Why the linear combination of eigenfunctions is not a solution of the TISE

• I
In summary: The TDSE is what "describes" the system in general, in the sense that any possible state of the system will be a solution of the TDSE.The TISE is used to figure out what the "stationary states" of the system are. Strictly speaking, these states are not actually "stationary"--they still do change with time. But the change with time is a pure phase factor, as @Nullstein said. That means that no observables change with time, since a pure phase factor has no effect on any observables. One can think of the solutions of the TISE as describing the stationary states of the system at time ##t = 0##, when the phase factor ##e^{i E t
TL;DR Summary
The linear combination of the eigenfunctions gives solution to the Schrodinger equation. For a system with time independent Hamiltonian the Schrodinger Equation reduces to the Time independent Schrodinger equation(TISE), so this linear combination should be a solution of the TISE. It is not always true(when eigenvalues are not equal). Why is it so?
The linear combination of the eigenfunctions gives solution to the Schrodinger equation. For a system with time independent Hamiltonian the Schrodinger Equation reduces to the Time independent Schrodinger equation(TISE), so this linear combination should be a solution of the TISE. It is not always true(when eigenvalues are not equal). Why is it so?

Please tell me where am I going wrong.

The Schroedinger equation is ##i\hbar \partial_t \Psi(t,\vec x) = (\hat H(t) \Psi)(t,\vec x)##. If ##\hat H(t) = \hat H## is time-independent, the equation is still time-dependent through ##\partial_t##. An eigenfunction of ##\hat H## satisfies the time-independen Schroedinger equation ##(\hat H \psi)(\vec x) = E \psi(\vec x)## and doesn't depend on ##t##. So if we insert it into the time-dependent Schroedinger equation, we get ##i\hbar\partial_t \psi(\vec x) = 0 = (\hat H \psi)(\vec x) = E\psi(\vec x)##, which is false, because ##\psi(\vec x)\neq 0##.

If you have a solution ##\psi(\vec x)## to the time-independen Schroedinger equation ##(\hat H \psi)(\vec x) = E\psi(\vec x)##, you can turn it into a solution of the time-dependen Schroedinger equation by multiplying it by a phase factor: ##\Psi(t,\vec x) = e^{-\frac{i}{\hbar} E t}\psi(\vec x)##

Nullstein said:
The Schroedinger equation is ##i\hbar \partial_t \Psi(t,\vec x) = (\hat H(t) \Psi)(t,\vec x)##. If ##\hat H(t) = \hat H## is time-independent, the equation is still time-dependent through ##\partial_t##. An eigenfunction of ##\hat H## satisfies the time-independen Schroedinger equation ##(\hat H \psi)(\vec x) = E \psi(\vec x)## and doesn't depend on ##t##. So if we insert it into the time-dependent Schroedinger equation, we get ##i\hbar\partial_t \psi(\vec x) = 0 = (\hat H \psi)(\vec x) = E\psi(\vec x)##, which is false, because ##\psi(\vec x)\neq 0##.

If you have a solution ##\psi(\vec x)## to the time-independen Schroedinger equation ##(\hat H \psi)(\vec x) = E\psi(\vec x)##, you can turn it into a solution of the time-dependen Schroedinger equation by multiplying it by a phase factor: ##\Psi(t,\vec x) = e^{-\frac{i}{\hbar} E t}\psi(\vec x)##
I don't understand why the SE is time dependent even when the hamiltonian is time independent?

For a system with time independent Hamiltonian the Schrodinger Equation reduces to the Time independent Schrodinger equation(TISE)
No, it doesn't. They are still two different equations.

The time-dependent Schrodinger Equation for a time-independent Hamiltonian is

$$i \hbar \frac{\partial \psi}{\partial t} = \hat{H} \psi$$

The TISE is

$$\hat{H} \psi = E \psi$$

These equations are not the same.

I don't understand why the SE is time dependent even when the hamiltonian is time independent?
Because it has ##\partial_t \psi## on the LHS. That is time dependence.

I don't understand why the SE is time dependent even when the hamiltonian is time independent?
It's an evolution equation. Given an initial state ##\Psi(0,\vec x)##, you want to evolve it to a state ##\Psi(t,\vec x)##, so naturally, there must be a time-dependence somewhere. In the TDSE, it's in the ##\partial_t## time derivative. As an analogy, Newton's equations of motion are also time-dependent, even if the force isn't.

PeroK
PeterDonis said:
No, it doesn't. They are still two different equations.

The time-dependent Schrodinger Equation for a time-independent Hamiltonian is

$$i \hbar \frac{\partial \psi}{\partial t} = \hat{H} \psi$$

The TISE is

$$\hat{H} \psi = E \psi$$

These equations are not the same.
Does this mean that Hamiltonian being time independent does not imply that the system is described by TISE. I have confusion regarding when to use TISE and TDSE for a system. All this while i had thought that if a system has time independent Hamiltonian then it should be described by TISE, but your comment makes me think that this is wrong. Can you please tell how and when do we use TISE and TDSE, and what should be the properties of the system to use either of them. Thanks

I have confusion regarding when to use TISE and TDSE for a system.
The TDSE is what "describes" the system in general, in the sense that any possible state of the system will be a solution of the TDSE.

The TISE is used to figure out what the "stationary states" of the system are. Strictly speaking, these states are not actually "stationary"--they still do change with time. But the change with time is a pure phase factor, as @Nullstein said. That means that no observables change with time, since a pure phase factor has no effect on any observables. One can think of the solutions of the TISE as describing the stationary states of the system at time ##t = 0##, when the phase factor ##e^{i E t}## is just ##1##.

The reason knowing the stationary states is useful is that they provide a convenient basis in which to express general states; that is, any state of the system (that is, any solution of the TDSE) can be expressed as a linear combination of one or more stationary states.

PeterDonis said:
The TDSE is what "describes" the system in general, in the sense that any possible state of the system will be a solution of the TDSE.

The TISE is used to figure out what the "stationary states" of the system are. Strictly speaking, these states are not actually "stationary"--they still do change with time. But the change with time is a pure phase factor, as @Nullstein said. That means that no observables change with time, since a pure phase factor has no effect on any observables. One can think of the solutions of the TISE as describing the stationary states of the system at time ##t = 0##, when the phase factor ##e^{i E t}## is just ##1##.

The reason knowing the stationary states is useful is that they provide a convenient basis in which to express general states; that is, any state of the system (that is, any solution of the TDSE) can be expressed as a linear combination of one or more stationary states.
When the Hamiltonian is time independent, can't we use the TISE to describe the system in general?

Does this mean that Hamiltonian being time independent does not imply that the system is described by TISE. I have confusion regarding when to use TISE and TDSE for a system. All this while i had thought that if a system has time independent Hamiltonian then it should be described by TISE, but your comment makes me think that this is wrong. Can you please tell how and when do we use TISE and TDSE, and what should be the properties of the system to use either of them. Thanks
You always use the TDSE if you want to compute the time evolution. The TISE is not so much a physical equation, but rather a convenient tool for intermediate calculations, because knowing the eigenfunctions is quite helpful. (However, it still has physical content. The eigenvalues are relevant in measurement theory.)

When the Hamiltonian is time independent, can't we use the TISE to describe the system in general?
Go read my post again. It answers this question. In particular, read the last sentence of the second paragraph.

PeterDonis said:
Go read my post again. It answers this question. In particular, read the last sentence of the second paragraph.
I understand that the eigenfunctions act as basis to describe the wavefunctions which are general solution of TDSE. What I can't understand is the significance of Hamiltonian being time independent. When the Hamiltonian is time dependent, we use TDSE to describe the system, and when the Hamiltonian is time independent we use TISE to describe the system. However in the latter the wavefunctions are not the solution of the TISE. How can we use TDSE in a system where the Hamiltonian is time independent?

When the Hamiltonian is time dependent, we use TDSE to describe the system, and when the Hamiltonian is time independent we use TISE to describe the system.
Not true. We use the TDSE in both cases. When the Hamiltonian is time-independent, we may additionally use the TISE as a convenient tool for intermediate calculations.

What I can't understand is the significance of Hamiltonian being time independent.
If you can't understand it, then you should not make definitive statements like this...

When the Hamiltonian is time dependent, we use TDSE to describe the system, and when the Hamiltonian is time independent we use TISE to describe the system.
...particularly when they are wrong. And even more particularly when you have already been told they are wrong earlier in the thread.

The significance of the TISE, as has already been explained, is that it picks out a particular set of useful functions, the "stationary state" functions (more precisely, as I explained in my previous post, they are functions describing the "stationary states" at time ##t = 0##, when the phase factor ##e^{i E t}## is ##1##). That's all.

Don't cling to the terminology and try to understand the concepts (I'm mostly reiterating what has been already written above):

We have an equation which gives the time evolution of the state in all cases. This equation is what corresponds to Newton's second law in classical mechanics and it reads $\partial_t |\psi \rangle = \hat H |\psi \rangle$.

We also have an equation which gives the possible energies for an energy measurement and the states associated with these energies. According to the postulates of QM this is an eigenvalue equation and it reads $\hat H|E_n\rangle = E_n|E_n\rangle$.

A straightforward thing is to call the first equation "Schrödinger equation" (without qualifier) and the second equation "eigenvalue equation for the Hamiltonian". Since both equations involve the Hamiltonian, people sometimes use a different terminology with similar names for these very different equations (TDSE for the first and TISE for the second). If this terminology confuses you, I suggest to always think in a more straightforward terminology.

Now rephrasing your question using such a terminology: do you expect linear combinations of energy eigenvectors to be solutions of the eigenvalue equation of the Hamiltonian?

Last edited:
PeterDonis said:
If you can't understand it, then you should not make definitive statements like this......particularly when they are wrong. And even more particularly when you have already been told they are wrong earlier in the thread.

The significance of the TISE, as has already been explained, is that it picks out a particular set of useful functions, the "stationary state" functions (more precisely, as I explained in my previous post, they are functions describing the "stationary states" at time ##t = 0##, when the phase factor ##e^{i E t}## is ##1##). That's all.
The state of a system is described by the wavefunction which is a solution to the TDSE. These wavefunctions are linear combination of the eigenfunctions. When the Hamiltonian is independent of time, the eigenfunctions which are the basis functions don't change with time so it becomes easy to find a general solution to the TDSE which tells us how the system evolves with time. However when the Hamiltonian is time-dependent the eigenfunctions would also be time dependent and hence finding the general solution to the TDSE would be quite difficult. However the linear combination of these time dependent eigenfunctions would still give the solution to the TDSE.

vanhees71
That's about right, except that if the Hamiltonian is time-dependent, then the time-dependent eigenfunctions have no relevance at all (except maybe in some approximation schemes). The time-dependent eigenfunctions don't solve the TDSE.

Nullstein said:
That's about right, except that if the Hamiltonian is time-dependent, then the time-dependent eigenfunctions have no relevance at all (except maybe in some approximation schemes). The time-dependent eigenfunctions don't solve the TDSE.
so there exists no concept of stationary states in time dependent hamiltonian system?

You can still possibly have stationary states, but they aren't so easy to get. You want to solve the equation ##U(t)\psi = e^{i\phi(t)}\psi##, where ##U(t)## is the time-ordered exponential of ##H(t)##. I don't think the time-dependent eigenfunctions will help you here.

In the case of time-periodic Hamiltonians there is Floquet theory which talks about certain time-periodic states, as opposed to stationary states. This comes up a lot in periodically-driven quantum systems. Section 5.3 of this write-up talks about the mathematics of these time-periodic systems.

The state of a system is described by the wavefunction which is a solution to the TDSE.
Yes.

These wavefunctions are linear combination of the eigenfunctions.
They are linear combinations of solutions of the TISE with phase factors. These are eigenfunctions of the Hamiltonian, yes. But this is only true if the Hamiltonian is time-independent; if it is not, you can't use the TISE to begin with, so you can't find any solutions of it to use as a basis to express other wave functions.

When the Hamiltonian is independent of time, the eigenfunctions which are the basis functions don't change with time
Not quite; they do have phase factors ##e^{i E t}##, which are functions of time. What doesn't change with time in this case is the values of observables for the eigenfunctions.

when the Hamiltonian is time-dependent the eigenfunctions would also be time dependent
No. As above, when the Hamiltonian is time-dependent, there is no eigenvalue equation to solve to begin with, since you can't use the TISE.

Nullstein said:
the time-dependent eigenfunctions
What time-dependent eigenfunctions? If the Hamiltonian is time-dependent, the TISE is not valid to begin with.

PeterDonis said:
What time-dependent eigenfunctions? If the Hamiltonian is time-dependent, the TISE is not valid to begin with.
Why shouldn't the eigenvalue equation ##\hat{H} |\psi \rangle =E |\psi \rangle## not be valid, only because ##\hat{H}## is (explicitly) time-dependent? Of course in this case the eigenvectors are time-dependent.

Nullstein
PeterDonis said:
What time-dependent eigenfunctions? If the Hamiltonian is time-dependent, the TISE is not valid to begin with.
If ##H(t+\Delta t)## is a sufficiently mild perturbation of ##H(t)##, then the eigenvalues and eigenfunctions are analytic functions of ##\Delta t##, so you get time-dependent eigenfunctions. They are just not very useful, as I said.

vanhees71
From memory there is the wavefunction and what it interacts with - the Hamiltonian- like the infinite well problem. The wells structure doesn't change in time but it does with change in x- so the H is time independent. Not sure about relativity here.

Last edited:

1. Why is the linear combination of eigenfunctions not a solution of the TISE?

The linear combination of eigenfunctions is not a solution of the TISE because the TISE (Time-Independent Schrödinger Equation) is a linear differential equation, and the linear combination of eigenfunctions violates the linearity property. In other words, the TISE cannot be satisfied by simply adding together eigenfunctions.

2. Can't we use superposition to combine eigenfunctions and create a solution to the TISE?

While superposition can be used to combine eigenfunctions, it cannot create a solution to the TISE. This is because the TISE requires a single, unique solution that satisfies all of its conditions, and simply adding together eigenfunctions does not meet this requirement.

3. What is the significance of eigenfunctions in the TISE?

Eigenfunctions play a crucial role in the TISE as they represent the possible energy states of a quantum system. The TISE is used to find the allowed energy levels of a system, and the eigenfunctions correspond to the wavefunctions that describe these energy levels.

4. Is there any way to combine eigenfunctions and still satisfy the TISE?

No, there is no way to combine eigenfunctions and still satisfy the TISE. The TISE requires a single, unique solution, and combining eigenfunctions violates this requirement. However, eigenfunctions can be combined in other ways, such as through superposition, to describe different physical phenomena.

5. How does the non-linearity of the TISE affect our understanding of quantum mechanics?

The non-linearity of the TISE is a fundamental aspect of quantum mechanics and plays a crucial role in our understanding of the behavior of quantum systems. It highlights the unique properties of quantum systems, such as superposition and uncertainty, and allows us to make predictions about their behavior. Without this non-linearity, our understanding of quantum mechanics would be incomplete.

• Quantum Physics
Replies
3
Views
488
• Quantum Physics
Replies
3
Views
915
• Quantum Physics
Replies
3
Views
1K
• Quantum Physics
Replies
9
Views
2K
• Quantum Physics
Replies
13
Views
2K
• Quantum Physics
Replies
2
Views
2K
• Quantum Physics
Replies
6
Views
862
• Quantum Physics
Replies
14
Views
969
• Quantum Physics
Replies
4
Views
1K
• Quantum Physics
Replies
12
Views
2K