NMR Shift and Relaxation and the Electronic Spin of Superconducting Cuprates

Abstract

Very recently, by inspecting large sets of data across all families of superconducting cuprates, it became obvious that the prevailing nuclear magnetic resonance (NMR) interpretation of cuprate properties is not adequate, as it does not account for the differences between the families, as well as common characteristics beyond simple temperature dependence. From the most abundant planar Cu shift data, one concludes readily on two electronic spin components with different doping and temperature dependencies. Their uniform response that causes NMR spin shifts consists of a doping-dependent component due to planar O, and another due to spin in the planar copper 3d(x² − y²) orbital, where the latter points opposite the field direction. Planar Cu relaxation was found to be rather ubiquitous (except for La_2−xSr_xCuO₄), and Fermi liquid-like, i.e., independent of doping and material, apart from the sudden drop at the superconducting transition temperature, T_c. Only the relaxation anisotropy is doping and material dependent. We showed previously that one can understand the shifts within a two-component scenario, but we failed with a model to account for the relaxation. Here, we suggest a slightly different shift scenario, still based on the two components, by introducing different hyperfine couplings, and, importantly, we are able to account for the Cu nuclear relaxation and its anisotropy for all materials, including also La_2−xSr_xCuO₄. The results represent a solid framework for theory.

1 Introduction

Nuclear magnetic resonance (NMR) is a powerful local, bulk probe of material properties [1]. This concerns the chemical as well as electronic structure of materials, which can be studied locally at various nuclear sites in the unit cell. The changes in the NMR shifts and relaxation from the modification of the density of states due to the opening of a superconducting gap in conventional superconductors are famous examples [2, 3]. Not surprisingly, after the discovery of cuprate superconductivity [4], NMR experiments focused in particular on planar Cu and O in these type II materials (for reviews of cuprate NMR, see [5, 6]). However, with the early focus on only a few systems and a lack of established theory, the NMR data interpretation ceased to evolve with a number of questions unanswered. Fortunately, more and more NMR studies of different materials appeared in the literature over the years.

During the last 10 years, with special NMR experiments on La_1.85S_0.15CuO₄ [7], YBa₂Cu₄O₈ [8], and samples of the HgBa₂CuO_4+δ family of materials [9, 10], a cornerstone of the old interpretation was questioned and shown to be not correct: a single temperature-dependent electronic spin component, s(T), that follows from the uniform spin susceptibility, i.e., s(T) = χ(T) ⋅ B₀, in an external field B₀, is not capable of describing the temperature-dependent NMR spin shifts, ${~}^{n}{K}{{~}_{d}}(T)={{~}^{n}H}_{d}\cdot \chi (T)$. Since ${~}^{n}{K}{{~}_{d}}(T)$ can be measured at various nuclei (n), or for any orientation (d) of the external field with respect to the crystal axes, with ${{~}^{n}H}_{d}$ being the corresponding hyperfine constant, one demands from different experiments that ${\Delta }^{n}{K}{{~}_{d}} \propto {\Delta }^{m}{K}{{~}_{e}}$, which was clearly not observed in general, only in certain ranges of temperature [7,8,9,10].

Also during the last decade, the understanding of the charge sharing in the CuO₂ plane advanced significantly from a more qualitative [11] into a quantitative model [12, 27]. It became apparent that, e.g., the maximum temperature of superconductivity correlates with the sharing of the inherent hole between planar Cu and O, the higher the oxygen hole content the higher T_{c, max} [13]. Other cuprate properties depend on the charge sharing, as well. This mostly family-dependent behavior stimulated some of us to inspect also a larger body of NMR shifts and relaxation for material-dependent differences or common characteristics.

In the first step, all available ⁶³Cu NMR shifts that are rather abundant and reliable were gathered [14]. And, indeed, by just plotting these shifts a new phenomenology emerged, and points immediately to a more complicated uniform response that cannot be explained with a simple χ(T). In the second step, all available ⁶³Cu NMR relaxation rates were collected [15, 16] and, again, simple plots revealed a surprisingly different scenario. Here, a rather material- and doping-independent relaxation was revealed with spin fluctuations similar, but not in excess to what one expects from a simple Fermi liquid. Only the relaxation anisotropy depends on the materials and decreases with increasing doping.

Then, in a first attempt, we tried to reconcile these findings [15]. We could show that a two-component description, as introduced earlier [7] (with two spin components that couple with two different hyperfine coefficients, ${~}^{n}{H}{{~}_{1d}}$ and ${~}^{n}{H}{{~}_{2d}}$, to each nuclear spin, n), is indeed sufficient to understand the planar Cu shifts (with the La_2−xSr_xCuO₄ family being some kind of outlier, cf. Fig. 1). Two spin susceptibilities (χ₁,χ₂) demand a third term from a coupling between the two electronic spin components. That is, one has to write, χ₁ = χ₁₁ + χ₁₂,χ₂ = χ₂₂ + χ₂₁ (χ₁₂ = χ₂₁), and,

$$ {~}^{n}{K}_{\parallel, \perp} = {{~}^{n}H}_{1 {\parallel},\perp} \cdot(a+c) + {{~}^{n}H}_{2\parallel,\perp} \cdot(b+c) $$

(1)

with a = χ₁₁B₀, b = χ₂₂B₀, and c = χ₁₂B₀, and the magnetic field parallel and perpendicular to the crystal c-axis.

Fig. 1

a Total ⁶³Cu shifts vs. temperature, $\hat {K}_{\parallel ,\perp }(T)$, for 4 doping levels of La_2−xSr_xCuO₄ and two directions (c ∥ B₀, c⊥B₀) of the external field B₀ with respect to the crystal c-axis (adopted from [17]). K_∥ is T independent and similar for all doping levels. K_⊥ shows a much larger spread with doping and decreases rapidly near T_c. b Sketch of the spin shift K_⊥(T) vs K_∥(T) plot valid for other cuprates [14], with real data from the 4 doping levels of La_2−xSr_xCuO₄ (with temperature as an implicit parameter). The shaded area is where the rest of the many cuprates can be found (the shaded triangle has a hypotenuse of slope ≈ 1), only typical data are shown by crosses. Data lie on straight line segments (lines) with a few slopes only: a slope ≈ 1 (dashed lines); a very steep slope (vertical lines); a slope of 2.5. For example, a slope of 2.5 is typical for HgBa₂CuO_4+δ (at higher T), a steep slope for YBa₂Cu₄O₈, and for symmetry reasons we use a slope of 1 for some Tl-based compounds, as well as for overdoped HgBa₂CuO_4+δ at low T. La_2−xSr_xCuO₄ is a clear outlier with only the steep slope. In the simple two-component description, cf. (5), (6), a change in one of the components a,b, or the coupling c leads to the indicated slopes in the middle of (b); for the subtraction of the orbital shifts, see main text

In order to independently test the important conclusion of two spin components with different doping and temperature dependencies, we investigated the planar O data [20], very recently. We found that, indeed, planar O shifts demand two spin components, as well, where one of them is doping dependent. This encouraged us to search for a better understanding of the very reliable Cu data, in particular of the nuclear relaxation and its anisotropy, which we failed to deliver previously [15].

By plotting a large set of literature Cu relaxation data [15], we found generic behavior, as well, with the exception of just one family, La_2−xSr_xCuO₄. In fact, all other cuprates have rather similar relaxation rates, 1/T_1⊥ [16], i.e., if measured with the magnetic field perpendicular to the crystal c-axis, c⊥B₀, cf. Fig. 2. In particular, just above T_c the value of $1/T_{1\perp }T_{\mathrm {c}}\sim {20}{/Ks}$ for all cuprates, while T_c can be very different, or even close to zero for strongly overdoped systems. There is no particular doping dependence of 1/T_1⊥ as one might naively expect if electronic spin fluctuations beyond those of a more regular Fermi liquid were to increase toward lower doping levels (there are hardly data available at very low doping). In fact, a value of 20/Ks follows with the Korringa relation [22] from those cuprates with the highest shifts, i.e., the upper right corner of the shaded triangle in Fig. 1 [15], which suggests that the shifts have the tendency to be suppressed if the Korringa relation fails, and it is not due to an increased relaxation. The situation is somewhat different for 1/T_1∥ since for this direction of measurement (c ∥ B₀) the rates differ between families and have the tendency to increase with decreasing doping. However, it was demonstrated that the ratio, $\left [1/T_{1\perp }(T)\right ]/\left [1/T_{1\parallel }(T)\right ]$, is temperature independent for all cuprates, and is the same above and below T_c; i.e., both rates are proportional to each other [15, 16].

Fig. 2

a Planar ⁶³Cu relaxation rates of the cuprates (data from [16]); 1/T_1⊥ of La_2−xSr_xCuO₄ in comparison is about twice as high as that of other cuprates. b 1/T_1⊥ vs 1/T_1∥, which is ≈ 2.3 for La_2−xSr_xCuO₄ (highlighted), is very similar to what is found for other cuprates (data [16]). It is mostly 1/T_1∥ that changes with doping and material, but remains proportional to 1/T_1⊥ (at all temperatures)

Interestingly, the La_2−xSr_xCuO₄ family of materials is the only outlier to this phenomenology, cf. Fig. 2. However, the anisotropy ratio is also temperature independent and has a value of about 2.3, very similar to that of some other cuprates.

In our first attempt at reconciling shift and relaxation [15], we could only explain the shift suppression, but failed to present a microscopic model that also explains the relaxation and its anisotropy. Here, we discuss and modify our previously suggested two-component scenario [15] in that we introduce somewhat different hyperfine coefficients, together with a new notation, still based on the identification of the two electronic components as being due to planar Cu 3d(x² − y²) and likely planar O 2p_σ spin densities. Most importantly, we are able to present a simple, yet fundamental, model of nuclear relaxation in terms of these two components that fits all planar Cu relaxation data, even including the outlier La_2−xSr_xCuO₄.

2 Planar Cu Shifts

Dissimilar from our previous attempt to understand all planar Cu shift data [15], we will be using a somewhat different nomenclature and hyperfine coupling coefficients here. Therefore, we repeat the basic arguments leading to the description, now.

The total magnetic shift for planar Cu, ${~}^{63}{\hat {K}}{{~}_{\parallel , \perp }}$, is the sum of an orbital and spin shift component, and we have for the two orientations (c ∥ B₀, c⊥B₀) of the magnetic field B₀ with respect to the crystal c-axis,

$$ {\hat{K}}{{~}_{\parallel , \perp}}(T) = {K}_{L {\parallel} , \perp}+{K}_{\parallel,\perp}(T). $$

(2)

It is of particular use to plot the total shifts ${\hat {K}}{{~}_{\perp }}(T)$ vs. ${\hat {K}}{{~}_{\perp }}(T)$ [14] with temperature as an implicit parameter, i.e., one does not make assumptions about K_L∥,⊥. Such a plot brings out a number of remarkable trends [14, 15]. A sketch of such a plot is presented in Fig. 1b, and we repeat some conclusions [15], but also include new ones, below.

A fundamental assumption is [14, 15],

$$ {K}{{~}_{L\perp}} \approx 0.30<percent>, $$

(3)

since all cuprates show a rather similar low temperature shift for c⊥B₀ with ${\hat {K}}{{~}_{\perp }}(T\rightarrow 0) \approx 0.30\%$. Therefore, this value appears to be reliable. It is the same assumption made early on [5]. We note that this value is backed by first principle calculations [19] (while this is not the case for K_L∥).

Second, except for La_2−xSr_xCuO₄, all data points in that plot are found in the lower right triangle that has as hypothenuse a line of slope 1, i.e., ${\Delta } {\hat {K}}{{~}_{\perp }}/{\Delta } {\hat {K}}{{~}_{\parallel }}(T) \approx 1$. This line points immediately to an isotropic hyperfine coefficient, while the fact that ${\hat {K}}{{~}_{\parallel }}(T) > {\hat {K}}{{~}_{\perp }}(T)$ (all data in the lower, right triangle) demands a second, very anisotropic hyperfine coefficient that acts mostly for c ∥ B₀. Very similar arguments as put forward in the old literature let us choose A_⊥,∥ and B: note that there must be spin in the 3d(x² − y²) orbital, and it is very likely that there will also be an isotropic coupling term. Then, the NMR shifts demand, however, that the spin polarization in the 3d(x² − y²) orbital must be negative, as pointed out recently [14, 15], for we know that A_∥ is negative, and |A_∥|≫ A_⊥ [23].

Thus, we write with (1),

$$ {K}_{\parallel {\perp}} = A_{\parallel,\perp} \cdot (a+4c_{j}) + B \cdot 4(b_{j}+c_{j}). $$

(4)

For symmetry reasons, we take (b_j + c_j) from each of the 4 neighbors to be the same, i.e., from spin in the planar O 2p_σ orbitals, cf. Fig. 3. As before [15], we will neglect A_⊥ and simply write:

$$ \begin{array}{@{}rcl@{}} &&{K}{{~}_{\parallel}} = A_{\parallel} (a+4c_{j}) + B \cdot 4(b_{j}+c_{j}) \end{array} $$

(5)

$$ \begin{array}{@{}rcl@{}} &&{K}{{~}_{\perp}} \approx B \cdot4(b_{j}+c_{j}). \end{array} $$

(6)

This is a different notation from before [15] where we used b = 4b_j.

Fig. 3

In an external magnetic field B₀ two spin components a and b_j appear, originating from the planar Cu 3d(x² − y²) and the four surrounding O 2p_σ orbitals, respectively. Due to a coupling (c), the effective components are (a + 4c_j) and (b_j + c_j). While (b_j + c_j) is positive for the cuprates, (a + 4c_j) turns out to be negative. The hyperfine coefficients A_d and B lead to orientation-dependent (d) NMR shifts K_d = A_d(a + 4c_j) + B(b_j + c_j) at the Cu nucleus

A zero spin shift, our first, fundamental assumption means that ${\sum }_{j}(b_{j}+c_{j}) = 0$, and we have for the other orientation:

$$ {\hat{K}}{{~}_{\parallel}}(T \rightarrow 0) = {{K}}_{L \parallel} + A_{\parallel} (a+4c_j). $$

(7)

In order to estimate the orbital shift, K_L∥, for this orientation of the field, as before [15], the most reliable approach is to use (3) together with calculations of the orbital shift anisotropy, since the latter is mostly determined by matrix elements involving the orbital bonding wave functions of Cu and O [18, 19]. In fact, we use the suggested value of 2.4 from [19]:

$$ {~}^{63}{K}{{~}_{L\parallel}} = 2.4\cdot^{63}{{K}}_{L \perp} \approx 0.72<percent>. $$

(8)

Note that this value could vary between families, but since the orbital shift for c⊥B₀ does not change significantly between families, we do not expect a large effect for c ∥ B₀, as well. This is important as it means that most cuprates have a non-vanishing spin shift for c ∥ B₀ from a negative spin polarization in the 3d(x² − y²) orbital, even at the lowest temperatures.

As mentioned earlier, a few special slopes govern the shift-shift plot presented in their figure 7 [14], and we highlighted them in Fig. 2b, again. These are segments defined by temperature or doping for which the ratio of changes in both shifts is constant, ${\Delta {\hat {K}}{{~}_{\perp }}(T)}/{\Delta {\hat {K}}{{~}_{\parallel }}(T)} = \kappa ,$ and one finds 4 slopes, $\kappa \approx 0, 1, 2.5, \infty$. For example, κ = 1 denotes isotropic shift lines and readily follows from a mere change of b_j only, as it enters both terms in (5) and (6). Then, κ ≈ 0 in this approximation is realized by a change in a, only, since we neglected the rather small A_⊥. Note that term c operates on both shifts, K_⊥ and K_∥, and must be involved in the special slopes κ = 2.5 and $\kappa \approx \infty$. While not favored before [15], we believe that $\kappa \approx \infty$ is caused by a mere change in c. The reasoning is as follows: not a single material in the shift-shift plot shows a negative slope, i.e., a slope to the right of $\kappa \approx \infty$. This is remarkable and must mean that the component a cannot significantly be involved in shift changes.

With this assumption that c_j causes $\kappa \approx \infty$, we note that (5) and (6) require:

$$ A_{\parallel} \approx - B, $$

(9)

and we have with (5) and (6):

$$ \begin{array}{@{}rcl@{}} &&{K}{{~}_{\parallel}} \approx B (4b_{j}-a) \end{array} $$

(10)

$$ \begin{array}{@{}rcl@{}} &&{K}{{~}_{\perp}} \approx B 4(b_{j}+c_{j}). \end{array} $$

(11)

Note that in this approximation, c effectively acts only for c⊥B₀. Then, the slope of κ ≈ 2.5 is given by a concomitant change of b_j and c_j, e.g., Δb_j = 1.5Δc_j if both terms change proportionally.

To summarize, in the above model, the individual changes of a, b_j, and c_j correspond to slopes of κ = 0,1, and $\infty$, respectively, in Fig. 1 (if all b_j and c_j are the same). Even if this is not precisely what happens, we think that (10) and (11) still capture the fundamental aspects of the planar Cu shifts.

With these results in mind, we can look at the data for La_2−xSr_xCuO₄ again.

The high temperature shifts for La_2−xSr_xCuO₄, K_⊥, are much larger than what we expect from its K_∥ = B(4b_j − a) values. In one scenario, a larger a and larger b_j could position this family at larger K_⊥ (for given c_j). The action of a temperature-dependent c_j then leads to the $\kappa \approx \infty$ slope. Alternatively, c_j could be much larger for La_2−xSr_xCuO₄, i.e., much more positive, at high temperatures. This also leads to a much larger B(b_j + c_j). Again, a drop in c_j then makes (b_j + c_j) disappear.

To conclude, while La_2−xSr_xCuO₄ is an outlier in the shifts, the position in Fig. 1 can be understood within the two-component scenario, as well.

3 Planar Cu Relaxation

The nuclear relaxation rate 1/T_1∥ measures the in-plane fluctuating magnetic fields, ${\langle {h_{\perp }^{2}} \rangle }$, from electronic spin fluctuations, while 1/T_1⊥ is affected by both, in-plane, ${\langle {h_{\perp }^{2}} \rangle }$, as well as out-of-plane, ${\langle {h_{\parallel }^{2}} \rangle }$, fields (only fluctuating field components perpendicular to the nuclear quantization axis lead to nuclear spin flips, required for spin-lattice relaxation).

Phonons will cause nuclear relaxation for quadrupolar nuclei (I > 1/2, like Cu and O) as they modulate the electric field gradient, but it has been shown that the magnetic fluctuations dominate in most situations [24, 25], and the recent analysis of all Cu relaxation data shows that a simple magnetic mechanism appears to capture the overall behavior quite well [15, 16].

In a straightforward approach, one would assume nearly isotropic spin fluctuations filtered by the nuclear hyperfine coefficients, which can then lead to a relaxation anisotropy. The electronic correlation time (τ₀) of electronic spin fluctuations is expected to be very fast compared with the slow precession of the nuclei. Thus, the nuclear relaxation rates can be written as [18]:

$$ \begin{array}{@{}rcl@{}} &&\frac{1}{T_{1\parallel}} = \frac{3}{2} {\gamma^{2}} \cdot 2 {\langle {h_{\perp}^{2}} \rangle} \tau_{0} \end{array} $$

(12)

$$ \begin{array}{@{}rcl@{}} &&\frac{1}{T_{1\perp}} = \frac{3}{2} {\gamma^{2}} \left[ {\langle {h_{\perp}^{2}} \rangle}+ {\langle {h_{\parallel}^{2}} \rangle}\right]\tau_{0}, \end{array} $$

(13)

from which the relaxation anisotropy follows:

$$ \begin{array}{@{}rcl@{}} \frac{1/T_{1\perp}}{1/T_{1\parallel}} = \frac{1}{2} +\frac{{\langle {h}_{\parallel}^{2}} \rangle}{2{\langle {h}_{\perp}^{2}} \rangle}. \end{array} $$

(14)

Given that the shifts demand two different electronic spin components coupled to the nuclei through an anisotropic constant A_∥,⊥ and an isotropic constant B, one should allow for two different fluctuating spin densities α and $\upbeta = {\sum }_{j} {\upbeta }_{j}$, as well. Furthermore, since the fluctuations are caused by rapid exchange, the correlation time τ₀ should be the same for both components.

We thus write:

$$ \begin{array}{@{}rcl@{}} &&{\langle {h}_{\perp , \parallel}^{2} \rangle} \approx {\langle {(\sum\limits_{j}B{\upbeta}_{j}+A_{\perp,\parallel}\alpha)^{2}} \rangle} \end{array} $$

(15)

$$ \begin{array}{@{}rcl@{}} &&{\langle {h}_{\parallel}^{2} \rangle} \approx B^{2}{\langle {(\sum\limits_{j}{\upbeta}_{j}-\alpha)^{2}} \rangle}, \end{array} $$

(16)

$$ \begin{array}{@{}rcl@{}} &&{\langle {h}_{\perp}^{2} \rangle} \approx B^{2}{\langle {(\sum\limits_{j}{\upbeta}_{j}+f\alpha)^{2}} \rangle}, \end{array} $$

(17)

where we introduced f = A_⊥/B ≈−A_⊥/A_∥ if A_⊥α is not negligible (see below).

With these expressions for the fluctuating field components, we seek to explain a rather doping- and material-independent 1/T_1⊥ (it only increases marginally with decreasing doping) and a material- and doping-dependent 1/T_1∥ that explain the temperature-independent anisotropy (14), as well as the exceptional behavior found for La_2−xSr_xCuO₄.

In the first scenario, one might be interested to see what would be the consequences of totally uncorrelated spin fluctuations for the 5 spin components, i.e., ${\langle {{\upbeta }_{i}{\upbeta }_{j}} \rangle } = {\langle {{{\upbeta }_{0}^{2}}} \rangle }\delta _{ij}$, and 〈β_jα〉 = 0, cf. Fig. 4. We then have ${\langle {h_{\perp }^{2}} \rangle } = 4 {\langle {{{\upbeta }_{0}^{2}}} \rangle }$ and ${\langle {h_{\parallel }^{2}} \rangle }=4 {\langle {{{\upbeta }_{0}^{2}}} \rangle }+{\langle {\alpha ^{2}} \rangle }$, thus with (12) and (13) for uncorrelated (u) fluctuations:

$$ \begin{array}{@{}rcl@{}} &&\frac{1}{T_{1\parallel,u}} = \frac{3}{2} {\gamma^{2}} B^{2}\cdot 8 {\langle {{{\upbeta}_{0}^{2}}} \rangle} \tau_{0} \end{array} $$

(18)

$$ \begin{array}{@{}rcl@{}} &&\frac{1}{T_{1\perp, u}} = \frac{3}{2} {\gamma^{2}} B^{2} \cdot\left[ 8{\langle {{{\upbeta}_{0}^{2}}} \rangle}+ {\langle {\alpha^{2}} \rangle}\right]\tau_{0}, \end{array} $$

(19)

and it follows for the anisotropy:

$$ \frac{1/T_{1\perp,u}}{1/T_{1\parallel,u}} = 1+\frac{{\langle {\alpha^2} \rangle}}{8{\langle {{\upbeta}_0^2} \rangle}}. $$

(20)

Clearly, for ${\langle {\alpha ^{2}} \rangle } \lesssim {\langle {{{\upbeta }_{0}^{2}}} \rangle }$, we find near isotropic relaxation, and in order to explain the largest anisotropy of about 3.3 [16], we conclude ${\langle {\alpha ^{2}} \rangle } \approx 18.4 {\langle {{{\upbeta }_{0}^{2}}} \rangle }$. This implies, however, rather large changes of α and β for meeting the experimental observations, i.e., the change in relaxation between materials and different doping levels, which appears to be difficult to meet in this approach (we do notice that a large α could be present, which demands that we do not neglect A_⊥ for the modeling of nuclear relaxation).

Fig. 4

Fluctuating spins α and β₀, respectively located in the Cu 3d(x² − y²) and O 2p_σ orbital. a, all 5 spin components fluctuate independently, i.e., ${\langle {\alpha {\upbeta }_{0}} \rangle }=0, {\langle {{\upbeta }_{i}{\upbeta }_{j}} \rangle }= {{\upbeta }_{0}^{2}} \delta _{ij}$, b, the fluctuations are fully correlated, i.e., ${\langle {\alpha {\upbeta }_{0}} \rangle }=\alpha {\upbeta }_{0}, {\langle {{\upbeta }_{i}{\upbeta }_{j}} \rangle }= {{\upbeta }_{0}^{2}}$

In the second scenario, cf. Fig. 4, we assume that all spins are aligned; i.e., the 5 fluctuating spin components are correlated, with ${\langle {{\upbeta }_{i}{\upbeta }_{j}} \rangle } = {\langle {{{\upbeta }_{0}^{2}}} \rangle }$ and 〈β_jα〉 = ±αβ₀. We note that the field fluctuations ${\langle {h}_{\perp }^{2} \rangle } \approx B^{2}{\langle {({\sum }_{j}{\upbeta }_{j}+f\alpha )^{2}} \rangle }$ that enter (12) and ${\langle {h}_{\parallel }^{2} \rangle } \approx B^{2}{\langle {({\sum }_{j}{\upbeta }_{j}-\alpha )^{2}} \rangle }$ that determine (13) are both quadratic in the resulting local spin densities. Therefore, in order to find a rather flat dependence for the relaxation for c⊥B₀ on β₀, as demanded by the experiment, we need to be close to its minimum, while at the same time, the parabola must be shifted by a negative α compared with the other parabola in order to meet a smaller but varying relaxation rate for c ∥ B₀. The results of simple calculations according to (12), (13) with (16) and (17) are shown in Fig. 5. We observe that there is only a special region with solutions that fit the experiments, for β₀/α = 0.04 to 0.11 according to anisotropies ranging from 3.3 to 1.0, respectively, cf. Fig. 5.

Fig. 5

a Calculated nuclear relaxation rates for c ∥ B₀ (1/T_1∥) and c⊥B₀ (1/T_1⊥) as a function of the ratio of the two spin components β₀ ≡β_j and α (β₀/α), according to (12), (13) with (16) and (17), in arbitrary units. b The anisotropy of the relaxation (14) varies between 4 and 0.5 in the same range of β₀/α. Corresponding line segments for La_2−xSr_xCuO₄ with an anisotropy of about 2.3 are indicated, as well

Furthermore, an increase of α by a factor of about 1.3, at an anisotropy ratio of 2.3, increases the relaxation rates in both directions by about a factor of 2, cf. Fig. 5, which readily explains the data found for La_2−xSr_xCuO₄. We thus conclude that the two components β₀ and α are crucial for the cuprates, but appear to be very similar for most of the materials.

4 Discussion

It seems out of question that a two-component scenario describes the shifts and relaxation in the cuprates quite well. It has spin density located in the Cu 3d(x² − y²) orbital, which couples to the nucleus through the rather anisotropic hyperfine constant A_∥,⊥, and, most likely, the planar O 2p_σ orbital, leading to an isotropic hyperfine interaction given by (4).

The spin density α is much larger than β₀, as one expects from the overall material properties; however, the uniform response of both spins is quite different, also due to the coupling term c_j.

The special slopes observed in the shift-shift plot, cf. Fig. 1, are caused by changes of the individual spin components as a function of doping or temperature, except for the slope κ ≈ 2.5 that must stem from a concomitant change of b_j and c_j. This leads to the simple conclusion that A_∥≈−B (while A_⊥≈ 0.15A_∥ [23]), and it leaves us with a straightforward description of the spin shifts of the cuprates in terms of (10) and (11), i.e., K_∥≈ B(4b_j − a) and K_⊥≈ B4(b_j + c_j) (in these equations, we also adopted a different notation in terms of b_j compared with our earlier analysis [15]). We note that the conclusion that A_∥≈−B has a similar origin as in the old interpretation.

Looking again at Fig. 1b, the cuprates are sorted in this shift-shift plot effectively by the high-temperature b_j, the component that grows with increasing doping (toward the upper right in Fig. 1b). As the temperature is lowered, at a given temperature, which can be above or at T_c, this term begins to disappear due to the action of c_j (both components c_j and b_j can fall together, as well). It is the coupling to a that sets c_j (and effectively couples different a terms, as well). The component a appears to be temperature independent. It emerges that either b_j or c_j can be exhausted independently indicated by changes in slope at lower temperatures. However, all cuprates seem to reach the same (b_j + c_j) = 0, which we define as zero spin shift. Importantly, there is no evidence that there is a different mechanism as one passes through T_c if the shift began to change already far above T_c (NMR pseudogap), but c_j can traverse the region below T_c at a much higher rate for given steps in temperature.

The earlier conclusion [14, 15] that the spin shift for c ∥ B₀ does not disappear at low temperatures, here takes the formulation that (4b_j − a)≠ 0 and says that the positive spin density b_j and the negative spin density a_j can remain temperature independent for systems with $\kappa \approx \infty$, or b_j can also drop with c_j as for the systems with slope κ ≈ 2.5; i.e., it does not change in the condensed state, while relaxation ceases.

In terms of a simple fluctuating field model, we can explain the cuprate relaxation rather well. Fast electronic, Fermi liquid-like spin fluctuations act through two different hyperfine coefficients with two different electronic spin densities on the Cu nucleus (or, these densities are part of that ubiquitous fluid). The corresponding fluctuations from the 5 locations must be correlated, as one might have guessed due to the close proximity. The on-site 3d(x² − y²) spin (α) is about 10 times as large as that due to one O neighbor (β₀). The spin density α appears to be the same for all cuprates; except for the La_2−xSr_xCuO₄ family, it is 30% larger. The spin β₀ varies with doping and between materials and leads to the change in 1/T_1∥ observed in the data. For large doping the relaxation anisotropy is about 1 and it increases to about 3.3 for YBa₂Cu₄O₈ (corresponding to a change in β₀ of about 3). For La_2−xSr_xCuO₄ the anisotropy is 2.3 and thus also β₀ is about a factor of two larger.

Since the ⁶³Cu relaxation begins to disappear only at T_c, for all cuprates, the electronic, Fermi liquid-like spin fluctuations freeze out and the relaxation disappears. Thus, the pseudogap in the relaxation is just due to the correlations that for planar Cu do not change α and β₀ spin alignment. For planar O the situation is different as the nucleus couples to two α spins at adjacent Cu nuclei and their coupling changes, which leads to the pseudogap in the relaxation for nuclei that are affected by different a spins [26].

The relation between the spin densities α,β₀ and the uniform response of the system in terms of a, b_j, and c_j is not known. It appears that the response of α is rather small compared to that of β, which may not be surprising since different a should favor antiferromagnetic alignment (that is somehow affected by β).

It appears that the doping dependent spread in K_∥ varies among the cuprates. This reminds us of the way the charge carriers enter the CuO₂ plane [12]. For the La_2−xSr_xCuO₄ family, the doped charges x enter almost exclusively the 2p_σ orbital (n_p) while for other systems the Cu 3d(x² − y²) (n_d) is affected as well (x = Δn_d + 2Δn_p [12, 27]), and the spread in doping appears to grow with Δn_d. The maximum achievable T_c, however, is set by the sharing of the parent material’s hole content, i.e., $n_{d}^{*} + 2n_{p}^{*} =1$ and $T_{\text {c, max}} \propto n_{p}^{*}$ [12, 13, 28]. Materials with the highest T_c appear to adopt κ ≈ 2.5, only. However, the jumping between different slopes κ in different regions of the shift-shift plot that involves b_j and/or c_j below T_c is absent for optimally doped systems, which probably means that b_j and c_j are matched at optimal doping.

Finally, one may argue that the intra cell charge variation between neighboring planar O atoms that appears to be ubiquitous and that can respond to the external magnetic field [21, 29] could be involved in the two component scenario.

5 Conclusions

Two spin densities were shown to reside in the planar Cu 3d(x² − y²) and likely the planar O 2p_σ orbitals, respectively, with hyperfine constants A_∥,⊥ and B ≈−A_∥. They connect the Cu nuclear spins with a rather ubiquitous Fermi liquid-like bath. The relaxation anisotropy is predominantly due to changes in the planar O spin density that increases with doping. Near T_c, these electronic fluctuations freeze out and the relaxation disappears.

The uniform response a and b_j of the two electronic spins on Cu and O is special in the sense that a is negative while b_j points along the field. The coupling term c_j between a and b_j sets the temperature dependence of the shift above (NMR pseudogap) and below T_c. Interestingly, at the lowest temperatures, 4(b_j + c_j) approaches the same value for all cuprates, probably zero, but a remains and most of b_j, as well, resulting in a non-vanishing spin shift for c ∥ B₀, K_∥≈ (4b_j − a)≠ 0.

The coupling term c_j must be related to a coupling between different spin components a_i on different Cu nuclei, and it is argued that the pseudogap phenomenon for planar O nuclear relaxation, and that of Y, is just a consequence of the temperature dependence of c_j, an effect that cannot be there in the Cu relaxation data.

This simple two-component scenario appears to fit all cuprates, in particular also the only outlier family so far, La_2−xSr_xCuO₄, which must make it a reliable framework for theory.