- #1

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- Thread starter athrun200
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- #1

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- #2

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- #3

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In fact that is the hint.

So can you also explain the hint?

And I still don't know how to get the answer:(

- #4

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If both elements in the linear combination are individually solutions to the differential equation then is it also true that the two elements combined in a linear combination is then a solution?

- #5

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However, all I know is how to calculate. I am not fully understand the meaning behind it.

Do you mean that I need to prove the solution for fourth eqtn is real.

i.e. [itex]i(\psi-\psi*)[/itex] is real?

- #6

ideasrule

Homework Helper

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For a linear system of equations (not necessarily differential equations), if you know that A and B are both solutions, all linear combinations of A and B must also be solutions. Here, you need to prove that if ψ is a solution to Schrodinger's equation, so is ψ*. Then you'll know that ψ+ψ* and i(ψ−ψ∗) are also solutions to Schrodinger's equation. But these are both real, which completes the proof.

- #7

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For a linear system of equations (not necessarily differential equations), if you know that A and B are both solutions, all linear combinations of A and B must also be solutions. Here, you need to prove that if ψ is a solution to Schrodinger's equation, so is ψ*. Then you'll know that ψ+ψ* and i(ψ−ψ∗) are also solutions to Schrodinger's equation. But these are both real, which completes the proof.

I wonder if i(ψ−ψ∗) is real, because it has a " i " in front of it.

- #8

vela

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What are [itex]\psi(x)+\psi^*(x)[/itex] and [itex]i(\psi(x)-\psi^*(x))[/itex] equal to?

- #9

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What are [itex]\psi(x)+\psi^*(x)[/itex] and [itex]i(\psi(x)-\psi^*(x))[/itex] equal to?

Oh! Thanks so much!!

Why I am so stupid that I didn't try to let [itex]\psi^*(x) = u(x) - i v(x)[/itex].

- #10

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Wait a minute, it is easy to solve because of the hint.

What are [itex]\psi(x)+\psi^*(x)[/itex] and [itex]i(\psi(x)-\psi^*(x))[/itex] equal to?

How about without the hint? How can I get the fourth eqtn?

Can I obtain it like this?

By mutiplying [itex]i[/itex] on eqtn 1 and 2, implys that [itex]i\psi(x) [/itex] and [itex]i\psi(x)^* [/itex] are soln to the eqtn.

By subtracting them get the fourth eqtn.

Am I right?

- #11

- 380

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Given

[tex] -\frac{{\hbar}^{2}}{2m}\frac{d^{2}{\psi}}{dx^{2}} + V{\psi} = E{\psi} [/tex]

Taking the complex conjugate:

[tex] -\frac{{\hbar}^{2}}{2m}\frac{d^{2}{\psi}^{*}}{dx^{2}} + V{\psi}^{*} = E{\psi}^{*} [/tex]

It's important to note that both V and E are real.

If we let [itex] {\psi}_{1} [/itex] and [itex] {\psi}_{2} [/itex] satisfy the first equation then so does some linear combination of them [itex] {\psi}_{3} [/itex] where:

[tex] {\psi}_{3} = c_{1}{\psi}_{1} + c_{2}{\psi}_{2} [/tex]

If we were to sub [itex] {\psi}_{3} [/itex] into the first equation we would expand out, rearrange blah blah blah simplify back down to the first equation again just with [itex] {\psi}_{3} [/itex] in place of [itex] {\psi} [/itex]

So there we have proved that a linear combination will satisfy the equation. We also know that both [itex] {\psi} [/itex] and [itex] {\psi}^{*} [/itex] are independent solutions.

The linear combinations:

[tex] {\psi} + {\psi}^{*} [/tex]

and

[tex] i({\psi} - {\psi}^{*}) [/tex]

are solutions and are also

Thus we can conclude that from any complex solution we can construct two real solutions.

We can write:

[tex] {\psi} = \frac{1}{2}[({\psi} + {\psi}^{*}) -i( i({\psi} - {\psi}^{*}))] [/tex]

which we can simplify down to just [itex] {\psi} [/itex] we have thus proven that any [itex] {\psi} [/itex] can be written as a linear combination of two real solutions.

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