# Time reversibility Schrodinger Eq.

1. Oct 23, 2004

### da_willem

I heard nonrelativistic QM is time-reversible. How does this follow from the Schrodinger equation?

2. Oct 23, 2004

### Fredrik

Staff Emeritus
Using the Schrödinger equation

$$i\hbar\frac{\partial\psi(x,t)}{\partial t}=H\psi(x,t)$$

it's easy to show that

$$\psi(x,t)=e^{-iHt/\hbar}\psi(x,0)$$

This means that the operator

$$U(t)=e^{-iHt/\hbar}$$

is a time evolution operator. It acts on a wavefunction to produce the wavefunction of the same system at a later time. This time evolution is clearly "reversible", since the time evolution operator has an inverse:

$$U(t)^{-1}=e^{iHt/\hbar}$$

This time evolution is unitary (and hence "reversible"), but the standard formulation of QM is not. (The Schrödinger equation is not all there is to QM. We also have to consider measurements and the Born rule). When a measurement is performed, the wavefunction changes abrubtly to an eigenstate of the measured observable. The probability of a particular outcome $$|a\rangle$$ is given by the Born rule:

Probability$$=|\langle a|\psi\rangle|^2$$

The evolution of the wavefunction into an eigenstate is not unitary (and not "reversible").

One thing that I learned only recently is that there is a formulation of QM that treats both "directions" of time in a symmetric way. Instead of saying that if the system is in a certain state before the measurement the probability of a particular outcome is given by the Born rule, this formulation says that if the system is in a certain state before the measurement and in another state after the measurement, the probability of a particular outcome is given by another rule (a mathematical expression that's similar to the Born rule, but not identical to it). Unfortunately I don't remember that rule, and I'm too lazy too look it up right now.

3. Oct 23, 2004

### da_willem

That sounds interesting. But I'm sure it yields the same results as the Born rule and does not overthrow the collapse of the wave function?!

I know there's more to QM than the Schrodinger equation, but if QM is time reversible I guess it should be derivable from it. Ofcourse time revarsal does not apply to a situation with a collapsing wavefunction. I would like to see how with a certain (time independent?) potential the same wavefunction with the time reversed obeys the Schrodinger equation or something like that...

4. Oct 23, 2004

### Tom Mattson

Staff Emeritus
It's not time reversible.

Let T=time reversal operator, and start from Schrodinger's equaiton:

H&psi;(x,t)=(i*hbar)(&part;&psi;(x,t)/&part;t)

Now let T&psi;(x,t)=&psi;(x,-t), and you get:

H&psi;(x,-t)=(-i*hbar)(&part;&psi;(x,t)/&part;t)

Note the negative sign, which means that the Schrodinger equation is not invariant under this transformation. In order to recover quantum mechanics under time reversal, you have to do a transformation to the wavefunctions as well, namely:

T&psi;(x,t)=&psi;*(x,-t).

5. Oct 24, 2004

### Fredrik

Staff Emeritus
What Tom said is of course correct. The Schrödinger equation is not invariant under time-reversal. My interpretation of the original question was different from his. What I had in mind is that there's no information loss caused by the unitary time evolution implied by the Schrödinger equation, so if you know the state at a certain time you can calculate it at an earlier time. The Shrödinger equation predicts a determinstic evolution of the wavefunction in both directions of time, but I should have emphasized that it's not the same evolution.

Time evolution is "reversible" in the sense that there is predictability in both directions of time, but not in the sense that the time-evolution operator (or equivalently, the Schrödinger equation) is invariant under time-reversal. It is not.

6. Oct 25, 2004

### seratend

Time reversal T is an antilinear operator. Time reversibility directly comes from the Hamiltonian H: [H,T]=0 or not.
So if we take Hψ(x,t)=i*hbar[∂tψ](x,t)
we thus have THψ(x,t)=T(ihbar[∂tψ](x,t))=-i*hbarT[∂tψ(x,t)]=-i*hbar[∂tψ](x,-t) where we have used the antilinear property of T.

Then if [H,T]=0 => THψ(x,t)=HTψ(x,t)=Hψ(x,-t)

SO we have Hψ(x,-t)=-i*hbar[∂tψ](x,-t)

But [∂tψ]=-[∂(-t)ψ] thus:

Hψ(x,-t)=i*hbar[∂(-t)ψ](x,-t) which is the original schroedinger equation (we just have change parameter t into -t).

Thus Schoedinger equation is Time reversible if [H,T]=0 (it depends on the interaction within H: e.g. electroweak interaction does not respect this invariance, but coulombian interaction yes).

Seratend.

7. Oct 25, 2004

### Fredrik

Staff Emeritus
Seratend, what you did here is a nice exercise, but I don't think that this calculation can be used to argue that the Schrödinger equation is time reversible. You just multiplied both sides of an equation with an operator from the left, and showed that the equation still holds. That by itself can't prove anything of course.

However, you used both the antilinearity property of T (Ti=-iT) and the commutation relation [H,T]=0 to get the necessary result. If you had used one of those properties but not the other, you would have obtained a contradiction. That means that what your calculation really showed is that [H,T]=0 if and only if T is antilinear.

I wouldn't say that the result implies that the Schrödinger equation is "time reversible" or "invariant under a time reversal transformation". To me that would mean that if ψ is a solution, then so is Tψ, but that's not the case:

$$i\hbar\frac{\partial}{\partial t}T\psi(\vec x,t)=i\hbar\frac{\partial}{\partial t}\psi(\vec x,-t)=-i\hbar\frac{\partial}{\partial t}\psi(\vec x,t)=-H\psi(\vec x,t)=H(-\psi(\vec x,t))$$

This is clearly not always equal to

$$HT\psi(\vec x,t)=H\psi(\vec x,-t)$$

so Tψ is not always a solution to the Schrödinger equation.

8. Oct 26, 2004

### da_willem

I think I'm looking for an argument that shows that

$$H(T \Psi)=-i \frac{\partial T \Psi}{ \partial t}$$

is equivalent to the original Schrodinger equation.

9. Oct 26, 2004

### seratend

Ok, let's go to the basics. First, I have to precise some items for all the people looking at this thread (and also my previous post):

T is an antilinear operator (I'll explain later why). It is defined up to an isomorphism (“gauge freedom”: the basis invariant under its action).
So we have T|ψ(t)>=| ψ_rev(t)>.
When we explain | ψ_rev(t)> in the |x> basis we have <x| ψ_rev(t)>= ψ_rev(x,t) the wave function.

Therefore, we also have: <x| ψ_rev(t)>= <x|T|ψ(t)>= ψ*(x,-t) .
Note: for the ones interested in group theory, we have chosen a peculiar representation of the time reversal operator that leaves the |x> basis unchanged (“gauge freedom”: T|x>=|x>).

In my previous post, I’ve omitted the complex conjugate of ψ, because I usually work with vectors “i*hbar|ψ(t)>=H|ψ(t)>” in the Schroedinger representation rather than the components of the Hilbert vectors in the x basis.
I’m sorry for this little mistake (Tom Mattson has not made the mistake), but I think everyone may correct this omission and get the same result: if ψ(x,t) is solution of the Schroedinger equation in the |x> basis, Tψ= ψ*(x,-t) is also solution (provided that [H,T]=0).

Well, I think you should have a look at the symmetry group representation in Hilbert spaces in order to understand better what I write quickly in the previous post: quickly,
T is an antilinear operator due to the representation of the time reversal (t into –t ) symmetry in the Hilbert space of QM (Wigner’s theorem).
Thus T is defined by the following:
- T is an anti linear operator
- T x T+ = x
- T p T+=-p

The definition of T as an antilinear operator comes from the structure of the space time (projective representation of a discrete symmetry group). It does not imply that the quantum system (our example, a particle with an interaction described by H) has the time reversal symmetry.

The time reversal symmetry of the quantum system thus depends only on whether or not [H,T]=0. Thus if |ψ(t)> is solution of a time reversal symmetric SE (i.e. [H,T]=0), we should have |ψ’(t)>= T|ψ(-t)> also solution of the SE (i.e. T|ψ> evaluated at time –t).

In an equation view we have (independent of the vector basis) the SE of the quantum system:

i*hbar ∂t|ψ(t)>=H|ψ(t)> (1)

We may apply the T operator to (1) (T does not depend on parameter t):

- i*hbar ∂tT|ψ(t)>=TH|ψ(t)> (2)

now if we change the t parameter in (2) into –t, we have (we suppose H does not depend on t):

i*hbar ∂tT|ψ(-t)>=TH|ψ(-t)> (3) (we use ∂(-t)= -∂t)

So to get T|ψ(-t)> solution of the same SE (1) :

i*hbar ∂t{T|ψ(-t)>} = H{T|ψ(-t)>} (4)

we require that TH=HT <=> [H,T]=0: It is what we want to prove.
Equation (1) and (4) may be projected onto the |x> to get the wave SE with <x|ψ(t)>= ψ(x,t) and <x|T|ψ(-t)>= ψ*(x,-t).

You should explain T|ψ> evaluated at time –t solution of the same SE (i.e. 1).
So if I take your equation and if we define t’=-t.
We thus start by writing your equation (you have forgotten the complex conjugate of the wave ψ, an it is really important because you really work on the components of |ψ> rather than on the vector, as I did - the complex conjugate is a peculiar action of T operator in the x basis) :

A=i*hbar ∂t’Tψ(x,t’)=> A= i*hbar ∂t’ψ*(x,-t’)= -i*hbar∂tψ*(x,t) (t=-t’ our choice)

Now we use the SE: i*hbar*∂tψ(x,t)=H ψ(x,t):
i*hbar*∂tψ(x,t)= i*hbar ∂tψ(x,t) => -i*hbar ∂tψ*(x,t)= [H ψ(x,t)]* (complex conjugate of SE)

So we have:

A=-i*hbar∂tψ*(x,t)= [H ψ(x,t)]*

So if [H,T]=0 and T|x>=|x>, we have [H ψ(x,t)]*= Hψ*(x,t) (the components of matrix H are real in the |x> basis): it is the condition of the symmetry of the SE.

Therefore

A=-i*hbar∂tψ*(x,t)= Hψ*(x,t) <=> i*hbar ∂tψ*(x,-t)= Hψ*(x,-t) . (∂(-t)= -∂t)

Thus we recover in the |x> basis : if ψ(x,t) is solution of SE ψ* evaluated at time –t is also solution of the same equation if [H,T]=0 or if you prefer [H ψ(x,t)]*= Hψ*(x,t) in the |x> basis.

No. You are looking at the argument ψ(x,t) of the SE, then ψ*(x,-t) also solution of SE if [H,T]=0, ie:

i*hbar ∂tψ(x,t)= Hψ(x,t) <=> i*hbar ∂tψ*(x,-t)= Hψ*(x,-t) if [H,T]=0

Seratend.

we live in a world of mistakes, so, sorry, if an error is forgotten in this post ........... or not