About the Schrödinger equation

  • #1
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Hi everybody,

For solving the time dependent Schrödinger equation ##H|\psi(t) \rangle = i\hbar \frac{\partial}{\partial t}|\psi (t)\rangle##, I read in quantum mechanics books the assumption about the solution ##|\psi(t) \rangle## which is made of a linear combination of a complete set of the stationnary states ##|\phi(t) \rangle ## (that form a vectorial space) as: $$|\psi(t) \rangle = \sum_n c_n(t) |\phi_n(t) \rangle $$.

My question:
When it is not easy to find stationnary states of the Schrödinger equation, is it correct to use a linear combination of another complete set of function say ##|f(t) \rangle ## (that form a vectorial space) to construct the solution ##|\psi(t) \rangle##?

Thanks.
 

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  • #2
hilbert2
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When it is not easy to find stationnary states of the Schrödinger equation, is it correct to use a linear combination of another complete set of function say ##|f(t) \rangle ## (that form a vectorial space) to construct the solution ##|\psi(t) \rangle##?
In that case, it's not as easy as ##\displaystyle c_n (t) = e^{-iE_n t/\hbar}##. The time development will cause transitions between the basis states if they are not solutions of the time-independent Schrödinger equation.
 
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  • #3
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In that case, it's not as easy as ##\displaystyle c_n (t) = e^{-iE_n t/\hbar}##. The time development will cause transitions between the basis states if they are not solutions of the time-independent Schrödinger equation.
So, I assume you mean that what I said (about linear combination of another complete set of function that form a vector space) is "mathematically correct" .

But if this way of solving is correct, I have another question.
Now, let say, I want to solve a complicated Schrödinger equation of a helium atom and I choose to use the set of solutions of harmonic oscillator (a complete set of equation forming a vector space) to express its solutions (states).
Then, according to your explanation, this development will cause transitions between the states of harmonic oscillator.

My question
: How to interpret all of this physically? What does "states of harmonic oscillator" mean for the helium atom?

Thanks
 
  • #4
PeterDonis
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When it is not easy to find stationnary states of the Schrödinger equation, is it correct to use a linear combination of another complete set of function say ##|f(t) \rangle## (that form a vectorial space)
Your terminology here is a little confused. The vector space in question--the correct term for this particular kind of vector space is "Hilbert space"--is the space of all square integrable functions of position and time, not just the functions in the set you choose. The two sets of functions you are talking about--one being the set of functions which are stationary states, the other some other complete set which are not stationary states--are two different choices of basis for the same Hilbert space, like choosing different sets of axes in ordinary space that are rotated with respect to to one another.
 
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  • #5
PeterDonis
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according to your explanation, this development will cause transitions between the states of harmonic oscillator
You're confusing actual, physical transitions of the system, with transitions of the basis states in the basis you have chosen for the Hilbert space.

If the overall state of the actual system is not a stationary state, then it will change with time (that's what "not stationary" means), so if you make a measurement of the system's energy, the probabilities of getting different results will change with time--at time ##t = 0## there might be a 100% probability of measuring energy ##E_1##, but after some time has passed there might be a 50% probability of measuring ##E_1## and a 50% probability of measuring some other energy ##E_2##. So you can measure the energy at different times and get different results; this is called a "transition". And all of this is true regardless of what basis you choose to mathematically describe the state.

If the overall state of the actual system is a stationary state, but you choose to describe it, mathematically, using a basis other than the set of stationary states, then the state of the actual system will not change with time, but how the mathematical description appears will, because the basis states you have chosen are changing with time, so the coefficients of each basis state in the mathematical description of the actual system will also have to change with time (since the state of the actual system is not changing with time). So if you measure the energy of the system at different times, you will get the same result; but the details of how the mathematical computation predicts that result will be different at different times.
 
  • #6
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Your terminology here is a little confused. The vector space in question--the correct term for this particular kind of vector space is "Hilbert space"--is the space of all square integrable functions of position and time, not just the functions in the set you choose. The two sets of functions you are talking about--one being the set of functions which are stationary states, the other some other complete set which are not stationary states--are two different choices of basis for the same Hilbert space, like choosing different sets of axes in ordinary space that are rotated with respect to to one another.
Thank you for clarifying my confused terminology. Considering those clarifications my last question stands.
 
  • #7
PeterDonis
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What does "states of harmonic oscillator" mean for the helium atom?
Nothing, because a helium atom is not a harmonic oscillator. The Hilbert space for a helium atom is not the same as the Hilbert space for a harmonic oscillator, so there aren't even any states in the helium atom's Hilbert space that answer to the description "states of harmonic oscillator".
 
  • #8
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If the overall state of the actual system is a stationary state, but you choose to describe it, mathematically, using a basis other than the set of stationary states
No, this is not the case that I want to discuss.
I am talking about another case where the overall state of the actual system is not stationary state and I want to express it by using a basis other than the set of stationary states.
 
  • #9
PeterDonis
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I am talking about another case where the overall state of the actual system is not stationary state and I want to express it by using a basis other than the set of stationary states.
Typically you would choose a basis that matches up with what you want to measure. The stationary states are the basis that matches with measuring the energy. What other observable do you want to measure?
 
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  • #10
hilbert2
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No, this is not the case that I want to discuss.
I am talking about another case where the overall state of the actual system is not stationary state and I want to express it by using a basis other than the set of stationary states.
The states of a hydrogen atom can be represented in the basis of eigenstates of any other system with 3 degrees of freedom, including those of a 3D harmonic oscillator with Hamiltonian

##\displaystyle\hat{H}=\frac{\hat{p}_x^2 + \hat{p}_y^2 + \hat{p}_z^2}{2m} + \frac{1}{2}(k_x \hat{x}^2 + k_y \hat{y}^2 + k_z \hat{z}^2 )##

If the new basis differs only a little bit from the hydrogenic eigenstate basis, it's often possible to use time-dependent perturbation theory for the calculation of time evolution.
 
  • #11
PeterDonis
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The states of a hydrogen atom can be represented in the basis of eigenstates of any other system with 3 degrees of freedom
You're considering the hydrogen atom as a point particle in 3-space here. The OP was asking, in post #3, about a "helium atom", and it seemed to me that he was interested in the internal states of the atom, for example, solving for the electron energy levels. That's a different problem with a different configuration space (and we haven't even mentioned the spin degrees of freedom).
 
  • #12
hilbert2
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Oh, sorry for the typo, it should have been helium and 6 position space degrees of freedom in the CMS coordinate system.
 
  • #13
PeterDonis
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sorry for the typo, it should have been helium and 6 position space degrees of freedom in the CMS coordinate system
Ok, got it. Yes, as long as we restrict ourselves to the position-momentum space degrees of freedom, I agree that we could use any basis we like on the Hilbert space for 6 degrees of freedom, including the harmonic oscillator states. Obviously those states won't be eigenstates of the actual helium atom Hamiltonian, though, which means it will probably be very confusing to try to do things in the harmonic oscillator basis.
 
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  • #14
hilbert2
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And there's also the complication that scattering states aren't allowed in the harmonic oscillator, but those are not usually needed in quantum chemistry.
 
  • #15
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Typically you would choose a basis that matches up with what you want to measure. The stationary states are the basis that matches with measuring the energy.
Through your answer, I understand that there is no real "mathematical objection" on choosing basis states apart the confusion resulting from trying to find a physical meaning if the basis does'nt match with an observable...

What other observable do you want to measure?
So, if I want now to measure linear momentum, how to proceed?

Nothing, because a helium atom is not a harmonic oscillator. The Hilbert space for a helium atom is not the same as the Hilbert space for a harmonic oscillator, so there aren't even any states in the helium atom's Hilbert space that answer to the description "states of harmonic oscillator".
However, I read on books a kind of assumption as following:

Let ## H ## a Hamiltonian composed of two parts as ## H = H_0 + H_{int} ##. Solving the Schrödinger equation $$(H_0+H_{int}(t)) |\psi (t) \rangle =
i\hbar \frac{\partial}{\partial t}|\psi (t)\rangle $$ is not easy , so they use the set formed by the solution ##|\phi (t)\rangle## of the truncated Schrodinger equation
$$H_0 |\phi (t) \rangle = i\hbar \frac{\partial}{\partial t}|\phi (t)\rangle $$ to construct the solution $$|\psi(t) \rangle = \sum_n c_n(t) |\phi_n(t) \rangle$$
Taking into account what you explain, I can say that mathematically, the Hilbert space from ## H_0## is not the same as from ##H##, nevertheless they wrote.
Or, is it only an approximation?

Thanks.
 
  • #16
hilbert2
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However, I read on books a kind of assumption as following:

Let ## H ## a Hamiltonian composed of two parts as ## H = H_0 + H_{int} ##. Solving the Schrödinger equation $$(H_0+H_{int}(t)) |\psi (t) \rangle =
i\hbar \frac{\partial}{\partial t}|\psi (t)\rangle $$ is not easy , so they use the set formed by the solution ##|\phi (t)\rangle## of the truncated Schrodinger equation
$$H_0 |\phi (t) \rangle = i\hbar \frac{\partial}{\partial t}|\phi (t)\rangle $$ to construct the solution $$|\psi(t) \rangle = \sum_n c_n(t) |\phi_n(t) \rangle$$
Taking into account what you explain, I can say that mathematically, the Hilbert space from ## H_0## is not the same as from ##H##, nevertheless they wrote.
Or, is it only an approximation?
Almost invariably the eigenstates of ##\hat{H}_0## belong to the same Hilbert space as those of ##\hat{H}_{int}## or ##\hat{H}##. An example is a slightly anharmonic oscillator:

##\hat{H} = \frac{\hat{p}^2}{2m} + \frac{1}{2}k\hat{x}^2 + \beta \hat{x}^4##,

which would be exactly solvable without the ##\beta \hat{x}^4## term, and therefore that term is naturally treated as a small perturbation ##H_{int}##.

In quantum field theory, the eigenstates of ##\hat{H}_0## are usually states with some number of free particles having definite values of momentum, and the ##\hat{H}_{int}## is an interaction term that allows those free particles to "collide" and produce new particles.
 
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  • #17
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Almost invariably the eigenstates of ##\hat{H}_0## belong to the same Hilbert space as those of ##\hat{H}_{int}## or ##\hat{H}##
Is there a mathematical theorem that supports this statement or is this only a kind of "feeling" that the physicist use to progress in his calculus?
 
  • #18
hilbert2
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Is there a mathematical theorem that supports this statement or is this only a kind of "feeling" that the physicist use to progress in his calculus?
When you operate on some state with ##\hat{H}##,

##\hat{H}\left|\right.\psi\left.\right> = \hat{H}_0 \left|\right.\psi\left.\right> + \hat{H}_{int} \left|\right.\psi\left.\right>##,

you naturally expect both terms ##\hat{H}_0 \left|\right.\psi\left.\right>## and ##\hat{H}_{int} \left|\right.\psi\left.\right>## to be defined. Therefore you have to expect that any state ##\left|\right.\psi\left.\right>## you're interested in, belongs to the domain of both operators.
 
  • #19
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Thanks for all of your answer.
I have an ultimate question in the context of this thread:
For a given hamiltonian ## H ## depending or not in time: formally, is there always a solution ##|\phi\rangle## which verify its eigenvalue equation?
 
  • #20
PeterDonis
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if I want now to measure linear momentum, how to proceed?
Linear momentum has an issue that position also has: although the operator corresponding to it is easy to define (in the position representation it's ##- i \hbar \nabla## for momentum, and just ##x##, i.e., multiply by ##x##, for position), it has no eigenstates in the Hilbert space (because the functions which satisfy the Schrodinger Equation for these operators are not normalizable). The best method I'm aware of for addressing this is the rigged Hilbert space formalism, which is described in a number of more recent QM textbooks.

That said, formally speaking, it is easy to write down any state in terms of momentum or position eigenstates. Momentum eigenstates are functions ##e^{i k x}##, and position eigenstates are delta functions.

Is there a mathematical theorem that supports this statement
Strictly speaking, no. In fact, for quantum field theory, there is a mathematical theorem called Haag's Theorem which says the opposite--that the Hilbert space of the interacting theory cannot be the same as that of the free (non-interacting) theory.

However, if you're talking about the Schrodinger Equation, you're talking about non-relativistic QM, and in that case, although I'm not aware of a mathematical theorem that proves it in the general case, for most problems of interest it can be shown that the same Hilbert space contains appropriate free and interacting states.
 
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