How to calculate an operator in the Heisenberg picture?

[Haorong Wu]

Homework Statement

Consider a particle subject to a one-dimensional simple harmonic oscillator potential. Suppose that at $t=0$ the state vector is given by

$exp \left ( \frac {-ipa} { \hbar } \right ) \left | 0 \right >$

where $\left | 0 \right >$ is one for which $\left < x \right >= \left < p \right > =0$, $p$ is the momentum operator and $a$ is some number with dimension of length. Using the Heisenberg picture, evaluate the expectation value $\left < x \right >$ for $t \ge 0$.

Relevant Equations

$\frac {d A^{\left ( H \right )}} {dt} = \frac {1} {i \hbar} \left [ A^{\left ( H \right )} , H \right ]$

I have some problems when calculating the operators in Heisenberg picture.

First, $\frac {dx} {dt} = \frac {1} {i \hbar} \left [ x, H \right ] = \frac {p} {m}$.

Similarly, $\frac {dp} {dt} = \frac {1} {i \hbar} \left [ p, H \right ] = - m \omega ^ 2 x$.

These are coupled equations. I solved them and had

$x\left ( t \right ) = c_1 e^{-i \omega t} + c_2 e^{i \omega t}$, and $p\left ( t \right ) = I am \omega \left ( - c_1 e^{-i \omega t} + c_2 e^{i \omega t} \right ) + c_3$.

Suppose the initial conditions are $x\left ( 0 \right ) =x_0$ and $p \left ( 0 \right ) = p_0$.

Then I am still missing one extra condition to determine the three coefficients $c_1$, $c_2$ and $c_3$.

I am not sure what is the third condition. Maybe there just have to be one free coefficient?

Thanks!

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I am so sorry. I made a mistake. The coefficient $c_3$ should be deleted. How can I delete this post?

 

[PeroK]

Why not look at $\ddot x$?

 

[Haorong Wu]

Why not look at $\ddot x$?

Thanks, PeroK. I am sorry I made a mistake.

 

[George Jones]

First, $\frac {dx} {dt} = \frac {1} {i \hbar} \left [ x, H \right ] = \frac {p} {m}$.

Similarly, $\frac {dp} {dt} = \frac {1} {i \hbar} \left [ p, H \right ] = - m \omega ^ 2 x$.

These are coupled equations. I solved them and had

$x\left ( t \right ) = c_1 e^{-i \omega t} + c_2 e^{i \omega t}$, and $p\left ( t \right ) = I am \omega \left ( - c_1 e^{-i \omega t} + c_2 e^{i \omega t} \right ) + c_3$.

$x\left(t\right)$ and $p\left(t\right)$ are operators, but your equations for $x\left(t\right)$ and $p\left(t\right)$ do not seem to reflect this.

 

[Haorong Wu]

$x\left(t\right)$ and $p\left(t\right)$ are operators, but your equations for $x\left(t\right)$ and $p\left(t\right)$ do not seem to reflect this.

Hi, George Jones. I am not sure what properties should $x \left ( t \right )$ and $p \left ( t \right )$ reflect.

 

[George Jones]

Hi, George Jones. I am not sure what properties should $x \left ( t \right )$ and $p \left ( t \right )$ reflect.

For example, in

$x\left ( t \right ) = c_1 e^{-i \omega t} + c_2 e^{i \omega t}$

how is $c_1 e^{-i \omega t} + c_2 e^{i \omega t}$ an operator?

 

[PeroK]

For example, in
how is $c_1 e^{-i \omega t} + c_2 e^{i \omega t}$ an operator?

Perhaps in QM there should be a mantra: "you should keep your hat on"! To give @Haorong Wu a hint, what about:
$$\hat x(t) = \hat c_1 e^{-i \omega t} + \hat c_2 e^{i \omega t}$$

 

[George Jones]

Perhaps in QM there should be a mantra: "you should keep your hat on"! To give @Haorong Wu a hint, what about:
$$\hat x(t) = \hat c_1 e^{-i \omega t} + \hat c_2 e^{i \omega t}$$

Yes, we should take Joe Cocker's advice.

First, $\frac {dx} {dt} = \frac {1} {i \hbar} \left [ x, H \right ] = \frac {p} {m}$.

Similarly, $\frac {dp} {dt} = \frac {1} {i \hbar} \left [ p, H \right ] = - m \omega ^ 2 x$.

These are coupled equations. I solved them and had

$x\left ( t \right ) = c_1 e^{-i \omega t} + c_2 e^{i \omega t}$, and $p\left ( t \right ) = I am \omega \left ( - c_1 e^{-i \omega t} + c_2 e^{i \omega t} \right ) + c_3$.

@Haorong Wu, to find $\hat{c_3}$, plug your expressions for $x\left(t\right)$ and $p\left(t\right)$ into
$$\frac {dx} {dt} = \frac {p} {m}.$$
Also, to give clearer physical meaning, it probably is better to use $\left\{\sin\omega t , \cos\omega t \right\}$ rather than $\left\{e^{i\omega t} , e^{-i\omega t} \right\}$.

 

[PeroK]

Yes, we should take Joe Cocker's advice.
@Haorong Wu, to find $\hat{c_3}$, plug your expressions for $x\left(t\right)$ and $p\left(t\right)$ into
$$\frac {dx} {dt} = \frac {p} {m}.$$
Also, to give clearer physical meaning, it probably is better to use $\left\{\sin\omega t , \cos\omega t \right\}$ rather than $\left\{e^{i\omega t} , e^{-i\omega t} \right\}$.

Given that $\hat p = m \frac{d\hat x}{dt}$, I'm not sure where $\hat c_3$ would come from?

 

[George Jones]

Given that $\hat p = m \frac{d\hat x}{dt}$, I'm not sure where $\hat c_3$ would come from?

Let's wait for @Haorong Wu to have a go at this , but notice that there are two first-order differential equations, so only two constants (initial conditions) are needed.

 

[Haorong Wu]

For example, in
how is $c_1 e^{-i \omega t} + c_2 e^{i \omega t}$ an operator?

Thanks, @George Jones , @PeroK . $c_i$ are operators. I am not used to keep hats on operators, and I really do not see the differences. Normally, something having $x$, $p$, $L_i$, or $S_i$ would be operator. It seems unnecessary to keep their hats. But in some confusing cases like this problem, I should put hats on $c_i$.

And $c_3$ should be crossed out. I pluged $x\left ( t \right ) = c_1 e^{-i \omega t} + c_2 e^{i \omega t}$ into $\frac {dp} {dt} = - m \omega ^ 2 x$, and $c_3$ poped out. (Only God knows why I did this instead of the other easier way)

 

[George Jones]

And $c_3$ should be crossed out.

Good. Now express $c_1$ and $c_2$ in terms of the constant operators $x\left(0\right)$ and $p\left(0\right)$.

 

[Haorong Wu]

Good. Now express $c_1$ and $c_2$ in terms of the constant operators $x\left(0\right)$ and $p\left(0\right)$.

Thanks, @George Jones . The rest part is a piece of cake.

 

[vanhees71]

$x\left(t\right)$ and $p\left(t\right)$ are operators, but your equations for $x\left(t\right)$ and $p\left(t\right)$ do not seem to reflect this.

Since the harmonic oscillator leads to linear operator-valued ODE's for the Heisenberg-picture EoMs, there's formally no difference between oparator and "c-number" valued equations. That's why it's so easy to solve the harmonic oscillator in the Heisenberg picture (as well as the free particle and motion under a constant force).

Going beyond that makes it really difficult!