1524 A.J.Leggett
is discussed, for example, in [11]. The simplest for-
mulation is probably to start by ignoring the inter-
conduction electron interactions and solving for the
eigenstates of the single-particle terms in the Hamil-
tonian (kinetic plus lattice potential energy); in this
way one generates the usual Bloch-wave states,which
are eigenstates not of total momentum P but rather
of total quasimomentum K, and are labeled corre-
spondingly by a quasimomentum index k and spin
. The resulting system is the analog of the nonin-
teracting gas in the translation-invariant case. Then
one turns on the interaction and assumes that the re-
sulting groundstate and low-lying excited states can
be put into one–one correspondence with these of
the starting (“noninteracting”) state. It is important
to note that, because of the possibility of umklapp
processes,the interaction does not conserve the total
quasimomentum K; hence while one can still label
the resulting states by ın(k, ), k no longer has the
physicalinterpretationof truequasimomentum (and
in fact the operator
ˆ
K is nondiagonal in the ın(k, )
basis). This observation has some importance in the
context of the discussion of the “mid-infrared” peak
in the cuprates: see next section.
Is the normal state of the cuprates a Fermi liq-
uid (in the same sense as that in which the normal
state of
3
He can be said to be a Fermi liquid)? This is
one of the most hotly debated questions in the the-
ory of cuprate superconductivity,andto discuss it we
need to define it carefully. Let us first ask the ques-
tion: Suppose we could somehow cool the system
down to zero temperature while forbidding the su-
perconducting transition to occur. Would the result-
ing “groundstate” and low-lying excited states con-
form to the definition of a Fermi liquid? In the case
of liquid
3
He (where there is apparently no practical
way of suppressing the superfluid transition, so that
this really is only a thought-experiment) the answer
is almost certainly yes. In the case of the cuprates,
there are real-life possibilities of suppressing the su-
perconducting transition,in particular (a) by the ap-
plication of a sufficiently strong magnetic field and
(b) by doping the CuO
2
planes with certain types
of impurity such as Zn. Although for practical rea-
sons rather few systematic studies have been done of
the resultant state an answer seems to be emerging
which is rather intriguing [12]: There appears to be a
critical concentration p
c
∼ 0.19 of the hole-doping
parameter p (number of holes per CuO
2
units) such
that for p > p
c
the lowest-lying states indeed have
Fermi-liquid like character. However, to the left of
this point in the phase diagram (i.e. for p < p
c
)the
behavior appears to be qualitatively different, with
the system tending continuously towards an insulat-
ing state as the temperature falls towards zero.If this
is really the correct picture, then it is not entirely
clear that the question “does the normal state of the
cuprate superconductors behave as a Fermi liquid?”
has any clearly defined meaning. Of course one could
in any case argue that the system placed in a strong
(∼ 30–60 T)magnetic field, or doped substantially
with Zn impurities, is not “the same” system as a
true (superconducting) cuprate, and that therefore
no conclusions can be drawn from experiments of
the above type. In that case it is conceivable that the
true situation is the following: If one, by theoretical
fiat rather than by any realistically available experi-
mental means, could forbid the formation of Cooper
pairs in the (pure,zero-field) system while cooling it
to zero, then the “groundstate” and low-lying states
would indeed be of Fermi-liquid type, as in the cor-
responding thought-experiment with
3
He. However,
in both cases we should expect the Fermi-liquid de-
scription to break down badly when the temperature
becomes some (small) fraction of T
F
, let us say for
example 0.1T
F
.Now,inthecaseof
3
He, (T
c
/T
F
∼
10
−2
) this“breakdown”temperature is still some way
above T
c
, so that there is a fairly wide temperature
regime (3mK
<
∼
T
<
∼
50 mK) where a well-defined
Fermi-liquid behavior is seen. On the other hand, in
the cuprates the breakdown temperature could be
so little above the observed T
c
that no clear Fermi-
liquid regime is seen. It has to be said, however, that
this scenario, while perhaps plausible for the higher-
temperature compounds such as the T and Hg se-
ries,ismuchlesssoforcompoundssuchas(opti-
mally doped) Bi-2201 where T
c
is only in the range
10–30 K, and to defend it one has to argue that such
compounds are in some way untypical, even in the
normal state, of the cuprates as a whole.
A point which has been widely noticed is that
whileatoptimaldoping the resistivity ofthecuprates