# What is $\langle \phi | H | \psi \rangle$ in QM?

I know that $\langle \phi | \psi \rangle$ is the probability of going from the $\psi$-state to the $\phi$-state, and that $\langle \phi | H | \phi \rangle$ is the expectation value of the energy for the $\phi$-state.

But how should I interpret $\langle \phi | H | \psi \rangle$?

• Apr 5, 2016 at 18:46
• The given answer is perfect, but just a comment on $\langle \phi|\psi \rangle$. This is the probability amplitude of going from $\phi$ to $\psi$ in the Heisenberg picture. $<\phi_H|\psi_H> = <\phi_S| e^{iHt} | \psi_S >$. Apr 6, 2016 at 11:09

## 2 Answers

This is a scalar value that is a projection of the state $H|\psi \rangle$ on the state $|\phi \rangle$. The state $H|\psi \rangle$ results from the action of the operator $H$ on the state $|\psi \rangle$. If the state $|\psi \rangle$ is an eigenstate of the operator $H$, the expression can be rewritten as $E \langle\phi|\psi \rangle$. If the state $|\phi \rangle$ is also an eigenstate of the operator $H$, we have $E \delta_{\phi,\psi}$, meaning we get zero if the states are orthogonal and an expectation value of the energy if they are conjugate.

If both states are not eigenstates, we can elaborate on it using the resolution of identity in terms of the Hamiltonian eigenstates: $1=\sum_i |i \rangle \langle i|$. By multiplying the identity from both sides of the Hamiltonian, the resulted expression reads: $$\sum_{ij} \langle\phi|i \rangle \langle i|H |j \rangle \langle j|\psi \rangle=\sum_{ij} E_j \delta_{i,j} \langle\phi|i \rangle \langle j|\psi \rangle=\sum_{j} E_j \langle\phi|j \rangle \langle j|\psi \rangle$$

Thus we have a sum of products of the two states projections on all eigenstates of the Hamiltonian multiplied by a corresponding energy.

This is a supplement to freude's correct answer:

Hamiltonian is the infinitesimal generator of time translation defined as $$\mathrm{\hat{U}}(\mathrm dt)= 1- \frac{i}{\hbar} \mathrm{\hat{H}}(t)~\mathrm dt\;.$$

# Time-Evolution Operator:

Let the system be at $$|\phi\rangle\;.$$ Now, let's wait for some time.....

What is the probability amplitude of finding our system at $$|\chi\rangle\;?$$

It shouldn't be $$\langle\chi|\phi\rangle$$ as now we have waited for a certain time-interval; this delay must be taken into consideration.

Time-evolution operator $$\mathrm{\hat{U}}$$ then comes to rescue.

Suppose, the system is prepared at $$|\phi\rangle$$ at $$t_1$$. What is the probability amplitude of finding our system at state $$|\chi\rangle$$ at time $$t_2\;?$$

The required amplitude is written as $$\left\langle\chi\left|\mathrm{\hat{U}}(t_2,t_1)\right|\phi\right\rangle$$

Or if expanded over base states, this can be written as

$$\sum_{jk}\langle\chi |j\rangle\left\langle j\left|\mathrm{\hat{U}}(t_2,t_1)\right|k\right\rangle\langle k|\phi\rangle\;.$$

# The $$\mathrm{\hat U}$$ matrix:

The probability amplitude of finding our system at a different state at some later time after preparing it at another state can be written as $$|\psi(t+\Delta t)\rangle = \mathrm{\hat{U}}(t+\Delta t, t)|\psi(t)\rangle$$

Multiplying both sides by $$\langle j|$$, the base state, we get $$\langle j|\psi(t+\Delta t)\rangle = \left\langle j\left|\mathrm{\hat{U}}(t+\Delta t, t)\right|\psi(t)\right\rangle$$

Resolving our state vector $$|\psi(t+\Delta t)$$ to our concerned base states, we get

$$\langle j|\psi(t+\Delta t)\rangle =\sum_{k} \left\langle j\left|\mathrm{\hat{U}}(t+\Delta t, t)\right|k\right\rangle\langle k|\psi(t)\rangle\;.$$

$$\mathrm{U}_{jk}\equiv \left\langle j\left|\mathrm{\hat{U}}(t+\Delta t, t)\right|k\right\rangle$$ constitutes one of the elements of $$\rm{\hat{U}}$$ matrix.

Then we can reform the probability amplitude as: $$\langle j|\psi(t+\Delta t)\rangle =\sum_{k} \mathrm{U}_{jk} C_k(t)$$

where $$C_{k}(t)$$ represents the probability amplitude of finding our system in the base state $$|k\rangle$$ at time $$t\;.$$

What does this imply?

This means that the amplitude of finding the system at a certain base state at $$t+\Delta t$$ is proportional to all the other amplitudes $$C_k$$ at time $$t\;.$$

# The Hamiltonian:

We can write the probability amplitude as: $$C_j(t+\Delta t) =\sum_{j} \mathrm{U}_{jk} C_k(t)\;.$$

As $$\Delta t\to 0\,,$$ $$\mathrm{ U}_{jk}\to \delta_{jk} \;.$$

So, we can write $$\mathrm{U}_{jk}(t+\Delta t,t)= \delta_{jk}+ \left(\frac{-i}{\hbar}\right)\mathrm{H}_{jk}\,\Delta t$$ where $$\mathrm{H}_{jk}$$ is defined as

$$\mathrm{ H}_{jk}= \lim_{\Delta t\to 0}\,\frac{\mathrm{U}(t+\Delta t)_{jk}- \mathrm{U}(t)_{jk}}{\Delta t}\;.$$

Using this, we re-write our amplitude as:

$$C_j(t+\Delta t)= \sum_{k}\left[\delta_{jk}- \left(\frac{i}{\hbar}\right)\,\mathrm{H}_{jk}(t)~\mathrm dt\right]\, C_k(t)$$

The elements $$\mathrm{H}_{jk}$$ constitute the Hamiltonian matrix. $$\mathrm H$$s determine the time-variation of the state of the system; they include the "physics of situation" which cause the coefficients to change over time.

The physical situation can correspond to electric field, varying magnetic field- anything. $$\rm H$$s ascertain what will happen over time.

# tl;dr:

To translate the state over a time-interval or to know about the time-development of a system, we use time-evolution operator as

$$\mathrm{\hat U}(t_2,t_1)|\psi(t_1)\rangle= \exp\left[\frac{-i}{\hbar}\int_{t_1}^{t_2}\,\mathrm{\hat H}(t')\,\mathrm dt'\right] |\psi(t_1)\rangle\;.$$

Here, $$\rm{\hat H}$$ generates infinitesimal time-translation.

But how should I interpret $$\langle \phi | H | \psi \rangle$$?

It represents the probability amplitude of transition per unit time of finding our system at $$\phi$$ provided that the system was prepared at $$\psi\;.$$

# References:

$$\bullet$$ Lectures on Physics by Feynman, Leighton, Sands.

$$\bullet$$ A Modern Approach to Quantum Mechanics by John S. Townsend.