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If I consider a wavefunction that is the superposition of Hamiltonian eigenfunctions, for example like: $$\psi(x)=\frac{1}{\sqrt{2}}\psi_1(x)+\frac{1}{\sqrt{2}}\psi_2(x)$$ with $\hat{H}\psi_1(x)=E_1\psi_1(x)$ and $\hat{H}\psi_2(x)=E_2\psi_2(x)$ and I want to know the time evolution of the state $\psi(x)$, can I apply the time evolution op. $e^{-i\frac{\hat{H}}{\hbar}t}$ directly to $\psi(x)$ so that $\psi(x,t)=\frac{1}{\sqrt{2}}e^{-i\frac{E_1}{\hbar}t}\psi_1(x)+\frac{1}{\sqrt{2}}e^{-i\frac{E_2}{\hbar}t}\psi_2(x)$?

I have this doubt 'cause usually I apply the time evolution op. only on the kets. In the case of eigenfunctions instead of eigenkets what I need to do is "apply" the Hamiltonian in Schroedinger representation to $\psi_1(x)$ and $\psi_2(x)$.

Is this right?

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2 Answers 2

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Using kets, because the evolution operator is linear, we can write : $$|\psi(t) \rangle = \frac{1}{\sqrt 2}\left(e^{-iHt/\hbar} |\psi_1\rangle + e^{-iHt/\hbar}|\psi_2\rangle\right) = \frac{1}{\sqrt{2}} \left(e^{-iE_1t/\hbar} |\psi_1\rangle + e^{-iE_2t/\hbar}|\psi_2\rangle\right) $$

Taking the product of this equation with $\langle x|$, we get (usingsince $\psi(x,t) = \langle x|\psi(t)\rangle$ and linearity) : $$\psi(x,t) = \frac{1}{\sqrt{2}}\left(e^{-iE_1t/\hbar} \psi_1(x)+ e^{-iE_2t/\hbar}\psi_2(x)\right)$$

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  • $\begingroup$ Thanks, is $\psi(x,t)=e^{-i\frac{E_1}{\hbar}t}\psi(x,0)$ even if $\psi(x,0)$ isn't an Hamiltonian eigenfunction no more? Is it always possible to apply the time evolution operator to wavefunction instead of kets? $\endgroup$
    – Luigid
    May 24 at 15:55
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This is correct. You can easily derive it from bra-ket notation by projection unto a position state, let $$ |\psi(t_0) \rangle = c_1|\psi_1\rangle + c_2 |\psi_2\rangle $$

We have $\hat U(t)|\psi_n\rangle = \exp \left (-\frac{i}{\hbar }E_nt \right )|\psi_n\rangle$, so $$\begin{aligned} \hat U(t) |\psi(t_0) \rangle &= \hat U(t)(c_1|\psi_1\rangle + c_2 |\psi_2\rangle )\\ &= c_1\hat U(t)|\psi_1\rangle + c_2 \hat U(t)|\psi_2\rangle \\ &= c_1 \exp \left (-\frac{i}{\hbar }E_1t \right )|\psi_1\rangle + c_2 \exp \left (-\frac{i}{\hbar }E_2t \right )|\psi_2\rangle \\ \end{aligned}$$ Now project unto the position basis, $$\begin{aligned} \langle x|\hat U(t) |\psi(t_0) \rangle & = c_1 \exp \left (-\frac{i}{\hbar }E_1t \right )\langle x|\psi_1\rangle + c_2 \exp \left (-\frac{i}{\hbar }E_2t \right )\langle x|\psi_2\rangle \\ \langle x|\psi(t)\rangle = \psi(x,t) &= c_1 \exp \left (-\frac{i}{\hbar }E_1t \right )\psi_2(x) + c_2 \exp \left (-\frac{i}{\hbar }E_2t \right ) \psi_2(x) \\ \end{aligned}$$

Answer to question in comments:

Lets look at the case where the ket is not an eigenstate of the Hamilton operator, we have $$ \hat U(t)|\psi\rangle = \exp(-\frac{i}{\hbar}\hat H t)|\psi\rangle $$ You can derive the expression in position basis by insertion of a resolution of the identity $\hat 1 = \int dx'|x'\rangle \langle x'| $ and projection, $$ \langle x|\hat U \hat 1|\psi\rangle =\int dx' \langle x|\hat U|x'\rangle \langle x'|\psi\rangle $$

we can simplify the expression once by inserting the matrix elements of the time evolution operator $\langle x|\hat U|x'\rangle $ with respect to the position basis. The matrix elements are $$ \langle x|\hat U|x'\rangle = \exp\left(-\frac{i}{\hbar}H(x)t\right)\delta(x-x') $$ with $H(x)= \frac{-\hbar^2}{2m}\nabla^2_x +V(x)$. Inserting this into our former expression yields $$\begin{aligned} \int dx' \langle x|\hat U|x'\rangle \langle x'|\psi\rangle &= \int dx' \exp\left(-\frac{i}{\hbar}H(x)t\right)\delta(x-x')\psi(x') \\ \langle x |\hat U(t)|\psi\rangle&=\exp\left(-\frac{i}{\hbar}H(x)t\right)\psi(x) \end{aligned}$$

Everything is still linear, but you cannot replace the Hamiltonian with the energy if your states aren't eigenfunctions. So a sum of states would look like this, $$ \psi(x, t) = c_1\exp\left(-\frac{i}{\hbar}H(x)t\right)\psi_1(x) + c_2\exp\left(-\frac{i}{\hbar}H(x)t\right)\psi_2(x) $$


Typically you can replace your propagator by $\exp\left(-\frac{i}{\hbar}H(x)t\right)$ in expressions like $\langle x |\hat U|\psi\rangle$ as long as your Hamilton operator is diagonal with respect to the position basis and time independent.

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  • $\begingroup$ What if the time evolution operator would be not a constant but still an operator? In other words if $\psi_i(x)$ had not been Hamiltonian's eigenfunctions, when project unto the position basis I'd have: $\langle x|\hat U(t) |\psi(t_0) \rangle = c_1 \langle x| \exp \left (-\frac{i}{\hbar }\hat{H}t \right )|\psi_1\rangle + c_2 \langle x| \exp \left (-\frac{i}{\hbar }\hat{H}t \right )|\psi_2\rangle$ And I can't go on I think. So, if $\psi(x)=\frac{1}{\sqrt{2}}\psi_1(x)+\frac{1}{\sqrt{2}}\psi_2(x)$ but $\psi_i(x)$ are not Hamiltonian's eigenfunctions can I still apply it on wavefunctions? $\endgroup$
    – Luigid
    May 24 at 15:50
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    $\begingroup$ @Luigid I have updated the answer to address your comment. $\endgroup$
    – Hans Wurst
    May 24 at 17:15
  • $\begingroup$ Thank you. So basically it's the same thing but now we have an operator acting on a wavefunction and in general we don't know the result while in the first case the result was a complex phase. $\endgroup$
    – Luigid
    May 24 at 17:37

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