Stationary schrodinger equation (basic problems regarding the basis being used)

In summary, the hamilotian operator (H) is represented in the energy basis, and the eigenstates (Y) therefore form a basis that span the hilbert space of the system. If the equation above is multiplied from the left by <x|, a bra representing one of the position states that form a position basis, then the equation is written as:<x|H|Y>=E*Y(x) BUT, it is also said that the hamilotian is represented in the energy basis, so why is "x" being the variable in use here? it seems as if the operator is in one basis
  • #1
19
0
i can't seem to understand something very basic about the stationary equation (a "simple" eigenvalue problem):

H|Y>=E|Y>

H - hamiltonian operator
Y - an eigenstate or an eigenfunction of the hamiltonian
E - the eigenvalue of the eigenstate

as far as i understand, the hamilotian operator (H) is represented in the energy basis, and the eigenstates (Y) therefore form a basis that span the hilbert space of the system.

now this is where i get fairly confused. if the equation above is multiplied from the left by <x|, a bra representing one of the position states that form a position basis, then the equation is written as
H*Y(x)=E*Y(x)
BUT, it is also said that the hamiltonian is represented in the energy basis, so why is "x" being the variable in use here?
it seems as if the operator is in one basis, but the state is in a different one, but i know that i am wrong i just can't figure this out

i hope someone can help me here...
 
Physics news on Phys.org
  • #2
If you multiply on the left by <x| you get:

[tex] <x|H|Y> = E<x|Y> = E Y(x) [/tex]

Note H can't be pulled out of the bracket since it's an operator. Now, H is an operator on the Hilbert space, not directly on the space of wavefunctions, but of course these spaces are really the same, so there should be a way of acting H on the wavefunction, and there is. Inserting a 1 in the form [itex]\int dx' |x'><x'|[/itex] we find:

[tex] <x|H|Y> = \int dx' <x|H|x'><x'|Y> = \int dx' H(x,x') Y(x') [/tex]

where I've defined [itex]H(x,x') = <x|H|x'>[/itex]. If you work this out, you'll find something like:

[tex] H(x,x') = -\frac{\hbar^2}{2m} \delta''(x-x') + V(x) \delta(x-x') [/tex]

And then inserting this into the integral you find:

[tex] <x|H|Y> = -\frac{\hbar^2}{2m} Y''(x) + V(x) Y(x)[/tex]

which gives you the Schrodinger equation in the position basis.
 
  • #3
oh very nice, this makes it a bit clearer, though i still have one more question.
i have covered the missing steps (finding the H acting on the position basis) though one step isn't clear to me. from what i know, the inner product of the two position basis vectors is:
<x|x'> = DiracDelta(x-x') (replacing the kronecker delta, moving from a discrete to a continuous space)

but in order to reach the final equation of H(x,x') which you have written down, this needs to be true:
<x|H|x'> = H<x|x'>
why is this true? i have tried entering a complete set in between, but i couldn't get the result above.

just to see if i actually get it now (hopefully), the state of the system is represented in the position basis, and the operator originally acts on the same hilbert space only under a different basis, and the idea is to find it's representation in the position base instead (did i get it right?)
 
Last edited:
  • #4
Again, you need to be careful about the distinction between states and wavefunctions. An equation like:

[tex] <x|H|x'> = H <x|x'> [/tex]

makes no sense: the LHS is a number (depending on x and x') while the RHS is an operator, ie, a matrix on the infinite dimensional Hilbert space (also depending on x and x').

Now, what you said at the end of your last post is true. Specifically, we can represent the Hilbert space in a number of ways, the most common of which is the position basis, where there is a one-to-one correspondence between states and (square integrable) complex valued functions on the real numbers. In this basis, we can also represent Hilbert space operators (like H), typically by differential operators. For example, we know from experience that the Hamiltonian has the representation:

[tex] H |\psi> \leftrightarrow \left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right) \psi(x) [/tex]

(note the thing in paranetheses on the RHS is not H, but a differential operator that represents it in the position basis). What this equation says is that if the state [itex]|\psi>[/itex] is represented in the position basis by the function [itex]\psi(x)[/itex], then the state on the LHS of the above equation is represented by the function on the RHS. To put the last sentence in equations:

[tex] <x|\psi> = \psi(x) \Rightarrow <x|H |\psi> = \left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right) \psi(x) [/tex]

In the case we're concerned with, since [itex]<x|x'> = \delta(x-x')[/itex] (in other words, the wavefunction of the state [itex]|x'>[/itex] is [itex]\delta(x-x')[/itex]), we have:

[tex] <x|H|x'> = \left(-\frac{\hbar^2}{2m} \frac{d^2}{dx^2} + V(x) \right) \delta(x-x') [/tex]

which is the formula I gave above.

You could also derive this from the more basic definition of the Hamiltonian (I'll put hats on things that are Hilbert space operators):

[tex] \hat{H} = \frac{\hat{p}^2}{2m} + V(\hat{x}) [/tex]

With a little work, using the commutation relations [itex] [\hat{x}, \hat{p}] = i \hbar [/itex] and the fact that [itex]|x'>[/itex] is an eigenstate of the operator [itex]\hat{x}[/itex] with eigenvalue x', you can derive the same formula without using the position basis at all.
 

1. What is the stationary Schrodinger equation?

The stationary Schrodinger equation is a fundamental equation in quantum mechanics that describes the time evolution of a quantum system. It is named after the Austrian physicist Erwin Schrodinger and is written as HΨ = EΨ, where H is the Hamiltonian operator, Ψ is the wave function, and E is the energy of the system.

2. What does it mean for a system to be stationary?

A stationary system is one in which the energy of the system does not change over time. In other words, the wave function of the system does not depend on time and remains constant. This is important in quantum mechanics because it allows us to make predictions about the behavior of a system at a particular energy level.

3. What are the basis states used in the stationary Schrodinger equation?

The basis states used in the stationary Schrodinger equation depend on the type of system being studied. In most cases, the basis states are the eigenstates of the Hamiltonian operator, which represent the different possible energy levels of the system. These basis states can be represented as a linear combination of wave functions, also known as the wave function expansion.

4. How is the stationary Schrodinger equation solved?

The stationary Schrodinger equation is typically solved using mathematical techniques such as separation of variables, perturbation theory, or numerical methods. The solution yields the wave function and the corresponding energy levels of the system. This allows us to make predictions about the behavior of the system at different energy levels.

5. What are some applications of the stationary Schrodinger equation?

The stationary Schrodinger equation has many applications in quantum mechanics, including calculating the energy levels and wave functions of atoms and molecules, predicting the properties of materials, and understanding the behavior of particles in a potential well. It is also used in developing technologies such as quantum computers and understanding the behavior of subatomic particles.

Suggested for: Stationary schrodinger equation (basic problems regarding the basis being used)

Replies
8
Views
1K
Replies
1
Views
994
Replies
2
Views
701
Replies
3
Views
1K
Replies
1
Views
718
Back
Top