Converting from hbar = 1 to physical units

[Kreizhn]

Hi,

Hopefully this should be a simple and quick question. I've seen many calculations done with [itex] \hbar = 1[/itex]. If I want to convert back into physical units, what is the corresponding transformation?

I know wiki gives a description of http://en.wikipedia.org/wiki/Natural_units#.22Natural_units.22_.28particle_physics.29". Does this give the transformation values? Namely, if we computed a time of 1 units when [itex] \hbar = 1[/itex] does this correspond to [itex] 6.58 \times 10^{-16} [/itex] seconds of physical time?

 

[Dale]

If h=1 then that is using some specified system of units, e.g. Planck units. If you need to convert to SI units then all you need to do is follow the usual unit conversion rules.

 

[Kreizhn]

Ok, so is there a canonical choice of those units? Do they correspond to the link that I posted above? Most articles that I've read that do this just say something to the effect of "WLOG, we take [itex] \hbar = 1 [/itex]," with no further details given.

 

[Dale]

No, there is not a standard set of units. For example, when setting c=1 you could be using units of Planck length and Planck time, but you could instead be using units of light years and years. Either way c=1. Similarly for Planck's constant.

 

[Kreizhn]

Okay, but then it's not contextually obvious what to do in my case. Essentially, I have the dimensionless operator-Schrodinger equation

[tex] \frac{d}{dt}U(t) = -iH(t)U(t), \qquad U(t) \in \mathfrak U(N) [/tex]

Let's say we know that after a time T, we get [itex] U(T) = U_f [/itex] for some specific unitary [itex] U_f [/itex], and I want to figure out what this time T is physically. How do I do this? There is absolutely no mention of units, because we have specifically chosen to make everything unitless (via arbitrarily setting [itex] \hbar = 1[/itex]).

 

[Kreizhn]

Is it actually much simpler than needing to do a conversion in this case? Had the hbar been included, all that changes is the time-factor right? Namely, instead of

[tex] U(T) = \mathcal T \exp\left[ -i\int_0^T H(t) \ dt \right] U(0) [/tex]
we would get
[tex] U(T) = \mathcal T \exp\left[ -\frac i\hbar\int_0^T H(t) \ dt \right] U(0) [/tex]
so then all this is is merely a time-scaling by a factor of hbar?

 

[Dale]

Okay, but then it's not contextually obvious what to do in my case. Essentially, I have the dimensionless operator-Schrodinger equation

[tex] \frac{d}{dt}U(t) = -iH(t)U(t), \qquad U(t) \in \mathfrak U(N) [/tex]

Let's say we know that after a time T, we get [itex] U(T) = U_f [/itex] for some specific unitary [itex] U_f [/itex], and I want to figure out what this time T is physically. How do I do this? There is absolutely no mention of units, because we have specifically chosen to make everything unitless (via arbitrarily setting [itex] \hbar = 1[/itex]).

I am not an expert in QM, but it is pretty clear that not everything is unitless in that equation. In fact, if U and H are unitless then this equation is dimensionally inconsistent since the left hand side has units of inverse time.

Regardless of the units of U, H must have units of inverse time for this equation to hold.

 

[Dale]

Is it actually much simpler than needing to do a conversion in this case? Had the hbar been included, all that changes is the time-factor right? Namely, instead of

[tex] U(T) = \mathcal T \exp\left[ -i\int_0^T H(t) \ dt \right] U(0) [/tex]
we would get
[tex] U(T) = \mathcal T \exp\left[ -\frac i\hbar\int_0^T H(t) \ dt \right] U(0) [/tex]
so then all this is is merely a time-scaling by a factor of hbar?

These cannot both be true if U and H have the same dimensions in both equations. The top equation could be true if H has the same dimensions of U and a factor of inverse time.

 

[Kreizhn]

Maybe there is something mathematically deep going on, but in a mathematical framework this equation is perfectly consistent so long as U is unitary and H is Hermitian. In that case, this is nothing but an integral curve of a right-invariant vector-field. More explicitly, [itex] -iH(t) U(t) = dR_{U(t)}(-iH(t)) [/itex] which is the pushforward of the group-action of right-translation on an element of the Lie algebra.

 

[Kreizhn]

What do you mean by dimension? The Lie-algebra and Lie-group must have the same dimension mathematically. That is, a unitary's evolution is always dictated by a skew-Hermitian Hamiltonian of exactly the same dimension.

Edit: Perhaps the integral sign is misleading. Try taking it out and realizing the evolution of U as the flow of a one-parameter subgroup generated by H.

 

[Dale]

Maybe there is something mathematically deep going on, but in a mathematical framework this equation is perfectly consistent so long as U is unitary and H is Hermitian.

Does U being unitary and H being Hermitian put any constraints on their units?

 

[Kreizhn]

Not so far as I know . I'm not sure what it means for an operator to have units in the first place.

 

[Dale]

What do you mean by dimension?

http://en.wikipedia.org/wiki/Dimensional_analysis

 

[Dale]

Not so far as I know . I'm not sure what it means for an operator to have units in the first place.

Well, if neither U nor H have units then the above equations are dimensionally inconsistent.

 

[Kreizhn]

Ah, that kind of dimension.

See, I think this is the physics and the math clashing. A parameterized curve [itex] U: \mathbb R \to \mathfrak U(N) [/itex] need not have a "dimension" of time.

Edit: The fact that [itex] U'(0) = H(0) [/itex] indicates that H(0) lives in the tangent space at U(0). I guess classically that would impart an inverse-time dimension to it? Like, if x is a position vector in [itex] \mathbb R^n [/itex] and we let v represent its instantaneous velocity, then x having units L will imply v has units [itex] LT^{-1} [/itex] right? But from a geometric point of view, we only need that [itex] v \in T_x \mathbb R^n \cong \mathbb R^n [/itex].

 

[Dale]

See, I think this is the physics and the math clashing. A parameterized curve [itex] U: \mathbb R \to \mathfrak U(N) [/itex] need not have a "dimension" of time.

No, but d/dt does and therefore dU/dt must have different units than U regardless of whatever the units of U are.

 

[Kreizhn]

That's just it, there is no mathematical reason for U to have units or for H to have units of inverse time. What does time even mean in this case? It is merely a parameter that defines the position of the flow of the infinitesimal generator.

 

[Rap]

You have to specify all of your units. MKS system defines, e.g., the kilogram, meter and second. You can choose to define unit speed, action, and mass as c, h, and me where c is the speed of light, h is Planck's constant, me is mass of electron.

When setting h=1, you are defining the unit of action to be the Planck constant. You have to know what two more units are before you can make dimensional sense of an equation. If you simply set h=1 in an MKS equation, it makes no sense until you explicitly state how the other two units have been modified. Are we talking meter, kilogram, action as our three new units? Kilogram, second, action?

 

[Kreizhn]

Maybe I'm missing something, but it does not make sense to talk about length or mass or even time when dealing in an operator framework. We are not discussing physical quantities: this equation describes the evolution of an operator on a differentiable manifold.

This is done quite often. For the sake of example, take a look at http://pra.aps.org/abstract/PRA/v65/i3/e032301" by Khaneja et al. written in 2002. The operator-Schrodinger equation is equation (1) introduced in the second paragraph. If it's clear to you what units he's using, by all means tell me.

 

[Dale]

Maybe I'm missing something, but it does not make sense to talk about length or mass or even time when dealing in an operator framework. We are not discussing physical quantities: this equation describes the evolution of an operator on a differentiable manifold.

An operator can certainly have units or change the units of the quantities on which it operates. A perfect example is the differentiation operator. Similarly with a matrix.

I can't access the paper. Perhaps a website or an arxiv link would work?

 

[Kreizhn]

Differentiation on a manifold (a la [itex] \frac\partial{\partial x^i} [/itex]) only makes sense if you've defined a coordinate system, otherwise one has to use the exterior derivative which is unitless. The definition of such a coordinate system may impart units to the elements, but the point is that those element exist independent of the definition of a system.

This should be the http://arxiv.org/abs/quant-ph/0106099" .

Edit: Fixed Link.

 

[Dale]

From that first equation it is unclear what the units of U are, but the units of H are clearly inverse time. This is confirmed on page 6 where J is explicitly identified as having units of inverse time and H/J is unitless.

 

[Rap]

Differentiation on a manifold (a la [itex] \frac\partial{\partial x^i} [/itex]) only makes sense if you've defined a coordinate system, otherwise one has to use the exterior derivative which is unitless. The definition of such a coordinate system may impart units to the elements, but the point is that those element exist independent of the definition of a system.

If I have a vector field V(r) as a function of vector position r, the divergence of that field is independent of coordinate system, yet does not have the same units as V.

 

[Kreizhn]

From that first equation it is unclear what the units of U are, but the units of H are clearly inverse time. This is confirmed on page 6 where J is explicitly identified as having units of inverse time and H/J is unitless.

It certainly seems that J has units 1/T, but where on page six is it stated that H/J is unitless?

If I have a vector field V(r) as a function of vector position r, the divergence of that field is independent of coordinate system, yet does not have the same units as V.

Certainly, we expect such properties to be invariant of coordinates. However, your calculation of the divergence still intrinsically requires a coordinate system. A vector field is by definition a smooth section of the projective mapping of the tangent bundle. One does not need a coordinate system to define such a thing. Further, the notion of divergence is not guaranteed to exist. We're lucky in that the unitary manifold is smooth and hence can be endowed with a Riemannian structure, allowing us to define a volume form and consequently a divergence, but any calculation requires the specification of a local coordinate system.

 

[Rap]

Certainly, we expect such properties to be invariant of coordinates. However, your calculation of the divergence still intrinsically requires a coordinate system. A vector field is by definition a smooth section of the projective mapping of the tangent bundle. One does not need a coordinate system to define such a thing. Further, the notion of divergence is not guaranteed to exist. We're lucky in that the unitary manifold is smooth and hence can be endowed with a Riemannian structure, allowing us to define a volume form and consequently a divergence, but any calculation requires the specification of a local coordinate system.

Yes, but my point is that the divergence does not require a coordinate system in order to be defined, and the divergence is not a dimensionless operator.

 

[Kreizhn]

Here I think is the problem. The operator [itex] \frac d{dt} [/itex] has units of [itex] T^{-1} [/itex] ONLY if t has units of time. Similarly, [itex] \frac d{dx} [/itex] has units of inverse length ONLY if x has units of length.

In general curves are parameterized by real, unitless values. Points can be represented in terms of unitless coordinates. In fact, if t,x, are both unitless then it makes perfect sense to say something like
[tex] \frac{ dx}{dt} + x = f(x) [/tex]
without demanding there is a dimensional violation of adding L/T to T. Now, we should expect H to have units of energy, since it represents the system Hamiltonian. This is why [itex] \hbar \frac d{dt} U = -i H U [/itex] makes sense. Why can we just set [itex] \hbar = 1[/itex] without performing a non-dimensionalizing process? I have no idea. What I think ultimately happens is that abstractly, when one casts the Schrodinger equation into a mathematical framework, hbar does not represent a unit-ful quantity but rather a unitless scaling property. Think about it, U has to be unitless since the wavefunction is unitless and
[tex] | \psi(t) \rangle = U(t) |\psi(0) \rangle [/tex]

Something fishy is going on here.

 

[Kreizhn]

Yes, but my point is that the divergence does not require a coordinate system in order to be defined, and the divergence is not a dimensionless operator.

On what grounds can you claim it is not dimensionless? In the Euclidean domain where you've identified points with length measures then sure, it's not dimensionless. However, this topic makes no allusion to Euclidean space, or a unit-ful identification to coordinates.

Edit: With the possible exception that originally, in the non-dimensionalized form we've associated the curve parameter to time. However, this is again on Euclidean space. I do not see how you could possibly push forward those coordinates onto the manifold.

I mean, at some point someone said "Hey, H has units of energy." But then I ask, what does it mean to exponentiate energy? What unit does that give you?

 

[Kreizhn]

Okay. Blah.

I think I have it.

Consider the dimension-ful expression
[tex] i\hbar \frac{dU}{dt} = HU [/tex]
when one solves this, you get
[tex] U(t) = \mathcal T \exp\left[ -\frac i\hbar \int_0^t H(t) \ dt \right] U(0) [/tex]
U is unitless, and in fact the quantity in the argument of the exponential must also be unitless right? Conversely, in the [itex] \hbar = 1 [/itex] case we have that the solution is
[tex] U(t) = \mathcal T \exp\left[ -i \int_0^t H(t) \ dt \right] U(0) [/tex]
and again the argument must be unitless, so that H(t) has units of inverse time. Since the argument must remain unitless, the conversion between these two forms in terms of the time parameter is simply a rescaling by [itex] \hbar [/itex]. Since hbar has units Energy x Time ( or if you want to be more precise and give it [itex] ML^2T^{-2} [/itex]), our choice of standardized energy is absorbed by the H. This means that the time-component is only affected by a scaling by hbar.

 

[Dale]

It certainly seems that J has units 1/T, but where on page six is it stated that H/J is unitless?

Where -iH/2J is used as the argument to exp. The argument to trigonometric functions is necessarily always unitless.

 

[Rap]

This is why [itex] \hbar \frac{d}{dt} U = -i H U [/itex] makes sense. Why can we just set [itex] \hbar = 1[/itex] without performing a non-dimensionalizing process?

We cannot. What you are doing is redefining H as well. In the original equation, H has units of action divided by time. When you say hbar=1, what you are doing is defining a unit of action, let's call it the Planck. Suppose we were working in the MKS system. We have to get rid of one of the fundamental units, because we have the new unit, the Planck, and we only need three.

Lets get rid of the meter. Now length is measured in (planck-sec/kg)^1/2. The original equation becomes [itex] \frac d{dt} U = -i H U [/itex] where U is in Planck/sec (energy) and H is in sec^-1. If you solve for U, you will get U in Planck/sec. Multiply U by hbar Joule-sec/planck and you will have U in Joules.

Lets get rid of the second. Now time is measured in kg m^2/planck. The original equation becomes [itex] \frac d{dt} U = -i H U [/itex] where U is in Planck^2/kg/m^2 (energy) and H is in kg m^2/planck (time). Multiply U by hbar^2 (Joule-sec/planck)^2 and you will have U in Joules.

Lets get rid of the kilogram - well, you get the idea.

Edit - I assumed the units of U were energy - probably a mistake (I've been doing thermo lately). Anyway, the analysis can be carried out the same way.

 

[Kreizhn]

Hmm...Well, U has be to unitless since
[tex] | \psi(t) \rangle = U(t) |\psi(t_0) \rangle [/tex]
and hence must preserve whatever units we give to the wave function. I'm not sure if that affects your transformations. Secondly, I might not have mentioned this, but we're only interested in how time scales. Namely, if I fix my H (which corresponds to changing the system) and solve it in the hbar = 1 and hbar \neq 1 system, then converting time still corresponds to a scaling by hbar.

 

[Dale]

I don't know if the QM literature is similar to the GR literature, but in GR you are expected to know the units for each variable. The factors of c and G are dropped everywhere, and if you ever need them then you just put them back in whatever combination is needed to make the units work correctly.

In this case I think that H is supposed to have dimensions of energy, and U is supposed to be dimensionless (not certain about that). So you just put in the factors of h and c wherever needed in order to make the equations consistent.

 

[Rap]

I don't know if the QM literature is similar to the GR literature, but in GR you are expected to know the units for each variable. The factors of c and G are dropped everywhere, and if you ever need them then you just put them back in whatever combination is needed to make the units work correctly.

In this case I think that H is supposed to have dimensions of energy, and U is supposed to be dimensionless (not certain about that). So you just put in the factors of h and c wherever needed in order to make the equations consistent.

No. G has units L^3/m/t^2 (L is length, m is mass, t is time). c has units L/t. Suppose I get a result T that I know is time. What combination do I put G and c back into make the units of T come out to time? You cannot use G because that will introduce a mass you cannot get rid of. You cannot use just c because that will introduce a length you cannot get rid of.

 

[Dale]

Huh? If T is already time then to get units of time you obviously just leave it alone, i.e. G^0 c^0. You only need to put in factors of G and c if you want to convert T to some other dimensions besides time.

 

[Rap]

Huh? If T is already time then to get units of time you obviously just leave it alone, i.e. G^0 c^0. You only need to put in factors of G and c if you want to convert T to some other dimensions besides time.

Ok, I think I see what you are saying - I don't know how to say this except with an example. Suppose I have a photon. The distance it travels in time t is x=c t so t=x/c. Ignore the c, and you get the time of travel as x meters. But you know it has units of time, so you divide by c to get the answer. Is that the way things are done?