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- #2

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

- #3

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

- #4

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

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Does this hold forUnitarity of ##U(t) =e^{-itH}## is trivial since ##U(t)^*=e^{-i^*tH^*}=e^{itH}=U(t)^{-1}##.

- #6

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Yes, in units where ##\hbar=1##. Without this restriction, the Schroedinger equationDoes this hold forany[itex]U[/itex]? Are they all expressed as ##U(t) =e^{-itH}##?linear time-development operator

##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.

- #7

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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!)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}##.

- #8

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- #9

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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)

- #10

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[..]the inner product between [itex]U(t)|i \rangle[/itex] and [itex]U(t)|j \rangle[/itex] should be 1 if [itex]|i \rangle=|j \rangle[/itex]. Why is this so? (Why is the product normalized?)

- #11

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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"?

- #12

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

- #13

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- #14

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Exactly! But

- #15

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No, none at all...

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

- #16

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Because of the Born rule. You want probabilities to stay probabilities.Exactly! Butwhyis the norm conserved?!

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- #18

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

- #19

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But the form [itex]\langle i|M|j \rangle[/itex] 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.Because of the Born rule. You want probabilities to stay probabilities.

- #20

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No, but the evolution operator returns a state.But the form [itex]\langle i|M|j \rangle[/itex] isn't a probability, is it?

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.

- #21

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Susskind writes:

And then:

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:

Take note of the "should be 1" part!

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 calledthe time-development operatorfor 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

He writes:

Susskind said:On the other hand, ifiandjare 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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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##!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.

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- #23

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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).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##!

- #24

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If the normThe 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).

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- #25

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Exactly! Butwhyis 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

- #26

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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.

- #27

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For such technical trivialities explained in a beginner-friendly way I highly recommend "Quantum Mechanics Demystified" by McMahon. (See also https://www.physicsforums.com/threads/the-ultimate-demystifier.871588/#post-5472646 )

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