# Proof of unitarity of time evolution in Susskind's book

• I
Gold Member
In "The Theoretical Minimum" of Susskind (p.98) it says that if we take any two basisvectors $|i \rangle$ and $|j \rangle$ of any orthonormal basis, and we take any linear time-development operator $U$, that the inner product between $U(t)|i \rangle$ and $U(t)|j \rangle$ should be 1 if $|i \rangle=|j \rangle$. Why is this so? (Why is the product normalized?) I can see how it is demonstrated that the inner product of $U(t)|i \rangle$ and $U(t)|j \rangle$ is 0 if $|i \rangle \neq |j \rangle$ (in fact, he assumes it). The reasoning is aimed to show that time evolution is unitary.

A. Neumaier
<##Ux|Uy##>=<##x|U^*Uy##>=<##x|y##> since U is unitary.

Gold Member
<##Ux|Uy##>=<##x|U^*Uy##>=<##x|y##> since U is unitary.
No, no, the reasoning is aimed to prove unitarity of U.

A. Neumaier
No, no, the reasoning is aimed to prove unitarity of U.
Unitarity of ##U(t) =e^{-itH}## is trivial since ##U(t)^*=e^{-i^*tH^*}=e^{itH}=U(t)^{-1}##.

Gold Member
Unitarity of ##U(t) =e^{-itH}## is trivial since ##U(t)^*=e^{-i^*tH^*}=e^{itH}=U(t)^{-1}##.
Does this hold for any linear time-development operator $U$? Are they all expressed as ##U(t) =e^{-itH}##?

A. Neumaier
Does this hold for any linear time-development operator $U$? Are they all expressed as ##U(t) =e^{-itH}##?
Yes, in units where ##\hbar=1##. Without this restriction, the Schroedinger equation
##i \hbar \dot\psi(t)= H\psi(t)## has the solution ##\psi(t)=e^{-itH/\hbar}\psi(0)=U(t)\psi(0)##
where ##U(t):=e^{-itH/\hbar}##.
If the Hamiltonian depends on time, the formula is more complicated (with Texp), and unitarity is easier to see by differentiating ##\langle \phi(t)|\psi(t)\rangle##, showing that it is constant.

entropy1
Gold Member
Yes, in units where ##\hbar=1##. Without this restriction, the Schroedinger equation
##i \hbar \dot\psi(t)= H\psi(t)## has the solution ##\psi(t)=e^{-itH/\hbar}\psi(0)=U(t)\psi(0)##
where ##U(t):=e^{-itH/\hbar}##.
Thank you. But still I don't get the proof Susskind is trying to give, so if anyone understands his line of reasoning on this, my question still stands! (peruse my OP!)

Strilanc
I'm guessing that you're muddling the argument somehow. Unitary matrices preserve preserve the inner product / perpendicularity (i.e. ##\langle a | b \rangle = \langle Ua | Ub \rangle##). Knowing that the linear time-development is unitary would allow you conclude length-preserving, and vice versa knowing the time-development was length-preserving would be important to showing unitarity, but from your quote it's not clear which is being proved from the other.

Gold Member
I'm guessing that you're muddling the argument somehow. Unitary matrices preserve preserve the inner product / perpendicularity (i.e. ##\langle a | b \rangle = \langle Ua | Ub \rangle##). Knowing that the linear time-development is unitary would allow you conclude length-preserving, and vice versa knowing the time-development was length-preserving would be important to showing unitarity, but from your quote it's not clear which is being proved from the other.
The postulate of U=unitary is not made! U=unitary is to follow from the postulates he gave in his book, which I tried to describe in my OP. (That is: it is to be proven that U=unitary)

Gold Member
The core of my question is:
[..]the inner product between $U(t)|i \rangle$ and $U(t)|j \rangle$ should be 1 if $|i \rangle=|j \rangle$. Why is this so? (Why is the product normalized?)

Strilanc
The postulate of U=unitary is not made! U=unitary is to follow from the postulates he gave in his book, which I tried to describe in my OP. (That is: it is to be proven that U=unitary)

Then you missed some of the postulates. Does the book give a definition for "linear time development operator"?

Gold Member
Does the book give a definition for "linear time development operator"?
No, none at all...

DrClaude
Mentor
Since ##U(t)|i\rangle## and ##U(t)|j\rangle## are the same state, the condition imposed is simply conservation of the norm.

Gold Member
Since ##U(t)|i\rangle## and ##U(t)|j\rangle## are the same state, the condition imposed is simply conservation of the norm.
Exactly! But why is the norm conserved?! (it doesn't follow from the proof)

Strilanc
No, none at all...

I'm not sure I believe you. Can you take a picture of the page?

entropy1
DrClaude
Mentor
Exactly! But why is the norm conserved?!
Because of the Born rule. You want probabilities to stay probabilities.

Demystifier and bhobba
Gold Member
I'm not sure I believe you. Can you take a picture of the page?
Unfortunately the pictures are out of focus...

Gold Member
Because of the Born rule. You want probabilities to stay probabilities.
But the form $\langle i|M|j \rangle$ isn't a probability, is it? At this point in the book, probabilities are formulated as the product of the inproduct of the eigenvector with the state and the the conjugate of the inproduct of the eigenvector with the state.

DrClaude
Mentor
But the form $\langle i|M|j \rangle$ isn't a probability, is it?
No, but the evolution operator returns a state.

Take a quantum system in state ##|\psi\rangle = |i\rangle##. What is the probability that it is in state ##|j\rangle##?
##\langle j | \psi \rangle = \langle j | i \rangle = \delta_{ij}##.

Now, after some time ##t##, what is the probability that the system is in state ##| \phi\rangle \equiv U(t)|j\rangle##?
##\langle \phi | U(t) | \psi \rangle = \langle \phi | U(t) | i \rangle = \langle j | U^\dagger(t) U(t) | i \rangle##

Now, if ##| i \rangle = | j \rangle##, then ##U(t) | \psi \rangle = | \phi \rangle##, so setting
##\langle \phi | U(t) | \psi \rangle = \langle \phi | \phi \rangle = 1##
is the same as taking ##U(t)## to be unitary.

Jilang and entropy1
Gold Member
Susskind writes:
Susskind said:
The state at time t is given by some operation that we call ##U(t)##, acting on the state at time zero. Without further specifying the properties of ##U(t)##, this tells tells us very little except that ##|ψ(t) \rangle## is determined by ##|ψ(0) \rangle##. Let's express this relation with the equation ##|ψ(t) \rangle = U(t)|ψ(0) \rangle##. The operation U is called the time-development operator for the system.

And then:
Susskind said:
Conventional quantum mechanics places a couple of requirements on ##U(t)##. First, it requires ##U(t)## to be a linear operator.

Then he mentions ##U(t)## requires to obey the conservation of distinctions, which is that ##U(t)## operating on orthogonal states remains orthogonal. He illustrates this by taking two different vectors ##|i \rangle## and ##|j \rangle## of an arbitrary orthonormal basis.

He writes:
Susskind said:
On the other hand, if i and j are the same, then so are the output vectors ##U(t)|i \rangle## and ##U(t)|j \rangle##. In that case, the inner product between them should be 1.

Take note of the "should be 1" part!

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Gold Member
No, but the evolution operator returns a state.

Take a quantum system in state ##|\psi\rangle = |i\rangle##. What is the probability that it is in state ##|j\rangle##?
##\langle j | \psi \rangle = \langle j | i \rangle = \delta_{ij}##.

Now, after some time ##t##, what is the probability that the system is in state ##| \phi\rangle \equiv U(t)|j\rangle##?
##\langle \phi | U(t) | \psi \rangle = \langle \phi | U(t) | i \rangle = \langle j | U^\dagger(t) U(t) | i \rangle##

Now, if ##| i \rangle = | j \rangle##, then ##U(t) | \psi \rangle = | \phi \rangle##, so setting
##\langle \phi | U(t) | \psi \rangle = \langle \phi | \phi \rangle = 1##
is the same as taking ##U(t)## to be unitary.
I fail to see why ##\langle \phi | \phi \rangle = 1## if ##| \phi\rangle = U(t)|j\rangle##. Why is the norm conserved? For instance: ##U(t)|j \rangle## could be twice as long as ##|j \rangle##!

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DrClaude
Mentor
I fail to see why ##\langle \phi | \phi \rangle = 1## if ##| \phi\rangle = U(t)|j\rangle##. Why is the norm conserved? For instance: ##U(t)|j \rangle## could be twice as long as ##|j \rangle##!
The choice of normalizaton is arbitrary, but we should expect that once that choice is made, the equations are consistent and that the norm doesn't change (especially since the norm is related to probabilities).

Gold Member
The choice of normalizaton is arbitrary, but we should expect that once that choice is made, the equations are consistent and that the norm doesn't change (especially since the norm is related to probabilities).
If the norm is conserved, I can see why ##U## is unitary. It seems to me Susskind hasn't made that conclusive. I have to take your word on that it is necesary to preserve probabilities (perhaps it should be a postulate then?). Thanks for your replies! (and the others )

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bhobba
Mentor
Exactly! But why is the norm conserved?! (it doesn't follow from the proof)

To make the math easier so the Born rule can be used it is desirable that states of unit length remain unit length. This implies it must be unitary.

There is a deep theorem associated with this Susskind doesn't mention because its advanced (you will learn of it in Ballentine) called Wigners Theroem:
https://en.wikipedia.org/wiki/Wigner's_theorem

But don't worry about it for now. As I often say in QM the full story sometimes isn't told at the start - you need to build up to it.

Thanks
Bill

entropy1 and DrClaude
In an earlier chapter on quantum states (p.40),
Susskind said:
$$\langle A|A \rangle = 1$$
This is a very general principle of quantum mechanics that extends to all quantum systems: the state of a system is represented by a unit (normalized) vector in a vector space of states. Moreover, the squared magnitudes of the components of the state-vector, along particular basis vectors, represent probabilities for various experimental outcomes.

entropy1, Demystifier and DrClaude
Demystifier