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In Quantum Mechanics (QM), the Planck's constant has the S.I. units of

(energy) x (time). So, how should one interpret that?!

How should I understand the physical meaning of the following relations?

$\frac E \nu = h$

as the "energy content" divided by the "fastness'' of oscillation equals h.

or in terms of the de Broglie definition

$ |\vec{p}| \times \lambda= h$

i.e. particle momentum times the ``size'' of wavelength equals h.

I know that the definitions are linked. However, for the sake of basic understanding, two questions

a) Should I understand that the size of the wavelength is $\underline{\text{fixed}}$ w.r.t the momentum of the particle and similarly its energy fixes the fastness of its oscillations? They are one and the same thing being looked at from different perspective?

b) Is there any classical version of a physical constant that has more than one definition like the Planck's constant?

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    $\begingroup$ Your first relation is wrong. The appropriate equation is $E = h \nu$. $\endgroup$
    – J. Murray
    May 24, 2020 at 3:30
  • $\begingroup$ Ah! thank you. I shall make the typo correction. $\endgroup$ May 24, 2020 at 3:51
  • $\begingroup$ Planck's constant has units of action. See en.wikipedia.org/wiki/Action_(physics) $\endgroup$
    – PM 2Ring
    May 26, 2020 at 8:32

2 Answers 2

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There are no two definitions of Planck's constant. In fact, the two relations you gave above are connected. According to Special Relativity, the energy of an object is $$E^2 = (\vec{p} c)^2 + (mc^2)^2$$ But a photon has no mass, so you get $E = \vec{p}c$. And the wavelength of the photon is connected to the frequency ($\nu$). Actually, the wavelength $\lambda$ is equal to ${c}/{\nu}$, where $c$ is the speed of light (and conversely, $\nu = c / \lambda$). So equating, you get: $$h = \frac{E}{\nu} = \frac{\vec{p}c}{c/\lambda} = \vec{p}{\lambda}$$

As for a), actually the relation is $E = h \nu$ and $\lambda = h / {p}$. So the wavelength and momentum are proportional, but not fixed w.r.t. each other. In fact the De-Broglie wavelength, shows that the wavelength of a particle depends on the momentum. Same for the relation between energy and frequency. The energy of a particular photon is proportional to it's frequency, and $h$ is just the proportionality constant between them.

Quick Note:- How the Planck's constant gets the unit $(energy) * (time)$. The unit of the frequency is $1 / T$, where time is the period of the wave measured in seconds. Then you have energy, having SI units Joules. So by $\frac{E}{\nu} = h$, you get $\frac{E}{1 / T} = ET$, hence the units $(energy) * (time)$.

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  • $\begingroup$ Wow! yes, this makes so much sense to me now. Using the special relativity here makes it easier to see the connection. What prompted my question was that i wanted to see energy x time along the lines of how we understand distance traveled as velocity of the particle x duration of time. So, now can we say that the for all the particles energy x time interval stays constant? or would it be wrong to say that time is inversely proportional to energy? $\Delta E = \hbar/ \Delta T$. Then, Planck's constant is just a proportionality constant! isn't it? $\endgroup$ May 24, 2020 at 7:33
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This is not unusual. Many quantities have two separate, common expressions in terms of two different sets of other quantities.

For example, the permeability of free space $\mu_0$ has units of $(\text{inductance})/(\text{length})$ and also of $(\text{force})/(\text{current})^2$.

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  • $\begingroup$ Yes. But then isn't there the idea of derived units vs basic units or a matter of definition between different unitary systems? The permeability could also be given as the ratio of B to H field. However, i wanted to know in b) if the two definitions of the Planck's constant arise due to the interlinked physical quantities involved. This is just for a basic conceptual understanding. Thank you for your answer. I think that part b) is closed now. $\endgroup$ May 24, 2020 at 4:17
  • $\begingroup$ @Ishika_96_sparkle Neither energy nor momentum are basic units, so there's no difference between the two constants on that front. $\endgroup$ May 24, 2020 at 4:18
  • $\begingroup$ Thanks for the clarification. I made a mistake in understanding your point. $\endgroup$ May 24, 2020 at 4:37

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