The K state, an early intermediate of the bacteriorhodopsin photocycle, contains the excess free energy used for light-driven proton transport. The energy gain must reside in or near the photoisomerized retinal, but in what form has long been an open question. We produced the K intermediate in bacteriorhodopsin crystals in a photostationary state at 100 K, with 40% yield, and determined its X-ray diffraction structure to 1.43 Å resolution. In independent refinements of data from four crystals, the changes are confined mainly to the photoisomerized retinal. The retinal is 13- cis,15- anti, as known from vibrational spectroscopy. The C 13C 14 bond is rotated nearly fully to cis from the initial trans configuration, but the C 14–C 15 and C 15NZ bonds are partially counter-rotated. This strained geometry keeps the direction of the Schiff base N–H bond vector roughly in the extracellular direction, but the angle of its hydrogen bond with water 402, that connects it to the anionic Asp85 and Asp212, is not optimal. Weakening of this hydrogen bond may account for many of the reported features of the infrared spectrum of K, and for its photoelectric signal, as well as the deprotonation of the Schiff base later in the cycle. Importantly, although 13- cis, the retinal does not assume the expected bent shape of this configuration. Comparison of the calculated energy of the increased angle of C 12–C 13C 14, that allows this distortion, with the earlier reported calorimetric measurement of the enthalpy gain of the K state indicates that a significant part of the excess energy is conserved in the bond strain at C 13.