
  3. Jun 26, 2010 #2
    Have I done something wrong regarding my post?
    Does anyone know BCS theory?
  4. Jun 27, 2010 #3
    Hi IFNT, welcome to the forum.

    I think most text books are a little vague on the subject and I am not sure I have a consistent picture of this stuff myself. However I can try to explain what I think Tinkham refers to.

    1. Not sure what reduced Hamiltonian means in this context. Could you give us the page number where this term is used?

    2. Phase coherence means that the phase [itex]\varphi[/itex] as in Eq. (3.18a) in the BCS ground state is well defined, i.e. has a certain value.

    3. If the phase in the BCS ground state is not well defined, i.e. you have a superposition of all possible phases as in Eq. (3.18b) you have a number conserving state. For a number conserving state the average of [itex]b_k[/itex] will necessarily be zero. You can try to evaluate the average w.r.t. the ground state in Eq. (3.18b) directly and you will find that the integration over phase causes the expectation value to "average out".

    Basically a well defined phase in a superconductor is similar to having a well defined direction of spin in a ferromagnet. In the symmetry conserving state an evaluation of [itex]\langle\vec{S}\rangle[/itex] will average out since there is no preferred direction of the spin. In the symmetry broken phase a particular direction of spin (or value of the phase in a superconductor) is chosen arbitrarily.

    I should note that the vanishing of the expectation value [itex]\langle b_k\rangle[/itex] does not necessarily imply that there is no superconducting order. A more general way to characterize the superconducting state is through the concept of Off-Diagonal Long Range Order (ODLRO).

    Hope this helps somewhat.
  5. Jun 29, 2010 #4


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    Dear IFNT,

    jensa answered most of your questions better than I could, hence I only want to comment on your question 1. The BCS hamiltonian contains a term in which electrons with momentum k and k' are anihilated and electrons with k+q and k'-q are generated. The reduced hamiltonian retains only the terms with k'=-k. This corresponds to working in the subspace with electrons with k and -k being paired as stated in Tinkham. The use of this hamiltonian is believed to lead to the correct expression for e.g. the ground state energy, but obviously there is no possibility to describe e.g. states carrying a current where k' not equal -k.
  6. Jun 30, 2010 #5
    Jensa, thank you for the reply. I don't know what book you are referring to, as there are no (3.18a) and (3.18b) in my book and the BCS ground state is first giving in (2-14).

    I have also trouble understanding the concept of diagonalizing a Hamiltonian. In the book he uses Bogoliubov transformation to change basis, then he says that the Hamiltonian is diagonalized when it only contains terms proportional to [itex]\gamma_{k0}^{\dagger}\gamma_{k0}[/itex] and [itex]\gamma_{k1}^{\dagger}\gamma_{k1}[/itex].
    Do you know why that is so?
  7. Jul 2, 2010 #6


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    The gamma operators fulfill the usual relations for fermion anihilation operators. Hence the ground state fulfills e.g. gamma |0>=0 and excited states can be obtained by operating on |0> with the adjoints of gamma (the particle creation operators). The hamiltonian is diagonal in the basis spanned by these operators as [tex] \gamma_k^+ \gamma_k[/tex] is the number operator which counts the number of excitations of momentum k (0 or 1). Compare also the formalism of creation and anihilation operators for a harmonic oscillator and for a gas of free fermions.
  8. Jul 2, 2010 #7
    Dear DrDu,
    Thank you for the reply. I have found a site, http://jila.colorado.edu/pubs/thesis/milstein/appB.pdf [Broken], where the Bogoliubov Transformation is explained somewhat short and I am trying to understand it right now. I think I have trouble understanding the whole setup.
    Changing basis, but what vector space are the creation and annihilation operators basis to?

    I can conclude from this that the excited states are eigenstates of the hamiltonian. How do I come from this to the fact that the Hamiltonian is diagonal?

    Best regard, IFNT
    Last edited by a moderator: May 4, 2017
  9. Jul 3, 2010 #8


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    To your second question: If you know the eigenstates of the Hamiltonian then the Hamiltonian is diagnoal in the basis of these eigenstates.
  10. Jul 3, 2010 #9
    Okay, but don't you have to know that the eigenstates span the space? If so, what space are we talking about? Is it the Fock space? Is it infinite dimensional?

    I am still on my bachelor degree so I am unclear about a lot of stuff.
  11. Jul 5, 2010 #10


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    Well, the gound state is characterized by [tex] \gamma_{..} | 0 \rangle =0 [/tex]. This yields the BCS ground state wavefunction.
    The excited states spanned by operating will all possible combinations of the creation operators gamma spans an infinitely dimensional Fock space. Note that there is not only one Fock space but an infinity of them. Especially, the one we are considering here is different from the Fock space of free Fermions. And, as in the case of free Fermions, for any electron density there is another Fock space.
  12. Jul 6, 2010 #11
    Thanks alot for your replies DrDu, I imagine that Linear algebra in a infinite dimensional space is similar to finite dimensional space.

    One last question, if you got the time:
    Why do we want to make operators diagonal? It is easy to find the eigenvalues, but is that all?
  13. Jul 6, 2010 #12


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    ...yes, and not to forget the eigenvalues.
  14. Jul 14, 2010 #13

    I cannot see what equations you are referring to. If we have the BCS wavefunction, does this mean that the phase between the two electrons in one single cooper pair is fixed, but the phase between two cooper pairs are arbitrary?
  15. Jul 14, 2010 #14
    Sorry for not answering for a while. I've been on vacation and didn't check the forum. The equations I refer to are in Tinkhams second edition, I suppose you have the first one. You can probably find the equations using google books or the preview feature at amazon. If not, let me know and I''ll write them for you.

    The phase in a superconductor is intimately related to the number of cooper pairs. More precisely the number of cooper pairs and the phase are conjugate variables. The phase that is referred to is the relative phase between two many particle states with [itex]N[/itex] and [itex]N+1[/itex] number of cooper pairs. To be phase coherent it is important that this phase is well defined (does not fluctuate). This in turn means that the number of electron pairs in your system will not be well defined (fluctuates). Inversely, if the number of electrons is well defined then the phase will not be well defined.
  16. Jul 14, 2010 #15
    Thanks for replying Jensa, I did my last post only a couple hours before you replied so thank you for replying so fast.^^
    The Tinkham book doesn't have a preview feature on Amazon and google books shows only to page 30 and BCS theory starts at page 43 I think. It would be great appreciated if you could write the equations for me.

    ""The phase that is referred to is the relative phase between two many particle states with [itex]N[/itex] and [itex]N+1[/itex] number of cooper pairs. To be phase coherent it is important that this phase is well defined (does not fluctuate).""
    How do I see that this phase is well defined? Can I see it from the BCS ground state wavefunction? Why is this phace coherence so important?
    And if one of the most important things about a superconducter is its phasecoherent wavefunction then how can we derive so many results from it? I mean, in a real physical problem isn't the particle number is always well defined?
  17. Jul 14, 2010 #16

    Eq. (3.18a)


    Eq. (3.18b)

    |\psi_N\rangle=\int_0^{2\pi}d\varphi e^{-iN\varphi/2}\Pi_k(|u_k|+|v_k|e^{i\varphi}c_{k\uparrow}^*c_{-k\downarrow}^*)|\phi_0\rangle = \int_{0}^{2\pi} d\varphi e^{-iN\varphi/2}|\psi_\varphi\rangle

    Yes, in Eq. (3.18a) above you see that there is a fixed (and constant) phase [itex]\varphi[/itex]. It is assumed to be arbitrary but the same everywhere in the superconductor.

    You may think of the analogy with the spin orientation in ferromagnets, if the spin orientation at different sites (different positions) are oriented randomly without any relation to their neighbours, you do no not have a ferromagnet. However, if all spins are oriented in the same (random) direction, you do have a ferromagnet.

    In Eq. (3.18a) the choice of the phase is random, but it is the same for any state [itex](k\uparrow,-k\downarrow)[/itex]. If you were to let the phase depend on the state [itex]\varphi\rightarrow \varphi_k[/itex], where each [itex]\varphi_k[/itex] were random without any relation to each other, you would not have a superconductor.

    These are excellent questions. Ironically, the two people who are trying to help you (me and DrDu) have both asked related questions on this very forum :) So you see, I am probably not the most qualified person to answer this. Maybe DrDu has gained some knowledge that he could share.

    I would just like to point out the following things:

    1) The BCS wave function is an approximate ground state. This approximation becomes exact (I think) in the limit of infinitely many particles (and infinitely large volume), at which point the fact that it does not conserve number of particles becomes negligible.
    The fact that we need to resort to the limit of infinitely many particles/infinite volume is a general feature in descriptions of phase transitions and spontaneously broken symmetries.

    2) In contrast to, say the global spin orientation in ferromagnets, the global phase of the superconductor is itself not an observable quantity. Any observables are represented by number conserving operators and thus the global phase will never enter any expectation values of such observables. This also suggests that the fact that the BCS ground state is a superposition of different particle states is not really the crucial aspect of the wave function.

    Note: Somehow my post looks completely messed up when I preview it, so I hope it comes out correctly.
  18. Jul 15, 2010 #17


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    In some old article P. W. Anderson ( have to find the citation at home) notes that the BCS ground state which does not conserve particle number obviously also means that we are dealing with a superposition of states with different charge. These states thus will decohere extremely rapidly. To show this, one only has to calculate the difference in electrostatic energy between states differing in charge by one. However, he also shows that this effect won't limit the observability of the Josephson effect.
  19. Jul 15, 2010 #18
    Thank you so much for this thorough reply Jensa!
    As I understand it, the phase is the same for every cooperpair [itex](k\uparrow,-k\downarrow)[/tex]. Is this phase the same as the relative phase between two many particle states with [tex] N[/tex] and [itex]N+1[/itex] number of cooper pairs, the relative phase between [tex]u_k[/tex] and [tex] v_k[/tex]?
    Last edited: Jul 15, 2010
  20. Jul 16, 2010 #19
    Forget my last post, it made no sense.
    I have one last question: the BCS ground state is phase coherent, and there exist an energi gap, why does this make the material superconducting?

    When I apply an E-field to the sample for a short period, the cooper pairs are accelerated and changes momentum per pair from [tex]\vec{k}=0[/itex] to [tex]\vec{k'} \noteq 0[/itex]. Only a scattering processes where a cooper pair splits up into two individual electrons will results in the delay of the current. This is only possible when the kinetic energy of the pair is higher than the energy gap, thus we have the critical current density. Where does the phasecoherence come into the picture?
  21. Jul 16, 2010 #20


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    Well, in quantum mechanics it is not sufficient that there is an energy gap to guarantee the vanishing of the current. After all you don't have to excite the pair completely to get a current
    but it is sufficient to create a superposition containing just a little bit of an excited pair with momentum k. For sufficiently small perturbing fields, the admixture of the excited states can be calculated in second order of perturbation theory. The energy gap guarantees that the denominator remains finite. Coherence of the wavefunction yields the result that the numerator vanishes in the limit of k to 0. This is known as the stiffness of the wavefunction, an idea going back to the Londons in the early days of superconductivity. This is nicely explained in Ch. Kittel, Quantum theory of solids, my absolute favorite on the theory of superconductivity.
    There exists an alternative explanation by Stephen Weinberg in Quantum theory of fields, vol. II which rests on the fact that a superconductor is a system with spontaneously broken symmetry. The argument can be understood without knowledge of the rest of the books.
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