Converting from hbar = 1 to physical units

[Kreizhn]

Hmm...Well, U has be to unitless since
| \psi(t) \rangle = U(t) |\psi(t_0) \rangle
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?

 

[Dale]

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?

Yes, exactly. For example, you may see the gravitational time dilation formula in the Schwarzschild written as:

t_0=t_f \sqrt{1-\frac{2M}{r}}

So you are expected to know that t has units of time, M units or mass, r units of length. Then as written it is dimensionally inconsistent in SI units, but we can make it dimensionally consistent by adding terms of G and c like so:

t_0=t_f \sqrt{1-\frac{2GM}{c^2 r}}

 

[Rap]

Yes, exactly. For example, you may see the gravitational time dilation formula in the Schwarzschild written as:

t_0=t_f \sqrt{1-\frac{2M}{r}}

So you are expected to know that t has units of time, M units or mass, r units of length. Then as written it is dimensionally inconsistent in SI units, but we can make it dimensionally consistent by adding terms of G and c like so:

t_0=t_f \sqrt{1-\frac{2GM}{c^2 r}}

Tell me where I go wrong here. G has units L^3 m/t^2 where L is length, m is mass, t is time. C has units L/t. That means the units of the second term under the square root is

\frac{Gm}{c^2 r}=\left(\frac{L^3m}{t^2}\right) \left(m\right) \left(\frac{t^2}{L^2}\right)\left(\frac{1}{L}\right)=m^2

In other words, that term is not dimensionless, and it should be.

 

[Dale]

Tell me where I go wrong here. G has units L^3 m/t^2 where L is length, m is mass, t is time. C has units L/t. That means the units of the second term under the square root is

\frac{Gm}{c^2 r}=\left(\frac{L^3m}{t^2}\right) \left(m\right) \left(\frac{t^2}{L^2}\right)\left(\frac{1}{L}\right)=m^2

In other words, that term is not dimensionless, and it should be.

You put the mass in the wrong spot for G. The dimensions of G are:
G \to \left(\frac{L^3}{t^2m}\right)

 

[Rap]

You put the mass in the wrong spot for G. The dimensions of G are:
G \to \left(\frac{L^3}{t^2m}\right)

Oh, ok. Division is not the same as multiplication. I should write that down, that's worth remembering.

Sorry about that.

I am trying to fit this procedure into the dimensional analysis formalism in my mind, making sure it can always be done. Any physical equation can be expressed as f(\pi_1,\pi_2,...)=1 where the \pi_i[/tex] are dimensionless. You could express a \pi by \pi=L^\alpha m^\beta t^\gamma G^p c^q where, using the units of G and c, we must have p=\beta and q=-\alpha-3\beta in order for \pi to be dimensionless. If you just set c and G to unity, you get a dimensioned \piwhich can only be "fixed" by multiplying again by G^p c^q. So now I'm ok, as long as I remember the difference between multiplication and division. :)

 

[Dale]

we must have p=\beta and q=-\alpha-3\beta in order for \pi to be dimensionless.

Oh, very nice work! Thanks.

 

[Kreizhn]

Hey,

Sorry I disappeared for so long. So Rap, your post #30 makes perfect sense except it's uncertain what units we get rid of. Do you have any suggestions. Also

Lets get rid of the second. Now time is measured in kg m^2/planck. The original equation becomes \frac d{dt} U = -i H U 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.

I agree that time is kg m^2/planck, but then would H be Planck/(kg m^2)? Just because if we solve the equation, we would get U = exp(-iHt) and if t has units kg m^2/planck wouldn't that force H to be the reciprocal?

On the other hand, I wonder if we could apply what DaleSpam found in the paper I linked. Since J has units of hertz and we can write exp(H/J) then H has units of 1/time. Does this then force us into your regime of "getting rid of the meter"? If we've gotten rid of the metre though, does this imply there's no need to change anything with time, since it hasn't been scaled?

Alternatively, if we do the "plug constants back in until things work out dimensionally" approach this is what I get. I'll try to be as rigorous as possible so there are no confusions.

Firstly, we know that the operator U(t) is a time-propogator. That means that given an initial state |\psi_0 \rangle evolving under the same Hamiltonian, we can get to an arbitrary time |\psi(t) \rangle by

|\psi(t) \rangle = U(t) |\psi_0 \rangle

Since |\psi(t)\rangle and |\psi_0 \rangle must have the same units, this implies that U(t) must be unitless.

Okay, so in the equation
i\hbar \frac{d}{dt} U(t) = H(t) U(t)
we know that \hbar has units of energy seconds, we'll say E^1 T^1 ( you can make that into ML^2T^{-1} if you want but I don't think it's necessary). Furthermore, we expect H(t), the Hamiltonian, to have units of energy E. Therefore this equation is dimensionally correct because the left hand side is E^1 T^1 T^{-1} = E^1 which agrees with the right hand side.

Okay. Now in the "plug constants in until it works out" we assume that everything has its normal units and insert hbars and c's until everything is correct right? So
\frac{d}{dt} U(t) = -i H(t) U(t)
has units of 1/time on the LHS, and Energy on the RHS. To fix this all we do is insert a \frac1\hbar on the RHS or a \hbar on the LHS. But is this not equivalent to what I had said earlier about just rescaling?

Alternatively, the solution is U(t) = \exp\left[ -i H t \right]U(0) where I have assume H is time independent for simplicity. In the "plug stuff in until it works" regime, the exponential argument currently has units Energy-seconds. To make it unitless, I would just add a factor of \frac1\hbar into the argument. Again, is this not equivalent to my scaling above?

 

[Kreizhn]

Hey guys,

I might be beating a dead horse here, but something weird is going on. By following Rap's advice and defining a new unit call the Planck (with units ML^2T^{-1}) we get the following table

\begin{array}{|c|cccc|} \hline<br /> \text{ Replaced unit} &amp; M &amp; L &amp; T &amp; \text{Energy}\\ \hline<br /> M &amp; \frac {PT}{L^2} &amp; L &amp; T &amp; \frac PT\\<br /> L &amp; M &amp; \sqrt{\frac{PT}M} &amp; T&amp; \frac PT\\<br /> T &amp; M &amp; L &amp; \frac{ML^2}P &amp;\frac{P^2}{ML^2} \\ \hline<br /> \end{tabular}

But then \exp\left[ -i H t \right] will always have units of P in the argument, which is okay if we set \hbar = 1 since then P is unitless. So let us assume that we're in the unitless regime. Are we saying that time is unaffected by in the case where we've replaced M and L, but needs to be transformed when we have replaced T? If we use DaleSpam's example of just "plugging in units" then we would need to modify the exponential argument by \frac 1\hbar and it's unclear whether H or t is absorbing this factor.

This is all very poorly defined. I think I have a mathematical way of doing this which is much better, since there is no such ambiguity. However, if physically someone can make sense of this predicament that would be great.

 

[Rap]

Hey guys,

I might be beating a dead horse here, but something weird is going on. By following Rap's advice and defining a new unit call the Planck (with units ML^2T^{-1}) we get the following table

\begin{array}{|c|cccc|} \hline<br /> \text{ Replaced unit} &amp; M &amp; L &amp; T &amp; \text{Energy}\\ \hline<br /> M &amp; \frac {PT}{L^2} &amp; L &amp; T &amp; \frac PT\\<br /> L &amp; M &amp; \sqrt{\frac{PT}M} &amp; T&amp; \frac PT\\<br /> T &amp; M &amp; L &amp; \frac{ML^2}P &amp;\frac{P^2}{ML^2} \\ \hline<br /> \end{tabular}

But then \exp\left[ -i H t \right] will always have units of P in the argument, which is okay if we set \hbar = 1 since then P is unitless. So let us assume that we're in the unitless regime. Are we saying that time is unaffected by in the case where we've replaced M and L, but needs to be transformed when we have replaced T? If we use DaleSpam's example of just "plugging in units" then we would need to modify the exponential argument by \frac 1\hbar and it's unclear whether H or t is absorbing this factor.

This is all very poorly defined. I think I have a mathematical way of doing this which is much better, since there is no such ambiguity. However, if physically someone can make sense of this predicament that would be great.

\exp[iHt]=1/0!+iHt/1!-(Ht)^2/2!-i(Ht)^3/3!+...

This means that each term must have the same dimensions which means that iHt must be dimensionless. I have to think about whether this is the root of the problem.

 

[Kreizhn]

Hey Rap,

Thanks for coming back to this old thing. Yeah, I agree that iHt has to be dimensionless. I think the reason why it works in the hbar = 1 regime is because the Planck unit is dimensionless. Hence we can technically throw it away and things still work out. We've kept that P unit around for accounting reasons; indeed, when we transform back we need to get rid of it so that everything works. Would this not just be done by a rescaling?

 

[Kreizhn]

For the record, I think this might be a more mathematically precise way of dealing with the problem.

Let U = \mu \hat U and t = \tau \hat t for some set of characteristic unitary and time \mu, \tau. Substituting this into the Schrödinger equation
i\hbar \frac{dU}{dt} = H(t) U(t)
we get
\begin{align*} i \hbar \frac\mu\tau \frac{d\hat U}{d\hat t} &amp;= H(\tau \hat t) \mu \hat U \\ <br /> \frac{d\hat U}{d\hat t} &amp;= -\frac i\hbar \frac{\tau \mu}\mu \hat H(\hat t) \hat U &amp; \text{ where } \hat H(\hat t) = H(\tau \hat t) \\<br /> &amp;= -i \frac\tau\hbar \hat H(\hat t) \hat U<br /> \end{align*}<br />
Hence \mu can be arbitrarily chosen and we set mu =1 so that \hat U = U. To remove the \hbar term we set \tau = \hbar yielding the desired equation
\frac{dU}{d\hat t} = -i \hat H(\hat t) U

From here I think the appropriate conversions can be made. Namely, the units of U do not change, \hat t can be converted by scaling by \hbar, and similarly for \hat H(\hat t) albeit this may be more complicated.

Thoughts?