Planck constant

SI defining constant

Name	Symbol	Base units
Planck constant	h	kg m² s⁻¹
The Planck constant, symbol h, is a fundamental constant of nature. It is the proportionality constant that relates the energy carried by a photon to its associated wave frequency. The numerical value of the Planck constant, symbol h, is defined to be exactly 6.626 070 15 × 10⁻³⁴ when expressed in the unit joule second, J s, or kg m² s⁻¹.
$h = 6.626 \mspace{4mu} 070 \mspace{4mu} 15 \times 10^{-34} \mspace{6mu} \text{J} \mspace{4mu} \text{s}$

The Planck constant and energy

In quantum mechanics, the Planck–Einstein relation states that the energy, E, of a photon is directly proportional to the frequency, ν, of its associated wave:

$E \propto \nu$

When expressed in the form of an equation, the Planck constant, h, is the proportionality constant:

Using SI coherent units,

$E = h \nu$

where:

E is energy in joules, symbol J,
ν is frequency in hertz, symbol Hz,
h is the Planck constant, in joule seconds, symbol J s.

The relation between the Planck constant and frequency forms the basis for the definition of the unit of energy, the joule:

For a photon with an associated wave frequency equal to the caesium frequency, the energy of the photon has the exact value:

$h \mspace{4mu} \Delta \nu _{Cs} \mspace{6mu} = (6.626 \mspace{4mu} 070 \mspace{4mu} 15 \times 10^{-34})(9 \mspace{4mu} 192 \mspace{4mu} 631 \mspace{4mu} 770) \mspace{6mu} \text{J}$

Inverting this relation gives an exact expression for the joule in terms of the Planck constant, h, and the caesium frequency, Δν_Cs :

$1 \mspace{4mu} \text{J} \mspace{6mu} = \dfrac{1}{(6.626 \mspace{4mu} 070 \mspace{4mu} 15 \times 10^{-34})(9 \mspace{4mu} 192 \mspace{4mu} 631 \mspace{4mu} 770)} \mspace{6mu} h \mspace{4mu} \Delta \nu _{Cs}$

Equivalence of mass and energy

Einstein’s principle of the equivalence of mass and energy describes the relation between energy, E, and mass, m, where c is the speed of light in vacuum:

Using SI coherent units,

$E = m c^2$

where:

E is energy in joules, symbol J,
m is mass in kilograms, symbol kg,
c is the speed of light in vacuum, in metres per second, symbol m s⁻¹.

The Planck constant and mass

Combining the two above expressions for energy gives the relation between the Planck constant and mass:

Using SI coherent units,

$m c^2 = h \nu\\ \\ \\ m \mspace{14mu} = \dfrac{h \nu}{c^2}$

where:

m is mass in kilograms, symbol kg,
ν is frequency in hertz, or reciprocal seconds, symbol s⁻¹,
c is the speed of light in vacuum, in metres per second, symbol m s⁻¹,
h is the Planck constant, in kg m² s⁻¹.

This relation forms the basis for the definition of the unit of mass, the kilogram:

For a photon with an associated wave frequency equal to the caesium frequency, the mass equivalent to the photon’s energy has the exact value:

$\dfrac{h \mspace{4mu} \Delta \nu _{Cs}}{c^2} \mspace{6mu} = \dfrac{(6.626 \mspace{4mu} 070 \mspace{4mu} 15 \times 10^{-34})(9 \mspace{4mu} 192 \mspace{4mu} 631 \mspace{4mu} 770)}{(299 \mspace{4mu} 792 \mspace{4mu} 458)^2} \mspace{6mu} \text{kg}$

Inverting this relation gives an exact expression for the kilogram in terms of the Planck constant, h, the caesium frequency, Δν_Cs, and the speed of light in vacuum, c:

$1 \mspace{4mu} \text{kg} \mspace{6mu} = \dfrac{(299 \mspace{4mu} 792 \mspace{4mu} 458)^2}{(6.626 \mspace{4mu} 070 \mspace{4mu} 15 \times 10^{-34})(9 \mspace{4mu} 192 \mspace{4mu} 631 \mspace{4mu} 770)} \mspace{6mu} \dfrac{h \mspace{4mu} \Delta \nu _{Cs}}{c^2}$

The Planck constant and electricity

The relation between the Planck constant and the elementary charge forms the basis for the definition of the unit of electrical resistance, the ohm:

Expressed in SI units, the quotient ^h⁄_e² has the exact value:

$\dfrac{h}{e^2} \mspace{6mu} = \dfrac{6.626 \mspace{4mu} 070 \mspace{4mu} 15 \times 10^{-34}}{(1.602 \mspace{4mu} 176 \mspace{4mu} 634 \times 10^{-19})^2} \mspace{6mu} \Omega$

Inverting this relation gives an exact expression for the ohm in terms of the Planck constant, h, and the elementary charge, e :

$1 \mspace{4mu} \Omega \mspace{6mu} = \dfrac{(1.602 \mspace{4mu} 176 \mspace{4mu} 634 \times 10^{-19})^2}{6.626 \mspace{4mu} 070 \mspace{4mu} 15 \times 10^{-34}} \mspace{6mu} \dfrac{h}{e^2}$

The Planck constant and magnetism

The relation between the Planck constant and the elementary charge also forms the basis for the definition of the unit of magnetic flux, the weber:

Expressed in SI units, the quotient ^h⁄_e has the exact value:

$\dfrac{h}{e} \mspace{6mu} = \dfrac{6.626 \mspace{4mu} 070 \mspace{4mu} 15 \times 10^{-34}}{1.602 \mspace{4mu} 176 \mspace{4mu} 634 \times 10^{-19}} \mspace{6mu} \text{Wb}$

Inverting this relation gives an exact expression for the weber in terms of the Planck constant, h, and the elementary charge, e :

$1 \mspace{4mu} \text{Wb} \mspace{6mu} = \dfrac{1.602 \mspace{4mu} 176 \mspace{4mu} 634 \times 10^{-19}}{6.626 \mspace{4mu} 070 \mspace{4mu} 15 \times 10^{-34}} \mspace{6mu} \dfrac{h}{e}$

Macroscopic quantum phenomena

The magnitudes of quantized values of voltage and resistance, measured using the Josephson and quantum Hall effects respectively, are directly proportional to the value of the Planck constant.

These two macroscopic quantum phenomena allow a link to be established between the Planck constant and the macroscopic measurement of mass.

In 2019, this link enabled the kilogram to be redefined in terms of the Planck constant.

The Planck constant and spin angular momentum of light

The component of the angular momentum of light associated with a photon’s quantum spin and the rotation between its polarisation degrees of freedom is known as the spin angular momentum of light.

When a beam of light is circularly polarised, each of its photons has a spin angular momentum equal to ±ħ, where ħ is the reduced Planck constant.

The SI derived unit used to express the Planck constant, the joule second, symbol J s, is equivalent to the joule per hertz, symbol J Hz^-1.

The reduced Planck constant, ħ, relates to the Planck constant in the same way as the hertz relates to the radian per second – one hertz being equal to one complete cycle, or 2π radians, per second.

$1 \ \text{rad s}^{-1} = \dfrac{1}{2 \pi} \ \text{Hz}\\ \\ \\ \hbar \mspace{56mu} = \dfrac{h}{2 \pi}\\ \\ \\ \hbar \mspace{56mu} = \dfrac{6.626\ 070\ 15 \times 10^{-34}}{2 \pi} \ \text{J Hz}^{-1}\\ \\ \\ \hbar \mspace{56mu} = 1.054\ 571\ 817\ \text{...} \times 10^{-34} \ \text{J Hz}^{-1}$