NMR of
Correlated Electron Superconductors
Jürgen Haase
University of Leipzig
1
Content
• Introductory remarks
• NMR, classical superconductors, spin shift and relaxation
• NMR of the cuprates
• Spin shifts and the single-component picture
• With new NMR data we demonstrate:
• the failure of the single-component picture
• the new hyperfine scenario
• the three susceptibilities from NMR
• Conclusions
2
Nuclear Spin Levels
Energy
NMR:
Induce transitions among the nuclear levels
and measure the splitting and life-time.
m = -3/2
m = -1/2
m = +1/2
m = +3/2
ω n = γ n Bz
Time
The chemical and electronic properties of
the material influence the NMR parameters.
NMR is a bulk probe with atomic scale resolution.
3
Classical Superconductors
27Al spin
shift and relaxation in Al-metal above and below Tc
Spin-lattice relaxation rate in 27Al
Knight (Spin) Shift in 27Al
1.6
1.2
1.4
Hebel-Slichter Peak
proves BCS theory
(T1T)-1 (Ks)-1
KS(T)/KS(0)
1.0
0.8
0.6
0.4
1.2
1.0
0.8
Korringa behavior
0.6
0.4
0.2
Yosida function
0.2
0
0
0
0.2
0.4
0.6
0.8
1.0
1
1.2
10
100
1000
T (K)
T/Tc
The disappearance of the spin shift and 1/T1T below Tc are predicted by BCS theory
for S=0 Cooper pairs (above Tc Pauli susceptibility and Heitler-Teller relaxation).
NMR is a useful tool for studying superconductivity.
4
Cuprate Superconductors
Special circumstances:
• They are type II superconductors
(partial diamagnetism below Tc influences shift)
• We can investigate various nuclei (Cu, O, La,Y, Hg, ...).
• Most nuclei posses an electric quadrupole moment, so we have
a competing interaction and various resonance lines.
• Inhomogeneities can cause substantial line broadening
and overlapping lines.
Measurements can be very difficult.
5
NMR Spin Shift and
Electronic Spin Susceptibility
H = Ι ⋅ An ⋅ S
spin hyperfine term
An
H n = An I z Sz
µe = −γ e  Sz = χ S Bz
An
H n = γ n I z Bz
χS
2
γ nγ e 
H total
χS
⎛
⎞
An
= γ n I z ⎜ 1 +
χ S ⎟ Bz
2
γ nγ e 
⎝
⎠
Δω n
An
= K S,n =
χS
2
ω n,0
γ nγ e 
hyperfine constant
uniform electronic spin
susceptibility
NMR measures locally the
electronic spin susceptibility.
6
Early Remark - Single Component
If we compare the spin shift KS at different nuclear sites
(1 and 2) they must be proportional to each other:
A1
K S,1 =
χ
2 S
γ 1γ e
K S,2
A2
=
χS
2
γ 2γ e 
γ 2 A1
K S,1 =
K S,2
γ 1 A2
K S,1
0
0
K S,2
This proportionality is a critical test for single
component behavior.
7
Early NMR: Discovery of Spin-Gap
Alloul,Ohno, and Mendels, PRL 63, 1700 (1989)
The spin shift is temperature-dependent above Tc.
8
Early 17O and 63Cu Shift Results on YBa2Cu3O6.63
Cu and O electronic
spins seem to form a
single component.
The single-fluid picture has dominated our view of the cuprates.
9
Conflicting Evidence from NMR
• Pines et al. had to assume a revised hyperfine scenario to fit NMR data.
• Walstedt 1994 could not explain NMR relaxation in a single-component
picture.
• We could not fit NMR linewidths data for LaSrCuO with the hitherto used
hyperfine scenario (Haase et al. 2002).
• We found a break in the planar Cu vs. planar O spin shifts for LaSrCuO
but did not know the Meissner contribution.
Related:
• Johnston (1989) and Nakano et al. (1994) found two uniform susceptibilities.
We set out to acquire a new set of data on La1.85Sr0.15CuO4.
10
The shifts at three nuclei in La1.85Sr0.15CuO4
Apical Oxygen
K = K Core Diamagn + K L + K Q + K Meissner (T ) + K S (T )
Planar
Oxygen
Planar
Copper
desired term
T-independent
ΔB γ n ΔB
K Meissner (T ) =
=
independent on γ n
B
γ nB
1. Measure all possible shifts (field parallel and perpendicular to
crystal c-axis) at all temperatures.
2. This set of shifts allows us to remove the uncertainty
from the Meissner effect and perform a complete
3. Test for single-component behavior.
11
1. La1.75Sr0.15CuO4 Shifts vs. Temperature
80
60
40
Apical Oxygen
c⊥ B
Apical Oxygen
cB
Planar Oxygen
cB
Tc
20
0
250
Shifts (ppm)
200
150
100
50
0
1500
1000
500
0
4000
3000
Copper
2000
c⊥ B
1000
0
0
100
200
300
Temperature (K)
12
Here are the shifts.
2. Remove any Meissner Shielding
We allow for an anisotropic
Meissner shielding by taking the
appropriate differences:
0.3
F⊥ (%)
F⊥ = 63 K S (T ) − 17 A K S (T )
0.2
0.1
with c ⊥ B0
0
0.1
F =
17 P
K S (T ) − 17 A K S (T )
with c  B0
F (%)
0.05
0
Both quantities are obviously
not proportional to each other:
0
50
100
150
200
Temperature (K)
13
250
300
3. The Test for Single-Component Behavior
0.4
⊥
0.3
F⊥ (%)
c F⊥ ≠  F
c
Tc
0.2
0.1
0
0
0.05
0.1
Based entirely on
experimental data.
F (%)
This is not single-component behavior.
We have to resort to a two-component scenario.
14
Spin Shifts for Two Components
( )
( )
Two electronic spins
act on each nucleus.
K k = ak ⋅ χ a T + bk ⋅ χ b T
K K spin shift for particular nucleus and orientation
χ a , χ b two susceptibilities
ak , bk hyperfine coefficients for particular nucleus and orientation
Note that we expect:
χ a = χ AA + χ AB
With the uniform spin susceptibility
χ b = χ BB + χ AB
χ 0 = χ a + χ b = χ AA + 2 χ AB + χ BB
15
Experimental Observation
33300; !1130 (/1000)
Planar oxygen c||B
Copper c B
3
2.5
2
1.5
1
0.5
0
0
0.02
0.04
0.06
1.5
1
0.5
0
!0.5
!1
!1.5
!2
0
0.08
1.5
Planar oxygen c||B
2
3.5
Apical oxygen c B
0.05
0.1
0.15
0.2
0.25
0.3
1
0.5
0
!0.5
!1
!1.5
0
Apical oxygen c||B
0.02
0.04
0.06
0.08
Apical oxygen c B
4
3.5
3.5
3
2.5
2
1.5
1
3
2.5
2
1.5
1
0.5
0.5
0
0
0.35
Apical oxygen c||B
4
Copper c B
Copper c B
2500; 45 (/1000)
4.5
0.5
1
1.5
Planar oxygen c||B
0
0
Apical oxygen c B
Apical oxygen
c||B
16
0.05
0.1
0.15
0.2
0.25
0.3
0.25
0.2
0.15
0.1
0.05
0
0
0.02
0.04
0.06
0.08
Apical oxygen c B
0.1
Experimental Observation
33300; !1130 (/1000)
Planar oxygen c||B
Copper c B
3
2.5
2
1.5
1
0.5
1.5
Planar oxygen c||B
2
3.5
1.5
1
0.5
0
!0.5
!1
!1.5
1
0.5
0
!0.5
!1
Every shifts is a linear function of any other shift
Apical oxygen c B
Apical oxygen c||B
Apical oxygen c
( y = m x + n ) at higher temperatures (T > Tc).
0
0
0.02
0.04
0.06
!2
0
0.08
0.05
0.1
0.15
0.2
0.25
0.3
!1.5
0
0.02
0.04
0.06
0.08
B
4
3.5
3.5
3
2.5
2
1.5
1
3
2.5
2
1.5
1
0.5
0.5
0
0
0.35
Apical oxygen c||B
4
Copper c B
Copper c B
2500; 45 (/1000)
4.5
0.5
1
1.5
Planar oxygen c||B
0
0
Apical oxygen c B
Apical oxygen
c||B
16
0.05
0.1
0.15
0.2
0.25
0.3
0.25
0.2
0.15
0.1
0.05
0
0
0.02
0.04
0.06
0.08
Apical oxygen c B
0.1
Conclusion from the shift-shift plots
Shift k:
K k = ak ⋅ χ a T + bk ⋅ χ b T
( )
( )
Shift l:
( )
( )
2
1.5
1
Kk
K l = al ⋅ χ a T + bl ⋅ χ b T
substitute χ a (T )
0.5
0
!0.5
!1
!1.5
!2
0
0.05
0.1
0.15
0.2
0.25
0.3
Kl
⎧⎪
ak
ak ⎫⎪
K k T = K l T + ⎨bk − bl ⎬ χ b T
al
al ⎪⎭
⎪⎩
( )
( )
( )
( )
χ b T = χ AB + χ BB
not a function of T for T>Tc
Two susceptibilities χ AB , χ BB must be temperature independent
above Tc. The temperature dependence can only be in χ AA .
17
Determining the Unknowns
⎧⎪
ak
ak ⎫⎪
K k T = K l T + ⎨bk − bl ⎬ χ b
al
al ⎪⎭
⎪⎩
( )
( )
These two numbers can be read off the plots for T > Tc.
We can solve for bk χ b and thus calculate ak ⋅ χ a (T ) = K k (T ) − bk ⋅ χ b .
We note that we can determine a universal function for T > Tc:
( )
χa T
(
χ a 300K
)
=
( )
K k T − bk ⋅ χ b
(
)
K k 300K − bk ⋅ χ b
18
Normalized Susceptibility vs. Temperature
( )
χa T
(
χ a 300K
1.5
)
1
0.5
0
(for Cu, o, the noise is larger)
!0.5
!1
!1.5
!2
!2.5
!3
!3.5
0
50
100
150
200
250
300
Temperature (K)
1. All shifts show the universal scaling down to Tc: χ a (T ) / χ a ( 300K ) .
2. Near Tc this expression is strongly negative (is χ AB negative?).
19
Uniform Susceptibility - 2 Components
D.C. Johnston (1989) and later Nakano et al. (1994) showed
that the uniform susceptibility (for LaSrCuO) could be
decomposed into two contributions (for all doping levels):
χ ( T , x ) = χ1 ( T , x ) + χ 2 ( x )
temperature independent
⎛ T ⎞
F⎜
⎟
⎝ Tmax ⎠
χ1 (T , x ) = ⎡⎣ χ max (T = Tmax , x ) − χ 2 ( x ) ⎤⎦ F (T / Tmax ( x ))
Since we can only have χ AA as temperature-dependent
susceptibility we must conclude that χ AA (T ) = χ1 (T ).
20
Determination of χ AB
( )
χa T
(
1.5
χ a 300K
)
1
Use Johnston’s
universal function
0.5
0
!0.5
!1
0
50
100
150
200
250
300
Temperature (K)
( )
χa T
(
χ a 300K
)
≡
( )
χ AA T + χ AB
(
)
χ AA 300K + χ AB
=
( )
χ1 T + χ AB
(
)
χ1 300K + χ AB
We can calculate χ AB .
21
Summary of Decomposition
Above Tc we find the following numbers:
( )
( )
χ AA T = χ1 T
(
)
χ1 300K = 6.8 emu / mol
χ BB = 18.7 emu / mol
χ AB = −4.2 emu / mol
(negative!)
With these values we can now determine the
hyperfine coefficients.
22
Hyperfine Coefficients La2-xSrxCuO4
χa
χb
apical O c ⊥ B
Cu c ⊥ B
planar O c  B
2.6
8.7
0.79
0.24
mol / emu
14.3
48.0
4.4
1.3
kG / µB
21.4
4.6
1.2
0.38
mol / emu
26
6.9
2.1
kG / µB
120
apical O c  B
These are the coefficients that describe the
coupling of each nucleus to the two electronic spin
components (for two orientations).
23
Susceptibilities vs. Temperature
χ
20
(emu / mol)
χ b = χ AB + χ BB
15
10
χ a = χ AA + χ AB
5
0
χ AA = χ dd (copper d-contribution)
χ BB = χ pp (oxygen pσ -contribution)
!5
0
100
200
300
χ AB = χ pd (correlation d,pσ )
Temperature (K)
These temperature-dependencies follow from
the NMR spin shift data!
24
Conclusions
• A single-component picture is not appropriate for the
description of all cuprate superconductors.
• The spin shifts suggest a two-component scenario:
★
( )
( T>T ) >0
χ AA T is temperature-dependent already above Tc (pseudogap)
★
χ BB
★
χ AB T>Tc <0 and temperature-independent
★
c
(
and temperature-independent
)
below Tc all susceptibilities decrease to zero
each nucleus couples to two electronic spin components
with the hyperfine couplings presented.
★
25
Acknowledgement
Charlie Slichter
(still a hot collaboration)
David Pines (many enlightening discussions)
G.V.M. Williams, Wellington NZ (many samples)
N. Curro, J. Schmalian (two-component description in 115-materials)
V. Barzikin, D. Pines (sharing their thoughts about the cuprates with us)
26